JEE Main + AdvancedClass XIIModern PhysicsHigh weightage

Dual Nature of Radiation and Matter

Light behaves as particles and matter behaves as waves — the photoelectric effect, the photon, and the de Broglie wavelength that opened quantum physics

🎯 Overview Chapter hero + roadmap

🔬 Interactive 3D · The photoelectric apparatus in 3D: monochromatic light of frequency $\nu$ falls on the emitter plate C, photoelectrons are drawn to the collector A, and a microammeter reads the photocurrent. Change the intensity and the saturation current rises but the stopping potential does not; change the frequency and the stopping potential $V_0$ shifts, tracking $K_{max}=eV_0=h\nu-\phi_0$. Below the threshold frequency no current flows, no matter how bright the light. frequency of incident light, intensity of light, collector plate voltage

By the end of the nineteenth century the wave nature of light seemed beyond doubt: interference, diffraction and polarisation were all explained beautifully by treating light as an electromagnetic wave with a continuous distribution of energy. Yet at almost the same moment, experiments on the conduction of electricity through gases revealed the electron, and a puzzling new effect — light shining on a metal could eject charged particles from its surface. This chapter is the story of how that one stubborn effect, the photoelectric effect, forced physics to accept that light also behaves as a stream of particles, and how the same logic, run in reverse by de Broglie, gave matter a wave nature. It is the doorway to quantum physics. 🔉⇢

We begin with electron emission and the work function — the minimum energy an electron needs to escape a metal. Then we follow the experiments of Hertz, Hallwachs and Lenard, and read off the four experimental facts: the photocurrent grows with intensity, saturates with collector voltage, is switched off by a stopping potential that depends on frequency but not intensity, and vanishes entirely below a threshold frequency however bright the light. We show carefully why the wave picture cannot explain three of these facts, then let Einstein's photoelectric equation rescue every one of them with a single idea: light arrives in quanta of energy $h\nu$. Finally we treat the photon's energy and momentum, and de Broglie's matter waves, confirmed by the diffraction of electrons in the Davisson-Germer experiment. 🔉⇢

To set the historical stage: the closing decades of the nineteenth century were a time of spectacular experimental discovery about the microscopic world. Studies of electric discharge through gases at very low pressure revealed cathode rays; in 1897 J. J. Thomson measured their specific charge $e/m$ by deflecting them in crossed electric and magnetic fields, found the value $1.76\times10^{11}\,\text{C kg}^{-1}$ to be the same whatever the cathode metal or the gas, and concluded that these particles — electrons — are universal constituents of matter. In 1913 Millikan's oil-drop experiment showed that charge comes in integer multiples of a single elementary charge $1.602\times10^{-19}\,\text{C}$, so charge is quantised. These facts about the electron are the backdrop against which the photoelectric effect must be read. 🔉⇢

The heart of the chapter is a single stubborn experiment. When light of high enough frequency falls on a metal, electrons are ejected — the photoelectric effect, discovered by Hertz in 1887 and studied in detail by Hallwachs and Lenard. Four experimental facts emerge, and three of them are impossible to reconcile with the idea that light is a continuous wave: emission begins the instant the light arrives (no build-up time), there is a sharp threshold frequency below which nothing happens however bright the light, and the maximum kinetic energy of the electrons depends on frequency but not at all on intensity. A wave that dumps energy continuously and in proportion to its brightness cannot produce any of these. This is the crisis that quantum physics was invented to resolve. 🔉⇢

Einstein's resolution, in 1905, was as bold as it was simple: radiation is not merely emitted and absorbed in lumps, it actually travels as lumps — quanta, later called photons — each carrying energy $h\nu$. One photon is absorbed by one electron; if the photon's energy exceeds the work function $\phi_0$, the electron escapes with kinetic energy up to $K_{max}=h\nu-\phi_0$. Every one of the puzzling facts then follows immediately, including the straight-line graph of stopping potential against frequency whose slope $h/e$ Millikan measured — while trying to disprove Einstein — and which instead delivered Planck's constant and clinched the photon picture. The photon carries momentum $h/\lambda$ as well as energy, confirmed by Compton's X-ray scattering, so light is fully particle-like when it exchanges energy and momentum. 🔉⇢

The chapter's second act turns the logic around. If waves can behave as particles, de Broglie argued in 1924, symmetry suggests that particles should behave as waves, with a wavelength $\lambda=h/p$ set by their momentum. For everyday objects this wavelength is unimaginably small — a cricket ball's is about $10^{-34}\,\text{m}$ — which is why we never see matter diffract. But an electron accelerated through a few tens of volts has a wavelength comparable to atomic spacings, and in 1927 Davisson and Germer saw exactly the diffraction pattern de Broglie predicted when they scattered electrons off a nickel crystal. Matter waves are real, and they are the foundation of the electron microscope and of Schrodinger's wave mechanics. 🔉⇢

For the JEE this is a high-yield, high-reliability chapter. Numerically it rewards fluency with a small set of relations — $K_{max}=h\nu-\phi_0$, $eV_0=K_{max}$, $E=h\nu=hc/\lambda$, $p=h/\lambda$ and $\lambda=h/\sqrt{2mqV}$ — and with two shortcuts that save time under pressure: $hc=1240\,\text{eV nm}$ and $\lambda=1.227/\sqrt{V}\,\text{nm}$ for an electron accelerated from rest. Conceptually it rewards a clear separation you should repeat to yourself constantly: intensity controls how many electrons come off, frequency controls how energetic the fastest one is. Almost every trap in the chapter is a failure to keep those two ideas apart. 🔉⇢

There is a deeper reason this chapter carries such weight, beyond its yield of marks. It is the place in the syllabus where the classical world view visibly breaks and the quantum one takes over, and the break is forced by experiment, not by mathematics. The wave theory of light was not wrong about interference and diffraction; it was simply unable to account for what happens when light delivers its energy to a single electron. Three stubborn facts — that emission begins the instant the light arrives, that below a threshold frequency no light however bright will free an electron, and that the maximum energy of the electrons depends on colour but not on brightness — cannot be reconciled with energy spread smoothly across a wavefront. They demand that light arrive in indivisible lumps of energy $h\nu$. Learning this chapter well means learning to feel exactly why those three facts are fatal to the wave picture, so that Einstein's equation reads as the resolution of a crisis rather than a formula to be memorised. 🔉⇢

The chapter's second half completes the symmetry. Having granted light a particle nature, physics was almost obliged to ask whether particles have a wave nature, and de Broglie's answer — $\lambda=h/p$ for any particle — was confirmed within three years by the diffraction of electrons off crystals. The lesson to carry away is not that light is a wave or a particle, nor that matter is one or the other, but that both are quantum objects showing whichever face the experiment is built to reveal. Keep two concrete anchors in mind for scale: an electron through a hundred volts has a wavelength of about a tenth of a nanometre, comparable to atomic spacings and therefore diffractable, while a cricket ball's wavelength is so absurdly small that its wave nature, though real, can never be seen. That contrast, more than any formula, is what makes wave-particle duality concrete. 🔉⇢

A word on how to study the chapter for maximum return. Treat the photoelectric effect as a single story told through one experiment and one graph: the current-versus-voltage curve on one hand, and the stopping-potential-versus-frequency line on the other. Almost every numerical question is a request to read one feature off one of these two pictures — a saturation height, a cut-off voltage, a slope, an intercept — and almost every conceptual question is a test of whether you have kept intensity and frequency in their separate boxes. On the matter-wave side, memorise the single relation and the two anchor numbers, and practise deciding quickly whether a problem fixes speed, energy, or momentum. If you can move fluently between the two graphs and apply de Broglie's relation without hesitation, you will find this among the most reliable sources of marks in the whole physics syllabus. 🔉⇢

What you will master 🔉⇢

🧠 Concepts Map of the deep dives

This chapter breaks into the concepts below. Each has its own tab with interactive 3D, a full derivation, and a JEE worked example. The Contents tab has the complete prose.

Electron Emission an_0The Photoelectric EfPhotocurrent: IntensV_0▶Stopping Potential v_0Why the Wave Theory Einstein's PhotoelechDetermination of Plah/e▶The Photon: Energy aThe de Broglie Wavelp▶
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A map of the whole chapter. Each box is one concept with its governing relation, and the arrows are the recommended order to learn them in.

Click any box to jump straight to that concept’s tab.

🗺️ Concept map — click any concept to open its tab. Each box shows the concept and its governing formula; the path is the recommended learning order.

Electron Emission and the Work Function 🔉⇢

The work function $\phi_0$ of a metal is the minimum energy an electron must be given to escape from the metal surface. Electrons can be supplied this energy thermally (thermionic emission), by a strong electric field (field emission), or by light (photoelectric emission).

The Photoelectric Effect: Hertz, Hallwachs and Lenard 🔉⇢

The photoelectric effect is the emission of electrons from a metal surface when light of a suitable frequency falls on it. It was discovered by Hertz in 1887 and studied in detail by Hallwachs and Lenard, who found a current that appears the instant the light does and a threshold frequency below which nothing happens.

Photocurrent: Intensity, Saturation and Stopping Potential 🔉⇢

At fixed frequency the photocurrent rises linearly with the intensity of light and, as the collector voltage is raised, climbs to a saturation current where every emitted electron is collected. Reversing the voltage, the current falls to zero at the stopping potential $V_0$, which measures the maximum kinetic energy through $K_{max}=eV_0$.

Stopping Potential vs Frequency and the Threshold Frequency 🔉⇢

At fixed intensity, raising the frequency of the light makes the stopping potential more negative: the maximum kinetic energy rises linearly with frequency. Extrapolating to zero stopping potential gives the threshold frequency $\nu_0$, below which no emission occurs however intense the light.

Why the Wave Theory of Light Fails 🔉⇢

The classical wave picture treats light as a continuous distribution of energy whose intensity sets the energy delivered to each electron. It therefore predicts that the maximum kinetic energy should grow with intensity, that there should be no threshold frequency, and that dim light should act only after a long delay — all three contradicted by experiment.

Einstein's Photoelectric Equation 🔉⇢

Einstein proposed that radiation energy comes in quanta of energy $h\nu$. An electron absorbs a single quantum and, if $h\nu$ exceeds the work function, escapes with maximum kinetic energy $K_{max}=h\nu-\phi_0$. This one equation explains every feature of the photoelectric effect.

Determination of Planck's Constant from the Stopping-Potential Slope 🔉⇢

Einstein's equation predicts that the graph of stopping potential against frequency is a straight line of slope $h/e$, the same for every metal. Millikan measured this slope precisely and, with the known value of $e$, obtained Planck's constant $h$ — confirming the photon picture.

The Photon: Energy and Momentum 🔉⇢

Light in interaction with matter behaves as particles called photons. Each photon of frequency $\nu$ (wavelength $\lambda$) has energy $E=h\nu=hc/\lambda$, momentum $p=h\nu/c=h/\lambda$, travels at speed $c$, is electrically neutral, and conserves total energy and momentum in collisions.

The de Broglie Wavelength of Matter Waves 🔉⇢

De Broglie proposed that every moving particle of momentum $p$ has an associated wavelength $\lambda=h/p=h/mv$. This wave nature of matter, negligible for everyday objects but decisive for electrons, was confirmed by the diffraction of electrons in the Davisson-Germer experiment.

📖 Contents Full chapter — read straight through

The complete chapter in reading order. Each section is narratable; the concept tabs add interactive 3D and worked examples on top of this text.

Dual Nature of Radiation and Matter
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The chapter banner — the physics of this chapter in a single moving picture, with live symbols and values.

Every diagram below it is interactive: drag the controls and the numbers move with the drawing.

Electron Emission and the Work Function 🔉⇢

🎯 An electron sits in a well at the metal surface. Give it energy by heat, a field or light: while the energy supplied stays under the work function the electron cannot cross the surface and stays trapped. The moment the energy reaches the work function the electron escapes, and any surplus appears as kinetic energy. That surplus is what the photoelectric measurements read.
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K = E − φ0 when E ≥ φ0; otherwise the electron stays bound. The work function φ0 is the least energy needed to reach the surface.
What this shows

An electron sits in a well at the metal surface. Give it energy by heat, a field or light: while the energy supplied stays under the work function the electron cannot cross the surface and stays trapped. The moment the energy reaches the work function the electron escapes, and any surplus appears as kinetic energy. That surplus is what the photoelectric measurements read.

Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.

Definition: The work function $\phi_0$ of a metal is the minimum energy an electron must be given to escape from the metal surface. Electrons can be supplied this energy thermally (thermionic emission), by a strong electric field (field emission), or by light (photoelectric emission). 🔉⇢

A metal conducts electricity because it contains free electrons that move about inside it. Yet these free electrons do not simply pour out of the metal on their own. If an electron attempts to leave the surface, the metal is left with a net positive charge that pulls the electron straight back. The free electron is therefore held inside the metal surface by the attractive forces of the positive ions, and it can come out only if it is given enough energy to overcome that attractive pull. 🔉⇢

A certain minimum amount of energy is required to be given to an electron to pull it out from the surface of the metal. This minimum energy required by an electron to escape from the metal surface is called the work function of the metal. It is generally denoted by $\phi_0$ and is measured in electron volts (eV). One electron volt is the energy gained by an electron accelerated through a potential difference of one volt, so that $1\,\text{eV}=1.602\times10^{-19}\,\text{J}$ — the natural unit of energy in atomic and nuclear physics. 🔉⇢

The work function depends on the properties of the metal and on the nature of its surface: a clean surface and an oxidised one can have quite different values. Typical values run from about $2\,\text{eV}$ for the alkali metals such as caesium, sodium and potassium up to roughly $4.5$–$5\,\text{eV}$ for metals such as zinc, copper and silver. This is why the alkali metals respond even to visible light, while zinc and copper need the more energetic ultraviolet. 🔉⇢

There are three physical ways to hand an electron the energy it needs. In thermionic emission the metal is heated, and thermal energy imparted to the free electrons lets them come out of the surface — this is how the filament of a cathode-ray tube or an old valve works. In field emission a very strong electric field, of the order of $10^{8}\,\text{V m}^{-1}$, pulls electrons straight out, as at the sharp tip of a spark plug. In photoelectric emission, light of a suitable frequency illuminates the surface and the electrons absorb energy from the light and escape. 🔉⇢

The photoelectrons emitted by light are exactly the same particles — electrons — that J. J. Thomson identified in 1897 while studying cathode rays. Thomson measured the specific charge $e/m$ of the cathode ray particles by passing them through mutually perpendicular electric and magnetic fields; the accepted value is $e/m=1.76\times10^{11}\,\text{C kg}^{-1}$, and it is independent of the metal or the gas used. This universality is what tells us that the electron is a fundamental constituent of all matter. 🔉⇢

In 1913 R. A. Millikan measured the charge on the electron directly in his oil-drop experiment, finding that the charge on a droplet is always an integral multiple of an elementary charge $e=1.602\times10^{-19}\,\text{C}$. Millikan's experiment established that electric charge is quantised. Once $e$ was known, the value of the specific charge $e/m$ fixed the mass of the electron, $m_e=9.11\times10^{-31}\,\text{kg}$. 🔉⇢

It helps to picture the work function as the depth of an energy well. The free electrons sit inside a region of roughly constant, low potential energy; the surface is a wall of height $\phi_0$ above the energy of the least tightly bound electrons. To escape, an electron must be lifted over that wall. Any energy it is given beyond $\phi_0$ can appear as kinetic energy once it is outside — a fact that becomes the heart of Einstein's photoelectric equation. 🔉⇢

Not all the free electrons in a metal have the same energy. Like the molecules of a gas, they have an energy distribution, so the energy an electron needs to get out is different for different electrons; those already near the top of the distribution need the least extra energy. The work function is defined as the least energy required by any electron to come out of the metal — it is the energy needed by the most loosely bound electrons at the surface. 🔉⇢

Because the work function is an energy, it is often quoted equivalently as a threshold frequency $\nu_0=\phi_0/h$ or a threshold wavelength $\lambda_0=hc/\phi_0$. A caesium surface with $\phi_0=2.14\,\text{eV}$ has $\nu_0\approx5.16\times10^{14}\,\text{Hz}$, in the visible; a zinc surface with $\phi_0\approx4.3\,\text{eV}$ has a threshold well into the ultraviolet. Reading a work function as a frequency or wavelength is a standard exam step. 🔉⇢

The distinction between the three emission processes matters for problems: thermionic and field emission depend on temperature and applied field respectively and are described elsewhere, while this chapter is almost entirely about photoelectric emission, where the energy comes packaged in light. But the escape condition — supply at least $\phi_0$ — is common to all three, and it is the reason the work function is the first quantity to fix in any dual-nature problem. 🔉⇢

The free electrons in a metal move about within it but are held inside by the attraction of the positive ions of the metal; an electron trying to leave the surface is pulled back. The minimum energy needed to remove an electron from the surface of the metal is called the work function of the metal, and it is conveniently measured in electron volts. One electron volt is the energy gained by an electron when it is accelerated through a potential difference of one volt. The work function depends on the nature of the metal and on the state of its surface, and different metals have different work functions. 🔉⇢

There are several ways of supplying an electron with the energy it needs to overcome the work function and leave the surface. In thermionic emission the metal is heated so that the free electrons gain enough thermal energy to escape; in field emission a very strong electric field pulls electrons out of the surface; and in photoelectric emission light of a suitable frequency falling on the surface delivers the energy. In every case the electron must be given at least the work function of the metal before it can be emitted, and the photoelectric effect is the case in which that energy is delivered by light. 🔉⇢

Derivation 🔉⇢

  1. The escape condition. An electron at the surface sits in a potential well; let the least tightly bound electrons need energy $\phi_0$ to reach the zero of energy just outside the metal.
  2. Supply an amount of energy $E$ to such an electron by any means (heat, field, or a photon).
  3. If $E\lt\phi_0$ the electron cannot cross the surface barrier and stays bound; no emission occurs.
  4. If $E\ge\phi_0$ the electron escapes, and by conservation of energy the surplus appears as kinetic energy outside: $K=E-\phi_0$.
  5. For the most loosely bound electron this surplus is a maximum, $K_{max}=E-\phi_0$; more tightly bound electrons emerge with less. This single line becomes Einstein's equation once $E=h\nu$.
⚠️ JEE trap: Students often think the work function is 'the energy of the electrons in the metal' or that a brighter light lowers it. Neither is true: $\phi_0$ is a fixed property of the metal and its surface, the energy barrier at the surface, and it does not change with the intensity or even the frequency of the light you shine on it. Another slip is forgetting that $\phi_0$ is the MINIMUM (least) energy — the most loosely bound electrons — not an average. 🔉⇢

The Photoelectric Effect: Hertz, Hallwachs and Lenard 🔉⇢

🎯 Light strikes the emitter plate. When the frequency is at or above the threshold, electrons are ejected the instant the light arrives, and a brighter beam simply ejects more of them. When the frequency is below the threshold, no electrons come off at all, no matter how bright or how long the light shines. Colour, not brightness, decides whether emission happens.
light sourceemitter
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emission needs f ≥ f0 = φ0/h   below the threshold there is no current at any brightness (Hertz 1887; Hallwachs; Lenard). f is in units of 1014 Hz.
What this shows

Light strikes the emitter plate. When the frequency is at or above the threshold, electrons are ejected the instant the light arrives, and a brighter beam simply ejects more of them. When the frequency is below the threshold, no electrons come off at all, no matter how bright or how long the light shines. Colour, not brightness, decides whether emission happens.

Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.

Definition: The photoelectric effect is the emission of electrons from a metal surface when light of a suitable frequency falls on it. It was discovered by Hertz in 1887 and studied in detail by Hallwachs and Lenard, who found a current that appears the instant the light does and a threshold frequency below which nothing happens. 🔉⇢

The phenomenon of photoelectric emission was discovered in 1887 by Heinrich Hertz, during his experiments on the production of electromagnetic waves by a spark discharge — the very experiments that were establishing the wave nature of light. Hertz noticed that the high-voltage sparks across his detector loop were enhanced when the emitter plate was illuminated with ultraviolet light from an arc lamp. Light shining on the metal was somehow helping charged particles escape. 🔉⇢

When light falls on a metal surface, some electrons near the surface absorb enough energy from the incident radiation to overcome the attraction of the positive ions in the material, and after gaining sufficient energy they escape from the surface into the surrounding space. At the time the electron had not yet been named — that came with Thomson in 1897 — so the nature of the escaping particles was itself a discovery in progress. 🔉⇢

Wilhelm Hallwachs and Philipp Lenard investigated photoelectric emission in detail between 1886 and 1902. Lenard allowed ultraviolet light to fall on the emitter plate of an evacuated glass tube containing two electrodes and observed that a current flowed in the circuit — and stopped the instant the ultraviolet light was cut off. The obvious reading is that light falling on the emitter ejects electrons, which the electric field then sweeps across the tube to the collector, producing the current. 🔉⇢

Hallwachs, in 1888, connected a negatively charged zinc plate to an electroscope and found that the plate lost its charge when illuminated by ultraviolet light. An uncharged zinc plate became positively charged under ultraviolet light, and a positively charged plate became still more positive. Each observation points the same way: ultraviolet light drives negatively charged particles off the zinc. 🔉⇢

Two features stood out at once. First, there is a threshold: Hallwachs and Lenard found that no electrons were emitted at all when the frequency of the incident light was below a certain minimum value, however long or however brightly the surface was illuminated. This minimum depends on the material of the emitter. Second, different metals respond to different colours — zinc, cadmium and magnesium react only to ultraviolet, while the alkali metals lithium, sodium, potassium, caesium and rubidium respond even to visible light. 🔉⇢

The dependence on the material of the surface is exactly the work-function dependence of the previous section: a metal with a small work function has a low threshold frequency and so responds to lower-energy (visible) light, while a metal with a large work function needs the higher-energy ultraviolet. Selenium, for instance, is more sensitive than zinc or copper. The same substance responds differently to different wavelengths: ultraviolet gives photoemission from copper while green or red light does not. 🔉⇢

The particles set free by light were, after 1897, recognised as electrons, and were named photoelectrons. Their charge-to-mass ratio was found to be the same as that of cathode-ray particles, confirming that the photoelectron is nothing exotic — it is the ordinary electron, given a way out of the metal by light. The current they carry through the external circuit is the photocurrent, and the whole phenomenon is the photoelectric effect. 🔉⇢

It is worth pausing on how ordinary this looked at first and how strange it became. A wave of light carrying energy should, on the classical picture, be able to shake electrons loose from any metal if it is intense enough and shines long enough. The experimental fact that a dim blue light works where an intense red one does not — that colour, not brightness, decides whether emission happens — is the first crack in the wave picture, and it is entirely due to Hallwachs and Lenard's careful frequency studies. 🔉⇢

The apparatus Lenard used is the direct ancestor of the modern photocell and of the experimental arrangement in the next section: an evacuated tube, a photosensitive emitter, a collector, a source of monochromatic light whose frequency and intensity can be varied, and meters to read the voltage and the resulting current. Everything quantitative we say about the photoelectric effect comes from varying one of these controls at a time and watching the photocurrent respond. 🔉⇢

So the early experiments hand us a phenomenon with a sharp qualitative signature: a current that switches on and off with the light without any measurable delay, that requires the frequency to exceed a material-dependent threshold, and whose mere existence depends on colour rather than brightness. Making these observations quantitative — how the current depends on intensity, on collector voltage and on frequency — is the task of the experimental study that follows. 🔉⇢

The photoelectric effect was discovered by Hertz during his experiments on the production of electromagnetic waves by means of a spark discharge. Hertz noticed that when ultraviolet light fell on the metal electrodes, a spark passed more readily across the gap. Hallwachs and Lenard studied the effect in detail and found that when ultraviolet light is allowed to fall on a clean metal surface, electrons are emitted from the surface. These emitted electrons are now called photoelectrons, and the current they carry is the photocurrent. 🔉⇢

The early experiments established the basic conditions of the effect. Hallwachs found that a negatively charged zinc plate lost its charge when ultraviolet light fell on it, while a positively charged plate did not, showing that the light ejects negative charges from the metal. Lenard, studying the emission with an evacuated tube containing an emitter and a collector, showed that the emission depends on the frequency of the incident light and that light below a certain frequency produces no emission however intense it is. These observations set the stage for the careful measurements of photocurrent, stopping potential and threshold frequency that follow. 🔉⇢

Derivation 🔉⇢

  1. Set up: an evacuated tube with a photosensitive emitter C and a collector A, a battery to bias A, a microammeter for the photocurrent and a source of ultraviolet/visible light on C.
  2. Observation 1 (Lenard): current flows while the light is on and stops the instant it is switched off — the emission tracks the light with no measurable lag.
  3. Observation 2 (Hallwachs): a negatively charged plate loses charge under UV; an uncharged plate turns positive — negative particles (electrons) are being driven off.
  4. Observation 3 (threshold): below a material-dependent minimum frequency $\nu_0$ there is NO current, however intense or prolonged the light.
  5. Conclusion: light of frequency above $\nu_0$ ejects electrons from the surface; the phenomenon is the photoelectric effect and the particles are photoelectrons.
⚠️ JEE trap: A very common error is to think a bright enough light of any colour will eventually cause emission if you wait long enough. The early experiments say otherwise: if the frequency is below threshold, no waiting and no brightness produce a single photoelectron. Brightness (intensity) controls how many electrons come off once you are above threshold; it never controls whether emission happens. 🔉⇢

Photocurrent: Intensity, Saturation and Stopping Potential 🔉⇢

Definition: At fixed frequency the photocurrent rises linearly with the intensity of light and, as the collector voltage is raised, climbs to a saturation current where every emitted electron is collected. Reversing the voltage, the current falls to zero at the stopping potential $V_0$, which measures the maximum kinetic energy through $K_{max}=eV_0$. 🔉⇢

The experimental arrangement is an evacuated glass or quartz tube holding a photosensitive emitter plate C and a collector plate A. Monochromatic light of short enough wavelength passes through a quartz window and falls on C; the electrons it ejects are collected by A under the electric field of a battery whose voltage can be varied and whose polarity can be reversed by a commutator. A voltmeter reads the potential difference between the plates and a microammeter reads the photocurrent. We now change one control at a time. 🔉⇢

Full derivation, worked example and interactive 3D on the Photocurrent: Intensity, Saturation and Stopping Potential tab →

Stopping Potential vs Frequency and the Threshold Frequency 🔉⇢

🎯 Plot the stopping potential against the frequency of the light and the points fall on a straight line. Slide the frequency: above the threshold the stopping potential climbs in a straight line, so the maximum kinetic energy rises with frequency. Extend the line down to where the stopping potential is zero and you reach the threshold frequency; below it there is no emission at all.
fV\u2080
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V0 = (h/e)·f − φ0/e   a straight line of slope h/e; the intercept on the f-axis is the threshold f0. f is in units of 1014 Hz.
What this shows

Plot the stopping potential against the frequency of the light and the points fall on a straight line. Slide the frequency: above the threshold the stopping potential climbs in a straight line, so the maximum kinetic energy rises with frequency. Extend the line down to where the stopping potential is zero and you reach the threshold frequency; below it there is no emission at all.

Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.

Definition: At fixed intensity, raising the frequency of the light makes the stopping potential more negative: the maximum kinetic energy rises linearly with frequency. Extrapolating to zero stopping potential gives the threshold frequency $\nu_0$, below which no emission occurs however intense the light. 🔉⇢

Having seen that intensity leaves the stopping potential untouched, we now hold the intensity fixed and change the frequency of the incident light. Studying the photocurrent against collector voltage at several frequencies, we find the same saturation current each time (the intensity is fixed, so the same number of electrons come off) but a different stopping potential. The energy of the emitted electrons depends on the frequency of the incident radiation. 🔉⇢

Specifically, the stopping potential is more negative for higher frequencies. If the frequencies are in the order $\nu_3\gt \nu_2\gt \nu_1$, then the stopping potentials come in the order $V_{03}\gt V_{02}\gt V_{01}$ (in magnitude). Greater frequency means greater maximum kinetic energy of the photoelectrons, so a larger retarding potential is needed to stop them completely. Frequency, unlike intensity, controls the energy of the fastest electron. 🔉⇢

If we now plot the stopping potential $V_0$ against the frequency $\nu$ of the incident light for a given photosensitive material, the points fall on a straight line. Two things are immediately visible on this graph. First, the stopping potential varies linearly with the frequency of the incident radiation for a given material. Second, there is a certain minimum cut-off frequency for which the stopping potential is zero — extend the line down until $V_0=0$ and it meets the frequency axis at a definite value. 🔉⇢

That minimum cut-off frequency is called the threshold frequency, $\nu_0$, and it is different for different metals. Below $\nu_0$ the graph would demand a positive stopping potential, which is physically meaningless — it simply means no electrons are emitted at all. For a frequency lower than the threshold, no photoelectric emission is possible even if the intensity is very large. This is the sharp threshold the early experiments found, now read straight off a graph. 🔉⇢

The observations carry two implications, and it is worth stating them exactly as the experiments do. First, the maximum kinetic energy of the photoelectrons varies linearly with the frequency of the incident radiation, but is independent of its intensity. Second, for a frequency below the cut-off frequency $\nu_0$, no photoelectric emission is possible however intense the light. Together these fix the shape of the $V_0$-versus-$\nu$ line: a straight line, of a definite slope, starting from $\nu_0$. 🔉⇢

There is one more fact that classical physics finds hardest of all: if the frequency exceeds the threshold, photoelectric emission starts instantaneously, without any apparent time lag, even if the incident radiation is exceedingly dim. Measurements put the delay at $10^{-9}\,\text{s}$ or less. A dim light above threshold produces a small current, but it produces it at once — it does not need to shine for a while to 'charge up' the electrons. 🔉⇢

Different photosensitive materials give lines with the same slope but different thresholds. A low-work-function alkali metal has a small $\nu_0$ sitting in the visible; a high-work-function metal such as zinc has a large $\nu_0$ out in the ultraviolet. Selenium is more sensitive than zinc or copper. But every one of these lines climbs at exactly the same rate with frequency — a hint, which the next card makes precise, that the slope is a universal constant of nature. 🔉⇢

Summarising the experimental facts in one place: the photocurrent is proportional to intensity above threshold; the saturation current is proportional to intensity while the stopping potential is independent of it; there is a threshold frequency below which no emission occurs however intense the light, and above which the stopping potential (equivalently $K_{max}$) rises linearly with frequency; and emission is instantaneous. These four statements are the target every theory of the photoelectric effect must hit. 🔉⇢

For solving problems, the $V_0$-versus-$\nu$ line is a gift. Its intercept on the frequency axis is $\nu_0=\phi_0/h$, so reading off $\nu_0$ gives the work function. Its slope, as the next card shows, is $h/e$, a pure combination of fundamental constants. And any single measured pair $(\nu,V_0)$ together with the line lets you find the other. Recognising that a question is really asking you to read this graph is half the battle. 🔉⇢

The threshold can equally be expressed as a threshold wavelength $\lambda_0=c/\nu_0=hc/\phi_0$. Light of wavelength longer than $\lambda_0$ (lower frequency) cannot cause emission; light shorter than $\lambda_0$ can. Many exam questions give a wavelength and a work function and ask whether emission occurs — the answer is simply whether $\lambda\lt \lambda_0$, or equivalently whether the photon energy $hc/\lambda$ exceeds $\phi_0$. 🔉⇢

The stopping potential is the value of the retarding potential at which the photocurrent becomes zero. When the collector is made negative with respect to the emitter it retards the emitted electrons, and as the retarding potential is increased fewer electrons reach the collector, until at the stopping potential even the fastest electrons are turned back and the photocurrent falls to zero. The maximum kinetic energy of the emitted electrons is therefore the electronic charge times the stopping potential, and the stopping potential is a direct measure of that maximum kinetic energy. 🔉⇢

The stopping potential, and hence the maximum kinetic energy of the emitted electrons, depends on the frequency of the incident light and not on its intensity. For a given metal there is a minimum frequency of the incident light, called the threshold frequency, below which no electrons are emitted however intense the light. Above the threshold frequency the stopping potential increases with the frequency of the incident light. The threshold frequency is a characteristic of the metal, because it is fixed by the work function of the metal, and a metal of larger work function has a higher threshold frequency. 🔉⇢

Derivation 🔉⇢

  1. Hold intensity fixed; record photocurrent-vs-voltage curves at several frequencies $\nu_1\lt \nu_2\lt \nu_3$.
  2. Read off each stopping potential; find $|V_0|$ increases with $\nu$ while the saturation current is unchanged (fixed intensity).
  3. Plot $V_0$ against $\nu$: the data lie on a straight line, so $V_0$ is linear in $\nu$.
  4. Extrapolate to $V_0=0$: the line meets the axis at the threshold frequency $\nu_0$; below it, no emission occurs.
  5. Hence $K_{max}=eV_0$ is linear in $\nu$ and vanishes at $\nu_0$ — the empirical law Einstein's equation must reproduce.
⚠️ JEE trap: A frequent confusion is that a very intense beam below the threshold frequency should eventually cause emission. The graph says no: the line gives zero (and then meaningless negative) $K_{max}$ below $\nu_0$, so no intensity helps. Another slip is treating the threshold frequency as a universal constant — it is not; $\nu_0=\phi_0/h$ is different for every metal, even though the slope of the line is universal. 🔉⇢

Why the Wave Theory of Light Fails 🔉⇢

🎯 Turn up the intensity and watch the two bars. The observed maximum kinetic energy stays exactly where it is, because energy depends only on frequency. The wave theory, treating light as a continuous energy flow, predicts the bar should climb with intensity. That gap, together with the sharp threshold and the instant emission of even dim light, is why the wave picture fails.
incident light (intensity I)observedwave says
🔉⇢
observed: Kmax independent of I   the wave picture predicts Kmax should grow with I — three facts (flat K, a threshold, instant emission) break it. f is in units of 1014 Hz.
What this shows

Turn up the intensity and watch the two bars. The observed maximum kinetic energy stays exactly where it is, because energy depends only on frequency. The wave theory, treating light as a continuous energy flow, predicts the bar should climb with intensity. That gap, together with the sharp threshold and the instant emission of even dim light, is why the wave picture fails.

Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.

Definition: The classical wave picture treats light as a continuous distribution of energy whose intensity sets the energy delivered to each electron. It therefore predicts that the maximum kinetic energy should grow with intensity, that there should be no threshold frequency, and that dim light should act only after a long delay — all three contradicted by experiment. 🔉⇢

By the end of the nineteenth century the wave nature of light was well established: interference, diffraction and polarisation were all explained naturally by treating light as an electromagnetic wave, an oscillating electric and magnetic field with a continuous distribution of energy spread over the region the wave occupies. The question this card answers is whether that same, highly successful wave picture can account for the photoelectric observations of the previous cards. It cannot, and seeing exactly where it breaks is the whole point of the chapter. 🔉⇢

Start with intensity. In the wave picture the free electrons at the surface absorb the radiant energy continuously, and the greater the intensity the greater the amplitude of the electric and magnetic fields. So a more intense beam should hand each electron more energy, and the maximum kinetic energy of the photoelectrons should increase with intensity. Experiment flatly contradicts this: the maximum kinetic energy (the stopping potential) does not change with intensity at all. This is the first failure. 🔉⇢

Next consider frequency. On the wave picture, energy is energy — a sufficiently intense beam, given enough time, should always be able to supply an electron the energy it needs to escape, whatever the frequency. A threshold frequency therefore should not exist; a bright enough red light ought to work on any metal. Experiment says otherwise: below the threshold frequency $\nu_0$ there is no emission however intense the light, and above it emission occurs even for dim light. This is the second failure, and it is the sharpest one. 🔉⇢

Finally consider timing. In the wave picture the energy is spread continuously over the whole wavefront, and a huge number of electrons share it, so the energy absorbed per electron per unit time is small. Explicit calculation shows that it could take hours for a single electron to accumulate enough energy from a dim source to overcome the work function. Experiment contradicts this too: emission is instantaneous, beginning within about $10^{-9}\,\text{s}$ even for very dim light. This is the third failure. 🔉⇢

So three of the four experimental facts are not merely unexplained by the wave theory — they are the opposite of what it predicts. The wave theory expects $K_{max}$ to rise with intensity (it is flat), expects no threshold (there is a sharp one), and expects a long delay for dim light (there is none). In short, the wave picture is unable to explain the most basic features of photoelectric emission. 🔉⇢

It is important to be fair to the wave theory about what it does get right. The one observation it handles comfortably is that the photocurrent (the saturation current) grows with intensity: a stronger wave shakes loose more electrons, so a larger current is no surprise. The trouble is entirely with the ENERGY of the electrons and with the existence of a threshold and the timing — the features that involve single electrons rather than the bulk current. 🔉⇢

The reason the wave theory fails is that it distributes the light's energy smoothly in space and time. Every photoelectric surprise is a symptom of energy arriving instead in concentrated lumps: an all-or-nothing threshold makes sense only if an electron must receive its escape energy in one indivisible package; an instantaneous response makes sense only if that package arrives whole; and the independence of $K_{max}$ from intensity makes sense only if the size of each package is fixed by frequency, not by how many packages arrive. 🔉⇢

This is exactly the diagnosis Einstein made in 1905. The failure is not in the wave theory's mathematics but in its central assumption — continuous energy. Replace 'continuous distribution of energy' with 'discrete quanta of energy $h\nu$' and every failure turns into a success. That single replacement is the subject of the next card, and it is one of the founding moves of quantum physics. 🔉⇢

For the exam, the disciplined way to use this card is to be able to name, for each of the three failures, the wave prediction and the observed fact side by side: (i) $K_{max}$ — wave says rises with intensity, observed flat; (ii) threshold — wave says none, observed sharp $\nu_0$; (iii) timing — wave says long lag for dim light, observed instantaneous. A question that asks 'which of the following can the wave theory not explain' is answered by these three, while the intensity-dependence of the current is the one it can. 🔉⇢

There is a deeper lesson worth keeping. A correct, well-tested theory — Maxwell's wave theory of light — turned out to have a limited domain. It still describes interference and diffraction perfectly; it simply fails when light exchanges energy with single electrons. Physics did not throw the wave theory away; it recognised that light has a dual nature, wave in some experiments and particle in others. Learning where a good theory stops working is as important as learning the theory itself. 🔉⇢

On the wave picture the energy of the light is spread continuously over the wavefront, and the energy delivered to the metal should depend on the intensity of the light. A more intense wave should give the emitted electrons a larger maximum kinetic energy, and a faint light should give them a smaller one. Experiment shows the opposite: the maximum kinetic energy of the emitted electrons is independent of the intensity of the incident light and depends only on its frequency. The wave theory cannot explain why the maximum kinetic energy is fixed by frequency and not by intensity. 🔉⇢

The wave picture also predicts a time lag before emission begins, because on that picture an electron must absorb energy from the wave continuously until it has collected enough to escape, and for a faint light this would take a measurable time. Experiment shows that emission begins the instant the light falls on the metal, however faint the light, provided its frequency is above the threshold. And the wave theory offers no reason for a threshold frequency at all, since a wave of any frequency, if intense enough, should eventually supply the needed energy. Instantaneous emission, the independence of the maximum kinetic energy from intensity, and the existence of a threshold frequency are the three facts the wave theory cannot explain. 🔉⇢

Derivation 🔉⇢

  1. Wave assumption: light is a continuous energy distribution; energy delivered to an electron grows with intensity and accumulates over time.
  2. Prediction 1: $K_{max}$ increases with intensity. Observation: $K_{max}$ is independent of intensity — FAILS.
  3. Prediction 2: with enough intensity/time any frequency causes emission, so no threshold. Observation: a sharp threshold $\nu_0$ exists — FAILS.
  4. Prediction 3: dim light needs a long time (hours) to free an electron. Observation: emission within $10^{-9}\,\text{s}$ — FAILS.
  5. Only the growth of the saturation current with intensity survives. Conclusion: energy must arrive in discrete quanta, not continuously.
⚠️ JEE trap: Students sometimes say 'the wave theory fails completely'. It does not — it correctly predicts that the photocurrent grows with intensity, and it still explains interference and diffraction. It fails specifically on the maximum kinetic energy, the threshold frequency and the instantaneous timing. Naming exactly which three facts break the wave theory is a common exam requirement. 🔉⇢

Einstein's Photoelectric Equation 🔉⇢

🎯 Einstein's idea in one picture: a single photon delivers energy that equals Planck's constant times the frequency. Part of that energy pays the work function to get the electron out; whatever is left over becomes the electron's maximum kinetic energy. Slide the frequency and watch the red kinetic slice grow; drop below the work function and no electron leaves at all.
photon E = h·fenergy budget
🔉⇢
Kmax = h·f − φ0   one photon, one electron: the purple slice is spent escaping, the red slice is left as kinetic energy. f is in units of 1014 Hz.
What this shows

Einstein's idea in one picture: a single photon delivers energy that equals Planck's constant times the frequency. Part of that energy pays the work function to get the electron out; whatever is left over becomes the electron's maximum kinetic energy. Slide the frequency and watch the red kinetic slice grow; drop below the work function and no electron leaves at all.

Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.

Definition: Einstein proposed that radiation energy comes in quanta of energy $h\nu$. An electron absorbs a single quantum and, if $h\nu$ exceeds the work function, escapes with maximum kinetic energy $K_{max}=h\nu-\phi_0$. This one equation explains every feature of the photoelectric effect. 🔉⇢

In 1905 Albert Einstein proposed a radically new picture of electromagnetic radiation. Photoelectric emission, he said, does not take place by the continuous absorption of energy from a wave. Instead, radiation energy is built up of discrete units — quanta — each quantum carrying an energy $h\nu$, where $h$ is Planck's constant and $\nu$ the frequency of the light. Light of a given frequency is a stream of these energy packets, and its intensity is simply the number of packets arriving per second. 🔉⇢

In the photoelectric effect a single electron absorbs a single quantum of energy $h\nu$. If this energy exceeds the minimum energy needed for the electron to escape — the work function $\phi_0$ — the electron is emitted, and the surplus appears as kinetic energy. For the most loosely bound electrons the surplus is greatest, giving the maximum kinetic energy $K_{max}=h\nu-\phi_0$. More tightly bound electrons emerge with less. Equation (11.2) is known as Einstein's photoelectric equation. 🔉⇢

Watch how each experimental fact now falls out. The maximum kinetic energy $K_{max}=h\nu-\phi_0$ depends linearly on frequency and does not contain the intensity at all — exactly the observed behaviour. It does so because in Einstein's picture the effect is a one-photon, one-electron process: the energy an electron gets is the energy of the single quantum it absorbs, which is fixed by frequency, and is entirely unaffected by how many other quanta are arriving. 🔉⇢

The threshold frequency appears automatically. Because $K_{max}$ cannot be negative, emission is possible only if $h\nu\ge\phi_0$, i.e. $\nu\ge\nu_0$ where $\nu_0=\phi_0/h$. The greater the work function, the higher the threshold frequency. Below $\nu_0$ a single quantum simply does not carry enough energy to free an electron, and since electrons absorb quanta one at a time, no amount of extra quanta (greater intensity) can make up the shortfall. That is the sharp threshold, explained in one line. 🔉⇢

The dependence of the photocurrent on intensity is equally natural. Intensity is the number of quanta arriving per unit area per unit time. The more quanta, the more electrons that absorb one and escape (provided $\nu\gt \nu_0$), so the photocurrent is proportional to intensity. Intensity governs the number of electrons — the current — while frequency governs the energy of each — the stopping potential. The two controls are cleanly separated, just as observed. 🔉⇢

The instantaneous response is explained too. The basic process — absorption of one quantum by one electron — is itself instantaneous. It does not matter how dim the light is: dim light means few quanta per second, hence a small current, but each quantum that does arrive is absorbed whole and at once. Low intensity does not mean a delay in emission, because the elementary process is the same however many quanta there are. There is nothing to 'charge up'. 🔉⇢

Using $K_{max}=eV_0$ from the stopping-potential measurement, Einstein's equation can be written in the form that experiments test directly: $eV_0=h\nu-\phi_0$, valid for $\nu\ge\nu_0$. Rearranged, $V_0=\dfrac{h}{e}\nu-\dfrac{\phi_0}{e}$. This predicts that the graph of stopping potential against frequency is a straight line whose slope is $h/e$ and whose intercept on the frequency axis is $\nu_0=\phi_0/h$ — precisely the line measured in the previous card. 🔉⇢

It is worth appreciating how bold this was. Einstein was extending Planck's idea of energy quanta from the walls of a hot cavity to free radiation itself, and using it to make a sharp, testable prediction about a straight-line graph. The prediction contained no adjustable fudge factors: the slope had to be $h/e$, the same for every metal. It was for this explanation of the photoelectric effect, not for relativity, that Einstein was awarded the Nobel Prize in Physics in 1921. 🔉⇢

In solving problems, the equation is used in one of a few standard forms. If you are given the wavelength, use $K_{max}=\dfrac{hc}{\lambda}-\phi_0$. If you are given the stopping potential, use $eV_0=h\nu-\phi_0$. If you are given two frequencies or two wavelengths, subtract to eliminate $\phi_0$. Always convert the work function to joules, or work throughout in electron volts using $hc=1240\,\text{eV nm}$, a shortcut that turns many wavelength problems into one line. 🔉⇢

The wider significance is that light genuinely behaves as particles when it exchanges energy with matter. Einstein's equation is a statement of energy conservation for a single photon absorbed by a single electron. The same quantum $h\nu$ that this card treats as an energy will, in the next card, also be given a momentum — completing the photon as a particle and setting up the symmetry that de Broglie will exploit for matter. 🔉⇢

Einstein explained the photoelectric effect by proposing that radiation is not only emitted and absorbed in quanta but actually travels as quanta of energy, each quantum, or photon, carrying an energy equal to Planck's constant times the frequency of the light. In the photoelectric effect a single photon is absorbed by a single electron. If the energy of the photon exceeds the work function of the metal, the electron escapes, and the maximum kinetic energy of the emitted electron is the photon energy less the work function of the metal. This is Einstein's photoelectric equation. 🔉⇢

Every one of the experimental observations follows at once from this equation. Because the maximum kinetic energy is the photon energy less the work function, it depends only on the frequency of the incident light and not on its intensity. Because a photon must carry at least the work function before an electron can escape, there is a threshold frequency equal to the work function divided by Planck's constant, below which no emission occurs however intense the light. And because one photon is absorbed by one electron in a single step, emission is instantaneous. Increasing the intensity of the light increases the number of photons and hence the number of electrons emitted, which is why the saturation photocurrent is proportional to the intensity. 🔉⇢

Derivation 🔉⇢

  1. Postulate (Einstein 1905): radiation of frequency $\nu$ is a stream of quanta each of energy $E=h\nu$.
  2. One electron absorbs one quantum, gaining energy $h\nu$.
  3. Energy conservation at the surface: (absorbed) $h\nu$ = (escape cost) $\phi_0$ + (kinetic energy outside). For the least-bound electron the kinetic energy is maximal.
  4. Hence $K_{max}=h\nu-\phi_0$ (Einstein's photoelectric equation).
  5. Since $K_{max}\ge0$: emission needs $\nu\ge\nu_0=\phi_0/h$. Using $K_{max}=eV_0$: $V_0=(h/e)\nu-\phi_0/e$, a straight line of slope $h/e$.
⚠️ JEE trap: A widespread error is to write $K_{max}=h\nu$ and forget the work function, or to think that two photons can combine to free one electron below threshold. In the standard photoelectric effect each electron absorbs exactly one photon, so sub-threshold photons never add up. Also, $\phi_0$ and $h\nu$ must be in the same units — convert eV to joules (or use $hc=1240\,\text{eV nm}$) before subtracting. 🔉⇢

Determination of Planck's Constant from the Stopping-Potential Slope 🔉⇢

Definition: Einstein's equation predicts that the graph of stopping potential against frequency is a straight line of slope $h/e$, the same for every metal. Millikan measured this slope precisely and, with the known value of $e$, obtained Planck's constant $h$ — confirming the photon picture. 🔉⇢

Write Einstein's equation in the form the experiment actually measures. Starting from $eV_0=h\nu-\phi_0$ and dividing by $e$, $V_0=\dfrac{h}{e}\nu-\dfrac{\phi_0}{e}$. This is the equation of a straight line in the variables $V_0$ and $\nu$. It predicts that the $V_0$-versus-$\nu$ curve is a straight line with slope $h/e$, and — crucially — that this slope is independent of the nature of the material. Different metals shift the line sideways (different $\nu_0$) but never tilt it. 🔉⇢

Full derivation, worked example and interactive 3D on the Determination of Planck's Constant from the Stopping-Potential Slope tab →

The Photon: Energy and Momentum 🔉⇢

🎯 A photon is a particle of light with a definite energy and a definite momentum, both fixed by its wavelength. Make the wavelength shorter and both the energy and the momentum grow; make it longer and both shrink. Crucially, brightness changes only how many photons arrive, never the energy or momentum each one carries.
travels at c
🔉⇢
E = h·c/λ,  p = h/λ   all photons of a given wavelength carry the same energy and momentum whatever the intensity; p is shown in units of 10−27 kg m s−1.
What this shows

A photon is a particle of light with a definite energy and a definite momentum, both fixed by its wavelength. Make the wavelength shorter and both the energy and the momentum grow; make it longer and both shrink. Crucially, brightness changes only how many photons arrive, never the energy or momentum each one carries.

Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.

Definition: Light in interaction with matter behaves as particles called photons. Each photon of frequency $\nu$ (wavelength $\lambda$) has energy $E=h\nu=hc/\lambda$, momentum $p=h\nu/c=h/\lambda$, travels at speed $c$, is electrically neutral, and conserves total energy and momentum in collisions. 🔉⇢

The photoelectric effect showed that light, in interaction with matter, behaves as if it were made of quanta or packets of energy, each of energy $h\nu$. It is natural to ask whether such a quantum should be regarded as a particle. Einstein pointed out that the light quantum can also be associated with a momentum equal to $h\nu/c$. A definite value of energy as well as momentum is a strong sign that the light quantum can be associated with a particle. This particle was later named the photon. 🔉⇢

The particle-like behaviour of light was confirmed independently in 1924 by A. H. Compton's experiment on the scattering of X-rays by electrons, in which the X-rays behave exactly like particles carrying energy and momentum that collide with electrons and recoil. For his work on the photoelectric effect Einstein received the Nobel Prize in 1921, and for measuring the elementary charge and studying the photoelectric effect Millikan received it in 1923. 🔉⇢

We can summarise the photon picture of electromagnetic radiation in a few clean statements. In its interaction with matter, radiation behaves as though made of particles called photons. Each photon has energy $E=h\nu$ and momentum $p=h\nu/c$, and moves at the speed of light $c$. Using $c=\nu\lambda$, these can be written in terms of wavelength as $E=hc/\lambda$ and $p=h/\lambda$. 🔉⇢

All photons of light of a particular frequency $\nu$, or wavelength $\lambda$, carry the same energy $E=h\nu=hc/\lambda$ and the same momentum $p=h\nu/c=h/\lambda$, whatever the intensity of the radiation. Increasing the intensity of a beam of a given wavelength does not make the individual photons more energetic; it only increases the number of photons crossing a given area per second. Photon energy is independent of the intensity of the radiation — the same separation of 'how many' from 'how energetic' that runs through the whole chapter. 🔉⇢

Two more properties complete the picture. Photons are electrically neutral, so they are not deflected by electric or magnetic fields — unlike the electrons they eject. And in a photon-particle collision, such as a photon-electron collision, the total energy and total momentum are conserved, although the number of photons need not be: a photon may be absorbed, or a new photon created. Conservation applies to energy and momentum, not to photon number. 🔉⇢

The momentum of a photon, though tiny, is physically real and measurable in its effects. Light falling on or absorbed by a surface delivers momentum and therefore exerts a radiation pressure; a pulse of light absorbed by an object gives it an impulse equal to the total photon momentum absorbed. For a beam of power $P$ fully absorbed, the momentum delivered per second is $P/c$, so the force is $F=P/c$ — a favourite JEE setup. 🔉⇢

The number of photons in a beam follows from its power. If a source emits power $P$ as light of frequency $\nu$, and each photon carries energy $E=h\nu$, then the number emitted per second is $N=P/E=P/(h\nu)$. For a milliwatt of visible light this is astronomically large — of order $10^{15}$ photons per second — which is exactly why everyday light looks perfectly continuous and its graininess never shows. 🔉⇢

It is illuminating to compare a photon with a massive particle. A photon has energy and momentum but no rest mass, and it always moves at $c$; its energy and momentum are linked by $E=pc$, consistent with $E=h\nu$ and $p=h\nu/c$. A slow electron, by contrast, has $E_{kinetic}=p^{2}/2m$. This different energy-momentum relationship is why, for the same energy, a photon and an electron have very different wavelengths — a point the de Broglie card returns to. 🔉⇢

For problem solving, keep the three photon relations at your fingertips: $E=h\nu=hc/\lambda$, $p=h/\lambda=E/c$, and $N=P/(h\nu)$ for photon flux. Decide first whether the question is about a single photon (use $E$ and $p$) or a beam (use $N$ and radiation force $P/c$). As always, $hc=1240\,\text{eV nm}$ converts a wavelength straight to a photon energy in electron volts without a joule in sight. 🔉⇢

The photon closes the particle side of light's dual nature: energy $h\nu$ from the photoelectric effect, momentum $h/\lambda$ from Compton scattering and radiation pressure, and no rest mass. Having given light a full particle description, physics faced an obvious question of symmetry — if waves can behave as particles, can particles behave as waves? De Broglie's answer, and its experimental confirmation, is the final card of the chapter. 🔉⇢

In Einstein's picture radiation of frequency consists of photons, each of which carries an energy equal to Planck's constant times the frequency of the light. A photon travels through space at the speed of light and has no rest mass, yet it carries momentum. The momentum of a photon is its energy divided by the speed of light, which is the same as Planck's constant divided by the wavelength of the light. So a photon of higher frequency carries both more energy and more momentum than a photon of lower frequency. 🔉⇢

The number of photons falling on a surface each second fixes the intensity of the light, while the energy and momentum of each photon are fixed by the frequency. This is why increasing the intensity of a beam of a given frequency increases the number of photons, and hence the number of electrons emitted in the photoelectric effect, without changing the energy of any one electron. By contrast, increasing the frequency increases the energy and the momentum carried by each photon. The photon is thus a particle of radiation with a definite energy and a definite momentum, both proportional to the frequency of the light. 🔉⇢

Derivation 🔉⇢

  1. From the photoelectric effect, a quantum of light has energy $E=h\nu$.
  2. Einstein's association: the same quantum carries momentum $p=h\nu/c$ (a photon has $E=pc$, as for a massless particle).
  3. Use $c=\nu\lambda$ to rewrite in wavelength: $E=\dfrac{hc}{\lambda}$ and $p=\dfrac{h\nu}{c}=\dfrac{h}{\lambda}$.
  4. Beam of power $P$: photons per second $N=\dfrac{P}{E}=\dfrac{P}{h\nu}$; momentum delivered per second (full absorption) $=\dfrac{P}{c}$, giving force $F=\dfrac{P}{c}$.
  5. In any photon collision, total energy and total momentum are conserved (photon number need not be).
⚠️ JEE trap: A common slip is to write a photon's momentum as $p=mc$ (photons have no rest mass) or to think a brighter beam has more energetic photons. Photon energy and momentum depend only on frequency/wavelength; intensity changes only the NUMBER of photons. Also, for a photon $p=E/c$, whereas for a slow electron $p=\sqrt{2mK}$ — do not use the photon relation for matter. 🔉⇢

The de Broglie Wavelength of Matter Waves 🔉⇢

Definition: De Broglie proposed that every moving particle of momentum $p$ has an associated wavelength $\lambda=h/p=h/mv$. This wave nature of matter, negligible for everyday objects but decisive for electrons, was confirmed by the diffraction of electrons in the Davisson-Germer experiment. 🔉⇢

The dual, wave-particle nature of light emerged clearly from the photoelectric and Compton effects: light shows its wave nature in interference, diffraction and polarisation, and its particle nature when it exchanges energy and momentum with matter. In 1924 the French physicist Louis de Broglie asked the symmetric question. If radiation, long thought purely a wave, has a particle aspect, might the particles of nature — electrons, protons and the rest — also have a wave aspect? He reasoned that nature is symmetrical and that matter and radiation should share this dual character. 🔉⇢

Full derivation, worked example and interactive 3D on the The de Broglie Wavelength of Matter Waves tab →

Photocurrent: Intensity, Saturation and Stopping Potential 🔉⇢deep concept

Definition: At fixed frequency the photocurrent rises linearly with the intensity of light and, as the collector voltage is raised, climbs to a saturation current where every emitted electron is collected. Reversing the voltage, the current falls to zero at the stopping potential $V_0$, which measures the maximum kinetic energy through $K_{max}=eV_0$. 🔉⇢

🔬 Interactive 3D · Drive the photoelectric apparatus yourself. Raise the intensity and the saturation current climbs while the stopping potential is unmoved; raise the frequency and the stopping potential $V_0$ grows more negative, tracking $K_{max}=eV_0$. Below the threshold frequency the microammeter stays at zero however bright the source. frequency of incident light, intensity of light, collector plate voltage

The experimental arrangement is an evacuated glass or quartz tube holding a photosensitive emitter plate C and a collector plate A. Monochromatic light of short enough wavelength passes through a quartz window and falls on C; the electrons it ejects are collected by A under the electric field of a battery whose voltage can be varied and whose polarity can be reversed by a commutator. A voltmeter reads the potential difference between the plates and a microammeter reads the photocurrent. We now change one control at a time. 🔉⇢

First keep the frequency and the collector voltage fixed and vary the intensity of the light. The photocurrent is found to increase linearly with intensity. Since the photocurrent is proportional to the number of photoelectrons emitted per second, this means the number of photoelectrons emitted per second is directly proportional to the intensity of the incident radiation. Brighter light frees more electrons per second, but — as we shall see — not more energetic ones. 🔉⇢

Next keep the frequency and intensity fixed and make the collector A more and more positive. The current grows as more of the emitted electrons are swept to A, and then levels off: at a sufficiently large positive voltage every electron emitted by C is collected, and the current can grow no further. This maximum value of the photoelectric current is called saturation current. It corresponds to the case when all the photoelectrons emitted by C reach the collector A. 🔉⇢

Now reverse the polarity and make A negative with respect to C. The field now repels the electrons, and only the more energetic ones can reach A against it, so the current falls. As the retarding voltage is made more negative the current drops rapidly and reaches zero at a sharply defined critical value $V_0$. For a given frequency, this minimum negative voltage that just stops the most energetic electrons is called the cut-off or stopping potential. 🔉⇢

The stopping potential gives us a direct handle on the energy of the fastest electrons. When even the most energetic photoelectron is turned back, all of its kinetic energy has been spent climbing the potential hill $eV_0$. Hence the maximum kinetic energy of the emitted electrons is $K_{max}=eV_0$. This is the single most useful measurement in the whole chapter, because it converts a hard-to-measure electron speed into an easy-to-read voltage. 🔉⇢

Repeat the experiment at the same frequency but higher intensities $I_1\lt I_2\lt I_3$. The saturation currents rise in the same order, because more electrons are emitted per second. But the stopping potential is exactly the same for all three curves. In other words, for a given frequency the stopping potential — and therefore the maximum kinetic energy — is independent of the intensity of the light. Intensity changes how many electrons come off, never how fast the fastest one moves. 🔉⇢

This is worth stating as a slogan because JEE problems test it constantly: intensity controls the saturation current; frequency controls the stopping potential. The three photocurrent-versus-voltage curves for $I_1,I_2,I_3$ share one stopping potential on the negative-voltage axis but fan out to three different saturation currents on the positive side. Reading such a graph correctly is often the whole question. 🔉⇢

The photocurrent also responds instantly. Provided the frequency exceeds threshold, emission begins the moment the light arrives, with no measurable delay — even for very dim light the lag is of the order of $10^{-9}\,\text{s}$ or less. A photocell can therefore be used as a fast light-operated switch, and the instantaneous response is itself one of the facts the wave picture will fail to explain. 🔉⇢

It helps to see the collector voltage as a filter on electron energy. A positive collector accepts even slow electrons, so the current saturates once all of them arrive. A negative collector rejects the slow ones first, then the faster ones, and only when it is negative enough to reject even the fastest ($eV_0=K_{max}$) does the current vanish. The value of $V_0$ is thus set entirely by the fastest electron, whose energy depends only on the light's frequency and the metal's work function. 🔉⇢

Everything measurable in the photoelectric effect is contained in these curves: the linear current-intensity line, the saturation plateau, the sharp cut-off at $V_0$, and the way $V_0$ ignores intensity but (as the next card shows) responds to frequency. The apparatus in the scene above lets you move each control and watch the microammeter, the saturation current and the stopping potential change in real time — the fastest way to make these facts your own. 🔉⇢

It is worth being precise about the apparatus itself, because JEE questions sometimes hinge on a detail of it. The tube is evacuated so that emitted electrons travel to the collector without colliding with gas molecules. The emitter C is a photosensitive metal; the collector A is an ordinary metal plate. Light reaches C through a quartz (not ordinary glass) window, because quartz transmits the ultraviolet that many metals require. A battery with a potential divider sets the collector voltage, and a commutator lets us reverse the polarity so that A can be made either positive (accelerating) or negative (retarding) with respect to C. A voltmeter reads $V$ and a sensitive microammeter reads the photocurrent. 🔉⇢

The linear rise of photocurrent with intensity has a clean microscopic reading. In the photon picture, intensity is the number of photons striking the surface per unit area per unit time, and each absorbed photon (above threshold) frees one electron. So doubling the intensity doubles the photons per second, doubles the electrons freed per second, and doubles the saturation current. This is why the photocurrent is directly proportional to intensity, and why the number of photoelectrons emitted per second is directly proportional to the intensity of the incident radiation. Nothing in this argument touches the energy of an individual electron. 🔉⇢

Why does the current saturate rather than rise without limit as the collector voltage increases? Because the emitter produces only so many electrons per second. Once the accelerating field is strong enough to sweep every emitted electron to the collector before it can fall back to C, collecting them faster is impossible — there are no more to collect. The plateau height, the saturation current, therefore measures the emission rate, which is set by intensity; increasing the voltage beyond saturation does nothing. 🔉⇢

The electrons do not all emerge with the same energy, and this is the key to reading the retarding side of the graph. Electrons deep inside the metal lose energy on the way out, so they emerge slower; only those freed right at the surface, having spent exactly the work function, carry the maximum kinetic energy $K_{max}$. As the collector is made increasingly negative, it first turns back the slowest electrons, then progressively faster ones, so the current tapers off gradually. Only when the retarding potential reaches $V_0$, enough to stop even the fastest electron, does the current finally reach zero — which is why $V_0$ measures $K_{max}$ and not some average energy. 🔉⇢

A concrete numerical feel helps. Suppose caesium ($\phi_0=2.14\,\text{eV}$) is lit by light of photon energy $2.49\,\text{eV}$. The fastest electrons carry $K_{max}=2.49-2.14=0.35\,\text{eV}$, so the microammeter reading falls to zero at $V_0=0.35\,\text{V}$. Double the intensity and the saturation current doubles, but that $0.35\,\text{V}$ cut-off does not budge. Switch to bluer light and $V_0$ climbs, even at unchanged intensity. Rehearsing these moves on the live apparatus builds the intuition that the two axes of the graph measure two independent things. 🔉⇢

The instantaneous onset of the current is more significant than it first looks. However dim the light, the current appears within about $10^{-9}\,\text{s}$ of switching it on, with no perceptible lag. In the photon picture this is automatic: an electron either meets a photon of sufficient energy and leaves at once, or it does not; there is no slow accumulation. This immediate response is what makes the photocell a practical fast light-operated switch — in burglar alarms, in automatic doors, in the light meters of cameras — and, as the wave-theory card will show, it is one of the facts a continuous wave simply cannot reproduce. 🔉⇢

Historically, these current measurements were the patient work of Hallwachs and Lenard between 1886 and 1902. Hallwachs found that a negatively charged zinc plate lost its charge under ultraviolet light and that a neutral plate became positive, showing that negative particles were being driven off. Lenard traced how the photocurrent varied with the collector voltage and with the frequency and intensity of the light. It was their careful curves — the linear intensity law, the saturation plateau, the sharp stopping potential — that later demanded, and then confirmed, Einstein's quantum explanation. 🔉⇢

It is worth restating why an electron needs a minimum energy to leave the metal at all. The electrons in a metal are free to move about within it but are held inside by the attraction of the positive ions; an electron trying to escape the surface is pulled back. The minimum energy needed to remove an electron from the surface of the metal is its work function. This is why the photoelectric effect has a threshold built into the metal itself: unless the energy delivered to an electron exceeds the work function, the electron cannot escape and no photocurrent flows. 🔉⇢

The electrons emitted from the surface do not all have the same energy. Different electrons are bound with different energies, so when light of a given frequency falls on the metal the emitted electrons emerge with a range of kinetic energies from nearly zero up to a maximum. The maximum kinetic energy belongs to the electrons that were least tightly bound. This spread of energies is what the retarding potential measures: a small retarding voltage turns back only the slowest emitted electrons, and as the retarding voltage is increased the faster electrons are turned back in turn, until at the stopping potential even the fastest are stopped and the photocurrent falls to zero. 🔉⇢

The current-voltage curve therefore has a definite shape that is worth holding in the mind. When the collector is sufficiently positive it collects every emitted electron and the photocurrent reaches its saturation value. As the collector potential is reduced and then made negative, the photocurrent falls, and at the stopping potential it becomes zero. The saturation value of the photocurrent depends on the intensity of the incident light, because intensity fixes the number of electrons emitted per second; the stopping potential depends on the frequency of the incident light, because frequency fixes the maximum kinetic energy of the emitted electrons. 🔉⇢

The two features of the curve respond to two different properties of the light, and keeping them apart is the whole skill of the topic. Increasing the intensity of the light while keeping the frequency fixed raises the saturation photocurrent but leaves the stopping potential unchanged, because more photons emit more electrons without changing the energy of any one of them. Increasing the frequency while keeping the intensity fixed raises the stopping potential and hence the maximum kinetic energy of the electrons, but scarcely changes the saturation current, because the number of photons is the same though each carries more energy. 🔉⇢

The photocurrent responds to the incident light without any delay. However low the intensity, emission begins the instant the light of a frequency above threshold falls on the metal. There is no waiting time for an electron to accumulate energy. This instantaneous emission is one of the four basic observations of the photoelectric effect, and together with the intensity and frequency observations it fixes the character of the effect: the photocurrent is proportional to the intensity, the maximum kinetic energy depends on the frequency and not on the intensity, there is a threshold frequency below which no emission occurs, and emission is instantaneous. 🔉⇢

The apparatus that establishes these facts is a photoelectric cell. A photosensitive metal plate, the emitter, and a collector plate are sealed in an evacuated glass tube fitted with a quartz window that lets ultraviolet light through. Light of a chosen frequency and intensity falls on the emitter, and a battery with a sliding contact sets the potential of the collector with respect to the emitter. The photocurrent is read on a sensitive meter as the collector potential and the frequency and intensity of the incident light are varied in turn. 🔉⇢

The linear relation between the photocurrent and the intensity has a simple reading in Einstein's picture of light. The intensity of the light is the number of photons falling on the metal each second; each photon that is absorbed can eject one electron; so the number of electrons emitted each second, and hence the saturation photocurrent, is proportional to the intensity of the incident light. The energy of each emitted electron, however, is fixed by the energy of a single photon, which depends only on the frequency, so raising the intensity cannot make the electrons more energetic. 🔉⇢

Collecting the observations in one place makes the wave picture's difficulty plain. Classical wave theory would have the energy of the light spread smoothly over the wavefront, so a fainter light should give electrons of smaller energy and should need a waiting time to emit at all, and there should be no threshold frequency. Experiment finds the opposite: the maximum kinetic energy of the emitted electrons is independent of intensity, there is a sharp threshold frequency, and emission is instantaneous. These are exactly the observations that the next section shows the wave theory cannot explain and that Einstein's photon picture does. 🔉⇢

In summary, the current-voltage curves of the photoelectric effect carry two independent messages. The height of the saturation photocurrent measures how many electrons the light emits each second, and this is set by the intensity of the incident light. The stopping potential measures the maximum kinetic energy of the fastest emitted electrons, and this is set by the frequency of the incident light. Reading these two features correctly, and never confusing more electrons with more energetic electrons, answers almost every question the photoelectric effect can pose. 🔉⇢

One further point about the saturation of the photocurrent is worth making. Once the collector is positive enough to collect every electron emitted from the surface, making it still more positive cannot increase the photocurrent, because there are no more electrons to collect; the photocurrent has reached saturation. Raising the intensity of the incident light emits more electrons each second and so raises the saturation photocurrent, while raising the frequency of the light does not change how many electrons are emitted and so does not change the saturation value. The saturation photocurrent is thus a direct measure of the intensity of the incident light. 🔉⇢

Derivation from first principles 🔉⇢

  1. Fix $\nu$ and $V$; vary intensity $I$. Measured photocurrent $i\propto I$, because $i$ counts electrons per second and each is freed by one photon.
  2. Fix $\nu$ and $I$; raise the collector voltage $V$ positive. Current rises then saturates when every emitted electron is collected — this plateau is the saturation current $i_{sat}$.
  3. Reverse $V$ to retarding values. Only electrons with kinetic energy $K\ge eV$ now reach A, so the current falls.
  4. At $V=V_0$ even the fastest electron is stopped: $eV_0=K_{max}$, i.e. $K_{max}=eV_0$.
  5. Raising $I$ (same $\nu$) raises $i_{sat}$ but leaves $V_0$ unchanged — so $K_{max}$ is independent of intensity.
⚠️ JEE trap: Many students read the saturation current as a measure of electron speed and the stopping potential as a measure of how many electrons there are. It is the other way round: the saturation current counts electrons (set by intensity), while the stopping potential measures the maximum kinetic energy (set by frequency). Confusing the two axes of the current-voltage graph is the commonest mistake in this chapter. 🔉⇢

Worked example · JEE Advanced 🔉⇢

SITUATION In a photoelectric experiment the stopping potential for a certain metal and frequency is measured to be $V_0=1.5\,\text{V}$.
TARGET Find the maximum kinetic energy of the emitted photoelectrons, in eV and in joules.
STRATEGY Use $K_{max}=eV_0$. In electron volts the number equals $V_0$ directly; multiply by $1.6\times10^{-19}$ for joules.
EXECUTE $K_{max}=eV_0=e\times1.5\,\text{V}=1.5\,\text{eV}=1.5\times1.6\times10^{-19}=2.4\times10^{-19}\,\text{J}.$
REFLECT The neat trick $K_{max}(\text{in eV})=V_0(\text{in volts})$ follows because one electron volt is exactly the energy $e\times1\,\text{V}$. Raising the intensity would not change this $1.5\,\text{eV}$; only changing the frequency would.

Source: NCERT Exercise 11.3

Determination of Planck's Constant from the Stopping-Potential Slope 🔉⇢deep concept

Definition: Einstein's equation predicts that the graph of stopping potential against frequency is a straight line of slope $h/e$, the same for every metal. Millikan measured this slope precisely and, with the known value of $e$, obtained Planck's constant $h$ — confirming the photon picture. 🔉⇢

🔬 Interactive 3D · Sweep the frequency and watch the stopping potential trace out a straight line. Its slope is $h/e$ — the same for every metal — so measuring the slope and multiplying by $e$ delivers Planck's constant $h$. The intercept on the frequency axis marks the threshold $\nu_0=\phi_0/h$. frequency of incident light (the metal fixes only the intercept, not the slope)

Write Einstein's equation in the form the experiment actually measures. Starting from $eV_0=h\nu-\phi_0$ and dividing by $e$, $V_0=\dfrac{h}{e}\nu-\dfrac{\phi_0}{e}$. This is the equation of a straight line in the variables $V_0$ and $\nu$. It predicts that the $V_0$-versus-$\nu$ curve is a straight line with slope $h/e$, and — crucially — that this slope is independent of the nature of the material. Different metals shift the line sideways (different $\nu_0$) but never tilt it. 🔉⇢

The slope being a pure ratio of two fundamental constants, $h$ and $e$, is an extraordinary claim. It says that if you measure the stopping potential at several frequencies for sodium, and separately for zinc, and for caesium, the three lines you draw will be exactly parallel. Any theory that got the photoelectric effect merely qualitatively right would not force this; Einstein's equation does, and that makes it sharply testable. 🔉⇢

During 1906–1916 R. A. Millikan set out, in his own words, to disprove Einstein's photoelectric equation, which he found too radical to believe. He performed a long and careful series of experiments on the photoelectric effect, measuring the stopping potential as a function of frequency for alkali metals such as sodium over a wide range of frequencies, and determined the slope of the resulting straight line with high precision. 🔉⇢

Using the known value of the electronic charge $e$, Millikan turned his measured slope $h/e$ into a value of Planck's constant $h$. The number he obtained was close to the value of Planck's constant, $6.626\times10^{-34}\,\text{J s}$, that had been determined in an entirely different context — from the spectrum of black-body radiation. Two utterly different experiments, one on hot cavities and one on light hitting metals, agreed on the same $h$. 🔉⇢

The upshot was the opposite of Millikan's intention. In 1916, instead of disproving Einstein's equation, he had confirmed it with great precision, for a number of alkali metals over a wide range of frequencies. The successful explanation of the photoelectric effect using light quanta, together with this experimental determination of $h$ and $\phi_0$ in agreement with other methods, led to the general acceptance of Einstein's photon picture of light. 🔉⇢

For calculations, the slope method is direct: slope $=\dfrac{\Delta V_0}{\Delta\nu}=\dfrac{h}{e}$, so $h=e\times(\text{slope})$. A measured slope of $4.12\times10^{-15}\,\text{V s}$, for example, gives $h=(1.6\times10^{-19})(4.12\times10^{-15})=6.6\times10^{-34}\,\text{J s}$. The neatness is that you never need to know the work function to get $h$ — the intercept carries $\phi_0$, the slope carries $h$, and they are read off independently. 🔉⇢

The intercepts, meanwhile, give the work functions. Where the line crosses the frequency axis is $\nu_0=\phi_0/h$, so $\phi_0=h\nu_0$. Where it crosses the $V_0$ axis (extrapolated to $\nu=0$) is $-\phi_0/e$. Either intercept, combined with the value of $h$ from the slope, yields the work function of the particular metal used. A single well-drawn line thus delivers both a universal constant and a material property. 🔉⇢

It is useful to know a couple of standard numbers. $h/e=4.14\times10^{-15}\,\text{V s}$, so any measured slope should come out near this value; a slope far from it signals an error. And in convenient units $h=4.136\times10^{-15}\,\text{eV s}$, while $hc=1240\,\text{eV nm}$. Recognising these combinations lets you check an answer's plausibility at a glance, which is valuable under exam time pressure. 🔉⇢

The ratio-of-slopes trick appears often in JEE problems. Because every metal shares the slope $h/e$, if a question gives you the ratio of the slopes of the $V_0$-versus-$\nu$ lines for two metals, the answer is one — the slopes are equal. If instead it gives the ratio of threshold frequencies or of intercepts, that is the ratio of work functions. Keeping straight which feature of the line is universal (the slope) and which is material-dependent (the intercept) resolves a whole family of questions. 🔉⇢

Historically this card is the hinge of the chapter. It is where a beautiful theoretical idea — light quanta — became an accepted fact of physics, because a sceptic's precise experiment could not shake it. The determination of $h$ from the photoelectric slope, in agreement with the black-body value, is one of the cleanest examples in all of physics of a bold hypothesis surviving a determined attempt to kill it. 🔉⇢

The whole determination rests on reading a straight line. For light of each frequency, the experiment measures the stopping potential — the smallest retarding voltage that reduces the photocurrent to zero. When the stopping potential is plotted against the frequency of the incident light, the measured points fall on a straight line, exactly as Einstein's photoelectric equation predicts. The maximum kinetic energy of the emitted electrons equals the electronic charge times the stopping potential, and this maximum kinetic energy is the photon energy less the work function of the metal. 🔉⇢

The slope of that straight line is the same for every metal. Sodium, potassium, caesium and zinc are photosensitive metals with different work functions, yet each gives a line of the same steepness. Only the position of the line differs: a metal with a larger work function needs incident light of higher frequency before any electrons are emitted, so its line is shifted along the frequency axis. That the slope is the same for every photosensitive surface is a direct experimental statement of Einstein's quantum picture of light, in which the energy of a photon is proportional to its frequency. 🔉⇢

Reading Planck's constant from the graph is a matter of measuring that slope. Choosing two well separated points on the line and dividing the change in stopping potential by the change in frequency gives the slope, and multiplying the slope by the electronic charge gives Planck's constant. Millikan, using his own measured value of the electronic charge, obtained a value of Planck's constant from the photoelectric slope that agreed closely with the value Planck had obtained earlier from the radiation of a hot body. Two independent experiments giving the same value of the constant established it as a fundamental physical constant. 🔉⇢

The history behind the measurement is instructive. Millikan did not accept Einstein's photoelectric equation and set out to disprove it, working on the problem for several years. To make the measurement reliable he had to prepare clean metal surfaces, because a metal surface exposed to air changes its work function. When his careful experiments were complete, the measured points lay on the predicted straight line and gave a value of Planck's constant in agreement with the radiation experiments. Instead of disproving Einstein, Millikan had confirmed the photoelectric equation, and he was later awarded the Nobel Prize in physics. 🔉⇢

It is worth being clear why the stopping potential, and not the photocurrent, is the quantity used to find the constant. The photocurrent at a given voltage depends on how many electrons are emitted, which depends on the intensity of the incident light and on the state of the surface. The stopping potential depends only on the maximum kinetic energy of the fastest emitted electrons, which is fixed by the frequency of the incident light through the photoelectric equation. By measuring the voltage at which the current is reduced to zero, rather than the size of the current, the experiment isolates the energy of the emitted electrons from their number. 🔉⇢

The same straight line also gives the work function and the threshold frequency of the metal. The point where the line meets the frequency axis is the threshold frequency, the minimum frequency below which no electrons are emitted however intense the incident light. Multiplying the threshold frequency by Planck's constant gives the work function of the metal. So a single measured line contains three pieces of information at once: its slope gives Planck's constant, its meeting point with the frequency axis gives the threshold frequency, and the threshold frequency multiplied by Planck's constant gives the work function. 🔉⇢

A worked reading of the graph shows the method. Suppose the measured line passes through two points, so that between them the stopping potential rises by a known amount while the frequency rises by a known amount. Dividing the change in stopping potential by the change in frequency gives the slope, and multiplying the slope by the electronic charge gives Planck's constant close to its accepted value. Substituting one of the points back into the photoelectric equation then gives the work function of the metal in electron volts. A common examination question supplies two points and asks for the work function; finding the slope first and the work function second is the fastest route. 🔉⇢

Keeping the quantities and their units straight is essential in every numerical problem on this experiment. Frequency is measured in hertz, the stopping potential in volts, and the maximum kinetic energy in joules or in electron volts. The photon energy equals Planck's constant times the frequency of the incident light. Converting a given wavelength of the incident light into a photon energy, comparing that photon energy with the work function of the metal, and reading the maximum kinetic energy of the emitted electrons as the difference is the standard sequence of steps. 🔉⇢

The reliability of the measurement depends on controlling several experimental details. The incident light must be of a single frequency, so that the maximum kinetic energy of the emitted electrons is well defined and the stopping potential is sharp. The collector must not itself emit electrons when light falls on it, so it is shielded or made of a metal of high work function. And because the emitter and the collector are different metals, a fixed difference of potential exists between them; this shifts every reading of the stopping potential by the same amount, which is why the slope, and not the position of the line, is used to find Planck's constant. 🔉⇢

There is a clean reason the slope is the reliable quantity. Any fixed experimental difference of potential shifts the whole straight line up or down without changing its steepness. The position where the line meets the axis therefore carries that fixed shift and cannot be trusted for an absolute work function unless the shift is separately measured, but the slope, being the rate at which the stopping potential increases with frequency, is unaffected by any fixed shift. Extracting the fundamental constant from the slope of a measured line is the careful physicist's response to a fixed experimental error. 🔉⇢

Standing back, this experiment is where the photoelectric equation becomes a measuring instrument. The equation that the previous section justified physically now gives, through the slope of a measured straight line, a value for one of the fundamental constants of physics. A quantum idea predicted a straight line of a definite slope, the same for every photosensitive metal, and a careful experiment measured exactly that line. When a photoelectric graph appears in an examination, the task is to repeat this determination in miniature: read the slope for Planck's constant, read the frequency where the line meets the axis for the threshold frequency, and read any point on the line for the maximum kinetic energy of the electrons emitted at that frequency. 🔉⇢

A last observation ties this section to the rest of the chapter. The three experimental observations of the photoelectric effect — that the photocurrent is proportional to the intensity of the incident light, that the maximum kinetic energy of the emitted electrons depends on the frequency and not on the intensity, and that there is a threshold frequency below which no electrons are emitted — are all contained in the single straight line of stopping potential against frequency. Einstein's photoelectric equation, whose slope this experiment measures, is the one relation that accounts for every one of those observations at once. 🔉⇢

The apparatus that produces the graph is the same photoelectric cell used throughout the chapter. Monochromatic light of a chosen frequency is allowed to fall on a photosensitive metal plate, the emitter, sealed in an evacuated tube. A second metal plate, the collector, is held at a variable potential with respect to the emitter. When the collector is made positive it attracts the emitted electrons and a photocurrent flows; when it is made negative it retards them. The stopping potential is the negative collector potential at which even the fastest emitted electrons are turned back and the photocurrent falls to zero. 🔉⇢

Repeating the measurement for several frequencies of the incident light builds up the straight line. For each frequency the intensity of the light is kept fixed and the collector potential is varied until the photocurrent is reduced to zero, giving one value of the stopping potential. Plotting the stopping potential against the frequency of the incident light for a given photosensitive metal gives a straight line whose slope is the same for all metals and whose position depends on the work function of the particular metal used. 🔉⇢

The independence of the slope from the metal deserves emphasis because it is the whole point of the experiment. Two different photosensitive metals, illuminated by light of the same frequencies, give two parallel straight lines. The metal of smaller work function begins to emit electrons at a lower threshold frequency, so its line meets the frequency axis sooner; the metal of larger work function needs light of higher frequency. Yet the two lines rise equally steeply, because the increase in the maximum kinetic energy of the emitted electrons per unit increase in frequency is Planck's constant, the same for every metal. 🔉⇢

The electron volt is the natural unit of energy for this experiment. One electron volt is the kinetic energy gained by an electron accelerated through a potential difference of one volt, and it equals the electronic charge times one volt. Because the maximum kinetic energy of the emitted electrons equals the electronic charge times the stopping potential, the stopping potential in volts is numerically equal to the maximum kinetic energy in electron volts. Work functions of photosensitive metals, being of the order of a few electron volts, are conveniently expressed in the same unit. 🔉⇢

In short, the determination of Planck's constant is a triumph of careful measurement over expectation. Einstein's photoelectric equation predicted a straight line of a universal slope; Millikan, intending to refute the equation, measured that line precisely and found the predicted slope and the predicted value of the constant. The maximum kinetic energy of the emitted electrons rising linearly with the frequency of the incident light, with a slope equal to Planck's constant divided by the electronic charge, is the quantitative heart of the photoelectric effect and one of the foundational measurements of quantum physics. 🔉⇢

It is worth restating the central result in words, because examinations reward the student who can. The stopping potential increases linearly with the frequency of the incident light; the slope of that line, multiplied by the electronic charge, is Planck's constant; the frequency at which the line meets the axis is the threshold frequency of the metal; and that threshold frequency, multiplied by Planck's constant, is the work function. Every quantitative question about this experiment is answered by locating one of these features on the measured straight line of stopping potential against frequency. 🔉⇢

Derivation from first principles 🔉⇢

  1. Start from Einstein's equation with $K_{max}=eV_0$: $eV_0=h\nu-\phi_0$.
  2. Divide by $e$: $V_0=\dfrac{h}{e}\,\nu-\dfrac{\phi_0}{e}$ — linear in $\nu$.
  3. Identify slope $=\dfrac{h}{e}$ (universal) and frequency-intercept $\nu_0=\dfrac{\phi_0}{h}$ (material-dependent).
  4. Measure the slope from the graph: $h=e\times\text{slope}$.
  5. Check: with the known $e$, Millikan's slope gave $h\approx6.6\times10^{-34}\,\text{J s}$, matching the black-body value — confirming the photon picture.
⚠️ JEE trap: Students often think each metal has its own slope. It does not: the slope $h/e$ is universal, and only the intercept (threshold frequency, hence work function) differs between metals. Another error is trying to find $h$ from a single $(\nu,V_0)$ point without the work function — you need the SLOPE (two points), which cancels $\phi_0$ automatically. 🔉⇢

Worked example · JEE Advanced 🔉⇢

SITUATION In a photoelectric experiment the slope of the stopping-voltage-versus-frequency line is measured to be $4.12\times10^{-15}\,\text{V s}$.
TARGET Calculate Planck's constant.
STRATEGY The slope equals $h/e$, so $h=e\times\text{slope}$.
EXECUTE $h=(1.6\times10^{-19}\,\text{C})(4.12\times10^{-15}\,\text{V s})=6.6\times10^{-34}\,\text{J s}.$
REFLECT This matches the accepted $6.63\times10^{-34}\,\text{J s}$. Notice the work function never entered — the slope alone fixes $h$, which is exactly why Millikan's slope measurement was so convincing.

Source: NCERT Exercise 11.5

The de Broglie Wavelength of Matter Waves 🔉⇢deep concept

Definition: De Broglie proposed that every moving particle of momentum $p$ has an associated wavelength $\lambda=h/p=h/mv$. This wave nature of matter, negligible for everyday objects but decisive for electrons, was confirmed by the diffraction of electrons in the Davisson-Germer experiment. 🔉⇢

🔬 Interactive 3D · Accelerate electrons at a nickel crystal and sweep the voltage: a sharp peak in the scattered intensity appears at a particular angle, exactly as a wave of wavelength $\lambda=h/\sqrt{2meV}$ would diffract. This is the Davisson-Germer confirmation of de Broglie's matter waves. accelerating voltage of the electron beam, scattering angle of the detector

The dual, wave-particle nature of light emerged clearly from the photoelectric and Compton effects: light shows its wave nature in interference, diffraction and polarisation, and its particle nature when it exchanges energy and momentum with matter. In 1924 the French physicist Louis de Broglie asked the symmetric question. If radiation, long thought purely a wave, has a particle aspect, might the particles of nature — electrons, protons and the rest — also have a wave aspect? He reasoned that nature is symmetrical and that matter and radiation should share this dual character. 🔉⇢

De Broglie proposed that the wavelength $\lambda$ associated with a particle of momentum $p$ is $\lambda=h/p=h/mv$, where $m$ is the mass of the particle and $v$ its speed. Equation (11.5) is known as the de Broglie relation, and the wavelength $\lambda$ is called the de Broglie wavelength of the matter wave. The dual aspect of matter is written into the relation itself: on the left is $\lambda$, an attribute of a wave, and on the right is $p$, an attribute of a particle, with Planck's constant $h$ linking the two. 🔉⇢

The relation is really a hypothesis, testable only by experiment, but it is reassuring that it is already obeyed by a photon. For a photon $p=h\nu/c$, so $h/p=c/\nu=\lambda$ — the de Broglie wavelength of a photon is just the ordinary wavelength of the light of which it is a quantum. The relation therefore unifies radiation and matter under one formula: momentum determines wavelength, whatever the object. 🔉⇢

The single most important feature of the relation is the smallness of $h$. Because $h=6.63\times10^{-34}\,\text{J s}$ is so tiny, $\lambda$ is extremely small for a heavy or fast object and only becomes appreciable for a very light one. This is precisely why everyday objects show no detectable wave behaviour, while electrons do. The whole difference between the classical and quantum worlds hides in the size of one constant. 🔉⇢

Take the standard NCERT example of a cricket-ball-sized object: a ball of mass $0.12\,\text{kg}$ moving at $20\,\text{m s}^{-1}$ has momentum $p=mv=2.40\,\text{kg m s}^{-1}$ and de Broglie wavelength $\lambda=h/p=2.76\times10^{-34}\,\text{m}$. This wavelength is so far below the size of any object or aperture that no diffraction or interference could ever be observed — which is the reason macroscopic objects in daily life do not show wave-like properties. 🔉⇢

Now contrast an electron. An electron of mass $9.11\times10^{-31}\,\text{kg}$ moving at $5.4\times10^{6}\,\text{m s}^{-1}$ has momentum $p=4.92\times10^{-24}\,\text{kg m s}^{-1}$ and de Broglie wavelength $\lambda=h/p=0.135\,\text{nm}$. This is comparable to X-ray wavelengths and to the spacing of atomic planes in a crystal — so a crystal, which diffracts X-rays, should diffract electrons too. In the sub-atomic domain the wave character of particles is significant and measurable. 🔉⇢

A very useful special case is an electron accelerated from rest through a potential difference $V$. It gains kinetic energy $eV=\tfrac12 mv^{2}=p^{2}/2m$, so $p=\sqrt{2meV}$ and $\lambda=\dfrac{h}{\sqrt{2meV}}$. Putting in the constants gives the memorable form $\lambda=\dfrac{1.227}{\sqrt{V}}\,\text{nm}$ (with $V$ in volts). A $54\,\text{V}$ electron, for instance, has $\lambda\approx0.167\,\text{nm}$ — the value at the heart of the Davisson-Germer result. 🔉⇢

That experiment is the confirmation. In 1927 Davisson and Germer directed a beam of electrons, accelerated through a known voltage, at a nickel crystal and measured the intensity of electrons scattered in different directions. They found a sharp peak in the scattered intensity at a particular angle for a particular voltage — a diffraction maximum, exactly as a wave of the de Broglie wavelength $\lambda=h/\sqrt{2meV}$ should produce off the regularly spaced atomic planes. The measured wavelength matched the de Broglie prediction, establishing the wave nature of the electron. The scene above lets you sweep the accelerating voltage and watch the diffraction maximum appear. 🔉⇢

It is worth being careful about what does and does not appear in the de Broglie relation. The wavelength depends only on the momentum; it does not depend on the charge or on the specific nature of the particle. An electron, a proton and a neutron of the same momentum have the same de Broglie wavelength, even though their charges and masses differ. For the same kinetic energy, however, a heavier particle has larger momentum ($p=\sqrt{2mK}$) and hence a shorter wavelength — a distinction JEE problems test often. 🔉⇢

This card closes the chapter's central story. Light, once purely a wave, was given a particle nature by the photon; matter, once purely particles, was given a wave nature by de Broglie and Davisson-Germer. Neither description is complete on its own; which one is appropriate depends on the experiment. The de Broglie wavelength is what makes the electron microscope possible and what Schrödinger built into a full wave mechanics — the theory that governs the quantum world. 🔉⇢

De Broglie's hypothesis is that the wave-particle nature found in light extends to matter. Just as light, long regarded as a wave, was shown by the photoelectric effect to have a particle nature, de Broglie proposed that moving particles of matter have a wave nature. A moving particle of momentum equal to its mass times its velocity has associated with it a wavelength, the de Broglie wavelength, equal to Planck's constant divided by the momentum. The larger the momentum of the particle, the smaller its de Broglie wavelength. 🔉⇢

The relation comes by analogy with the photon. For a photon the energy is Planck's constant times the frequency, and the momentum is the energy divided by the speed of light, which is the same as Planck's constant divided by the wavelength. De Broglie assumed that the same relation between wavelength and momentum holds for a material particle as for a photon. So for any moving particle the wavelength associated with it is Planck's constant divided by the momentum of the particle, and this is the central relation of the wave nature of matter. 🔉⇢

Because Planck's constant is so small, the de Broglie wavelength of an everyday object is far too small to be measured. A moving object of ordinary mass has a very large momentum compared with Planck's constant, so its de Broglie wavelength is extremely small, many orders of magnitude smaller than the size of any atom. This is why the wave nature of matter is never noticed for macroscopic objects: the wavelength associated with them is immeasurably small, and no wave effect such as diffraction can be observed. 🔉⇢

For a particle as light as an electron the situation is quite different. Because the mass of the electron is very small, its momentum is small and the de Broglie wavelength associated with it is large enough to be measured. An electron moving at a modest speed has a de Broglie wavelength comparable to the spacing of atoms, so its wave nature can be revealed by experiment. The smallness of the electron's mass is what brings the wave nature of matter within reach of measurement. 🔉⇢

A particularly useful case is an electron accelerated from rest through a potential difference. The electron gains kinetic energy equal to its charge times the potential difference, and from the kinetic energy its momentum follows, since the kinetic energy is the square of the momentum divided by twice the mass. The de Broglie wavelength associated with the electron is then Planck's constant divided by that momentum. In this way the wavelength of the electron is fixed by the potential difference through which it has been accelerated, and larger accelerating potentials give the electron a larger momentum and hence a smaller wavelength. 🔉⇢

The de Broglie wavelength depends only on the momentum of the particle, not separately on its mass or its speed. Two particles that have the same momentum have the same de Broglie wavelength, whatever their masses. If instead two particles have the same kinetic energy, the heavier particle has the larger momentum and so the smaller wavelength, because for a given kinetic energy the momentum is the square root of twice the mass times the kinetic energy. And if two particles move with the same speed, the heavier particle again has the larger momentum and the smaller wavelength. 🔉⇢

The wave nature of matter proposed by de Broglie was confirmed by experiment. A beam of electrons of known energy, and therefore of known de Broglie wavelength, was directed at a crystal, and the electrons were found to be scattered in a way that showed the interference and diffraction characteristic of waves. The wavelength deduced from the diffraction agreed with the de Broglie wavelength calculated from the momentum of the electrons. This confirmation established that the de Broglie relation holds for material particles just as it does for photons. 🔉⇢

The dual nature of matter is thus placed on the same footing as the dual nature of radiation. Light shows a wave nature in interference and diffraction and a particle nature in the photoelectric and Compton effects; matter, through the de Broglie relation and the diffraction of electrons, shows a wave nature as well as its familiar particle nature. Neither the wave picture nor the particle picture alone is complete; which nature is revealed depends on the experiment performed. This is the wave-particle duality that lies at the heart of the quantum picture of nature. 🔉⇢

It is worth being careful about what the de Broglie wavelength is and is not. It is the wavelength associated with a moving particle of a given momentum, and it becomes shorter as the momentum of the particle increases. It is not tied to the charge of the particle: an uncharged particle of matter has a de Broglie wavelength given by the same relation, Planck's constant divided by the momentum. The relation applies to every material particle, from an electron to a macroscopic object, differing only in the size of the momentum and hence of the wavelength. 🔉⇢

The smallness of Planck's constant is the single fact that separates the quantum world of the electron from the everyday world of large objects. For an electron the momentum is small enough that the de Broglie wavelength is measurable and the wave nature of matter shows itself in diffraction. For a large moving object the momentum is so large that the de Broglie wavelength is unimaginably small and the wave nature can never be observed. One relation, Planck's constant divided by the momentum, governs both; only the value of the momentum decides whether the wave nature of matter can be seen. 🔉⇢

The de Broglie relation also explains why the wave nature of matter matters most for particles of small mass and low energy. A slow, light particle has a small momentum and therefore a large de Broglie wavelength, so its wave nature is easiest to observe; a fast, heavy particle has a large momentum and a small wavelength, so it behaves like an ordinary particle. This is why the wave nature of matter dominates the behaviour of electrons in atoms and is quite hidden in the motion of everyday objects. 🔉⇢

In summary, de Broglie extended the wave-particle duality of radiation to matter by proposing that every moving particle has a wavelength equal to Planck's constant divided by its momentum. The wavelength is large for particles of small momentum, such as slow electrons, and immeasurably small for macroscopic objects. The relation is confirmed by the diffraction of electrons, which shows the wave nature of matter directly. Together with the photon nature of light, it establishes that both radiation and matter have a dual, wave-particle nature, the value of the momentum deciding which nature is revealed in a given experiment. 🔉⇢

It helps to connect the de Broglie wavelength to the kinetic energy of the particle directly. For a particle of a given mass, the momentum is the square root of twice the mass times the kinetic energy, so the de Broglie wavelength is Planck's constant divided by the square root of twice the mass times the kinetic energy. A particle of larger kinetic energy therefore has a larger momentum and a smaller wavelength. This is why increasing the energy of a beam of electrons shortens the de Broglie wavelength associated with them, and why a slow electron of small kinetic energy has a longer wavelength than a fast one. 🔉⇢

The de Broglie relation makes a striking numerical contrast between an electron and an everyday object. An electron accelerated through a modest potential difference has a de Broglie wavelength comparable to the size of an atom, so its wave nature is easily revealed. An ordinary moving object, however massive and slow, has so large a momentum that its de Broglie wavelength is far smaller than any length that could ever be measured. The same relation, Planck's constant divided by the momentum, gives a measurable wavelength for the electron and a hopelessly small one for the everyday object; only the momentum differs. 🔉⇢

Historically the wave nature of matter came as a bold guess before it was seen. De Broglie put forward the hypothesis that a moving particle of matter has a wavelength associated with it purely by analogy with the photon, before any experiment had shown a wave effect for matter. The hypothesis was soon confirmed when a beam of electrons was found to be diffracted, and the wavelength deduced from the diffraction matched the de Broglie wavelength calculated from the momentum of the electrons. A daring analogy between matter and light had been vindicated by experiment. 🔉⇢

The wave nature of matter does not replace the particle nature; the two are complementary. An electron still has a definite charge and a definite mass and still behaves as a particle when it is detected, yet a beam of electrons shows diffraction and interference like a wave. The de Broglie wavelength tells us how strong the wave behaviour will be: when the wavelength is comparable to the size of the obstacle or aperture, the wave nature shows itself, and when the wavelength is far smaller, the particle behaves as an ordinary particle. This is the same complementarity that light shows between its photon nature and its wave nature. 🔉⇢

A useful way to fix the idea is to compare a photon and an electron of the same wavelength. A photon of a given wavelength has a momentum equal to Planck's constant divided by that wavelength, and by de Broglie's relation an electron of the same wavelength has exactly the same momentum. Their energies differ, because the energy of a photon is its momentum times the speed of light while the kinetic energy of the electron is the square of its momentum divided by twice its mass, but the wavelength and the momentum are shared. This is the sense in which the de Broglie relation puts the wave nature of matter on the same footing as the wave nature of light: one relation between wavelength and momentum serves both the photon and the moving particle of matter. 🔉⇢

Derivation from first principles 🔉⇢

  1. For a photon, $E=h\nu$ and $p=h\nu/c$, so $\lambda=h/p$. De Broglie postulates the SAME relation for matter: $\lambda=h/p=h/mv$.
  2. Everyday object (ball): $p=mv=0.12\times20=2.40\,\text{kg m s}^{-1}$, $\lambda=h/p=2.76\times10^{-34}\,\text{m}$ — unmeasurably small.
  3. Electron accelerated through $V$: $eV=p^{2}/2m\Rightarrow p=\sqrt{2meV}$.
  4. Hence $\lambda=\dfrac{h}{\sqrt{2meV}}=\dfrac{1.227}{\sqrt{V}}\,\text{nm}$ (with $V$ in volts).
  5. Davisson-Germer: electrons diffract off a nickel crystal with a maximum at the angle predicted for this $\lambda$ — confirming matter waves.
⚠️ JEE trap: Students often think the de Broglie wavelength depends on charge, or use $\lambda=h/mv$ with a photon (a photon has no rest mass — use $\lambda=h/p$). Another trap: for equal kinetic energy a lighter particle (electron) has a LONGER wavelength than a heavier one (proton), because $\lambda=h/\sqrt{2mK}$; for equal SPEED it is the reverse. Track whether the question fixes energy, momentum or speed. 🔉⇢

Worked example · JEE Advanced 🔉⇢

SITUATION An electron and a ball of mass $150\,\text{g}$ are considered. The electron moves at $5.4\times10^{6}\,\text{m s}^{-1}$ and the ball at $30.0\,\text{m s}^{-1}$.
TARGET Find the de Broglie wavelength of each and comment.
STRATEGY Use $\lambda=h/(mv)$ for both, with $m_e=9.11\times10^{-31}\,\text{kg}$ and $m_{ball}=0.150\,\text{kg}$.
EXECUTE Electron: $p=9.11\times10^{-31}\times5.4\times10^{6}=4.92\times10^{-24}$, $\lambda=\dfrac{6.63\times10^{-34}}{4.92\times10^{-24}}=0.135\,\text{nm}$. Ball: $p=0.150\times30.0=4.50$, $\lambda=\dfrac{6.63\times10^{-34}}{4.50}=1.47\times10^{-34}\,\text{m}.$
REFLECT The electron's wavelength is comparable to X-ray wavelengths and to atomic-plane spacings, so it diffracts off crystals; the ball's is about $10^{-19}$ times the size of a proton — utterly beyond measurement.

Source: NCERT Example 11.3

✍️ Worked Examples Polya 5-move · JEE tier

WE1 · Electron Emission and the Work Function · JEE Advanced 🔉⇢

SITUATION The work function of caesium is $\phi_0=2.14\,\text{eV}$.
TARGET Find the threshold frequency $\nu_0$ below which caesium shows no photoelectric emission.
STRATEGY At threshold the photon energy just equals the work function: $h\nu_0=\phi_0$. Convert $\phi_0$ to joules before dividing by $h$.
EXECUTE $\nu_0=\dfrac{\phi_0}{h}=\dfrac{2.14\times1.6\times10^{-19}\,\text{J}}{6.63\times10^{-34}\,\text{J s}}=5.16\times10^{14}\,\text{Hz}.$
REFLECT This lies in the visible (green-blue), which is why caesium responds to visible light. For frequencies below $5.16\times10^{14}\,\text{Hz}$ no photoelectrons are ejected however intense the beam.

Source: NCERT Example 11.2(a)

WE2 · The Photoelectric Effect: Hertz, Hallwachs and Lenard · JEE Advanced 🔉⇢

SITUATION Ultraviolet light ejects electrons from a zinc plate ($\phi_0\approx4.3\,\text{eV}$), but an intense beam of red light ($\lambda=650\,\text{nm}$) produces no current at all.
TARGET Explain why, using the photon energy of the red light.
STRATEGY Compare the energy of a single red photon with the work function of zinc; emission needs $h\nu\ge\phi_0$.
EXECUTE $E_{red}=\dfrac{hc}{\lambda}=\dfrac{(6.63\times10^{-34})(3\times10^{8})}{650\times10^{-9}}=3.06\times10^{-19}\,\text{J}=1.91\,\text{eV}$, which is less than $\phi_0=4.3\,\text{eV}$.
REFLECT Because one red photon carries only $1.91\,\text{eV}$, no single photon can free an electron from zinc; adding more red photons (greater intensity) only adds more too-weak photons. This is precisely the threshold effect the wave picture cannot explain.

Source: NCERT §11.3 (derived)

WE3 · Stopping Potential vs Frequency and the Threshold Frequency · JEE Advanced 🔉⇢

SITUATION The threshold frequency for a metal is $\nu_0=3.3\times10^{14}\,\text{Hz}$. Light of frequency $\nu=8.2\times10^{14}\,\text{Hz}$ falls on it.
TARGET Predict the stopping (cut-off) potential for the photoelectric emission.
STRATEGY Use $eV_0=h(\nu-\nu_0)$, so $V_0=\dfrac{h(\nu-\nu_0)}{e}$.
EXECUTE $V_0=\dfrac{(6.63\times10^{-34})(8.2-3.3)\times10^{14}}{1.6\times10^{-19}}=\dfrac{(6.63\times10^{-34})(4.9\times10^{14})}{1.6\times10^{-19}}=2.0\,\text{V}.$
REFLECT Only the frequency difference above threshold matters. Doubling the intensity would leave $V_0=2.0\,\text{V}$ unchanged; only shifting $\nu$ changes it.

Source: NCERT Exercise 11.6

WE4 · Why the Wave Theory of Light Fails · JEE Advanced 🔉⇢

SITUATION A student argues that shining a very intense infrared laser on a sodium surface ($\phi_0=2.3\,\text{eV}$) for a long time must eventually eject electrons.
TARGET Decide whether this is correct and state which wave-theory prediction it relies on.
STRATEGY Check whether a single infrared photon (say $\lambda=1000\,\text{nm}$) can supply $\phi_0$; recall that intensity and time do not add photon energies.
EXECUTE $E=\dfrac{hc}{\lambda}=\dfrac{(6.63\times10^{-34})(3\times10^{8})}{1000\times10^{-9}}=1.99\times10^{-19}\,\text{J}=1.24\,\text{eV}\lt 2.3\,\text{eV}$.
REFLECT The student is wrong. Each infrared photon delivers only $1.24\,\text{eV}$, less than $\phi_0$, and photons act one-at-a-time on single electrons, so no intensity or exposure time helps. The argument relies on the discredited wave prediction that energy accumulates continuously.

Source: NCERT §11.5 (derived)

WE5 · Einstein's Photoelectric Equation · JEE Advanced 🔉⇢

SITUATION Light of wavelength $\lambda=488\,\text{nm}$ from an argon laser falls on an emitter and the stopping potential is measured to be $V_0=0.38\,\text{V}$.
TARGET Find the work function of the emitter material.
STRATEGY Use $\phi_0=\dfrac{hc}{\lambda}-eV_0$. Compute the photon energy in eV with $hc=1240\,\text{eV nm}$, then subtract $eV_0=0.38\,\text{eV}$.
EXECUTE $E=\dfrac{1240\,\text{eV nm}}{488\,\text{nm}}=2.54\,\text{eV}$; $\phi_0=2.54-0.38=2.16\,\text{eV}.$
REFLECT About $2.16\,\text{eV}$ is a typical alkali-metal work function — consistent with a surface that responds to visible (blue) light. Using $hc=1240\,\text{eV nm}$ avoided any joule conversion.

Source: NCERT Exercise 11.9

WE6 · The Photon: Energy and Momentum · JEE Advanced 🔉⇢

SITUATION A laser produces monochromatic light of frequency $\nu=6.0\times10^{14}\,\text{Hz}$ and emits a power $P=2.0\times10^{-3}\,\text{W}$.
TARGET Find (a) the energy of one photon and (b) the number of photons emitted per second.
STRATEGY Use $E=h\nu$ for one photon and $N=P/E$ for the flux.
EXECUTE $E=h\nu=(6.63\times10^{-34})(6.0\times10^{14})=3.98\times10^{-19}\,\text{J}$; $N=\dfrac{P}{E}=\dfrac{2.0\times10^{-3}}{3.98\times10^{-19}}=5.0\times10^{15}\text{ per second}.$
REFLECT Five thousand million million photons a second from just two milliwatts — the beam's graininess is utterly invisible, which is why light seems continuous in everyday life.

Source: NCERT Example 11.1

WE7 · Photon energy and photon flux of a laser · easy 🔉⇢

SITUATION A monochromatic laser of frequency $6.0\times10^{14}\,\text{Hz}$ emits power $2.0\times10^{-3}\,\text{W}$.
TARGET Find the energy of one photon and the number of photons emitted per second.
STRATEGY Each photon has $E=h\nu$; the number per second is $N=P/E$.
EXECUTE $E=(6.63\times10^{-34})(6.0\times10^{14})=3.98\times10^{-19}\,\text{J}$; $N=\dfrac{2.0\times10^{-3}}{3.98\times10^{-19}}=5.0\times10^{15}\text{ s}^{-1}$.
REFLECT The enormous photon count is why laser light appears perfectly smooth and continuous.

Source: JEE-pattern

WE8 · Threshold frequency and wavelength of caesium · medium 🔉⇢

SITUATION Caesium has work function $\phi_0=2.14\,\text{eV}$; a stopping potential of $0.60\,\text{V}$ brings the photocurrent to zero.
TARGET Find the threshold frequency and the wavelength of the incident light.
STRATEGY Threshold: $h\nu_0=\phi_0$. Wavelength from $eV_0=hc/\lambda-\phi_0\Rightarrow\lambda=hc/(eV_0+\phi_0)$.
EXECUTE $\nu_0=\dfrac{2.14\times1.6\times10^{-19}}{6.63\times10^{-34}}=5.16\times10^{14}\,\text{Hz}$; $\lambda=\dfrac{(6.63\times10^{-34})(3\times10^{8})}{(0.60+2.14)\times1.6\times10^{-19}}=454\,\text{nm}$.
REFLECT The $454\,\text{nm}$ blue light lies above the threshold, so emission occurs and a $0.60\,\text{V}$ retarding potential just stops it.

Source: JEE-pattern

WE9 · Maximum kinetic energy, stopping potential and speed · medium 🔉⇢

SITUATION Light of frequency $6\times10^{14}\,\text{Hz}$ is incident on caesium ($\phi_0=2.14\,\text{eV}$).
TARGET Find the maximum kinetic energy, the stopping potential, and the maximum speed of the photoelectrons.
STRATEGY $K_{max}=h\nu-\phi_0$; $V_0=K_{max}/e$; $v_{max}=\sqrt{2K_{max}/m_e}$.
EXECUTE $h\nu=(6.63\times10^{-34})(6\times10^{14})=3.98\times10^{-19}\,\text{J}=2.49\,\text{eV}$; $K_{max}=2.49-2.14=0.35\,\text{eV}=5.6\times10^{-20}\,\text{J}$; $V_0=0.35\,\text{V}$; $v_{max}=\sqrt{2(5.6\times10^{-20})/9.11\times10^{-31}}=3.5\times10^{5}\,\text{m s}^{-1}$.
REFLECT Just above threshold the electrons come off slowly; a modest $0.35\,\text{V}$ suffices to stop them.

Source: JEE-pattern

WE10 · Does 330 nm light eject electrons from a 4.2 eV metal? · easy 🔉⇢

SITUATION A metal has work function $4.2\,\text{eV}$; light of wavelength $330\,\text{nm}$ falls on it.
TARGET Decide whether photoelectric emission occurs.
STRATEGY Compare the photon energy $hc/\lambda$ with $\phi_0$; emission needs $hc/\lambda\ge\phi_0$.
EXECUTE $E=\dfrac{1240\,\text{eV nm}}{330\,\text{nm}}=3.76\,\text{eV}\lt 4.2\,\text{eV}$.
REFLECT The photon energy is below the work function, so no emission occurs however intense the light — a pure threshold decision.

Source: JEE-pattern

WE11 · Planck's constant from the stopping-potential slope · medium 🔉⇢

SITUATION The slope of the cut-off-voltage-versus-frequency line is $4.12\times10^{-15}\,\text{V s}$.
TARGET Determine Planck's constant.
STRATEGY Slope $=h/e$, so $h=e\times\text{slope}$.
EXECUTE $h=(1.6\times10^{-19})(4.12\times10^{-15})=6.6\times10^{-34}\,\text{J s}$.
REFLECT Independent of the work function — the slope alone gives $h$, matching the black-body value.

Source: JEE-pattern

WE12 · de Broglie wavelength from an accelerating voltage · medium 🔉⇢

SITUATION An electron is accelerated from rest through a potential difference of $100\,\text{V}$.
TARGET Find its de Broglie wavelength.
STRATEGY $\lambda=h/\sqrt{2meV}=1.227/\sqrt{V}\,\text{nm}$.
EXECUTE $\lambda=\dfrac{1.227}{\sqrt{100}}=0.1227\,\text{nm}=0.123\,\text{nm}$.
REFLECT Comparable to atomic-plane spacings — which is why such electrons diffract off crystals (electron microscopy, Davisson-Germer).

Source: JEE-pattern

WE13 · Comparing de Broglie wavelengths of an electron and a proton · hard 🔉⇢

SITUATION An electron and a proton are accelerated through the same potential difference $V$.
TARGET Find the ratio of their de Broglie wavelengths.
STRATEGY $\lambda=h/\sqrt{2mqV}$; with the same $q=e$ and $V$, $\lambda\propto1/\sqrt{m}$.
EXECUTE $\dfrac{\lambda_e}{\lambda_p}=\sqrt{\dfrac{m_p}{m_e}}=\sqrt{\dfrac{1.67\times10^{-27}}{9.11\times10^{-31}}}=\sqrt{1836}\approx42.8$.
REFLECT The lighter electron has the much longer wavelength for the same accelerating voltage.

Source: JEE-pattern

WE14 · Radiation force of a fully absorbed light beam · hard 🔉⇢

SITUATION A beam of light of power $P=30\,\text{mW}$ is completely absorbed by a small object.
TARGET Find the force the light exerts on the object.
STRATEGY A fully absorbed beam delivers momentum $P/c$ per second, so $F=P/c$.
EXECUTE $F=\dfrac{30\times10^{-3}}{3\times10^{8}}=1.0\times10^{-10}\,\text{N}$.
REFLECT Radiation pressure is minute for ordinary beams but real — it is measurable and matters for solar sails and dust grains.

Source: JEE-pattern

WE15 · Ratio of de Broglie wavelengths of an electron and a photon of equal energy · JEE Advanced 🔉⇢

SITUATION An electron (mass $m$) and a photon have the same energy $E$, of a few eV.
TARGET Find the ratio of their de Broglie wavelengths.
STRATEGY Photon: $\lambda_{ph}=hc/E$. Electron (non-relativistic, $E=p^2/2m$): $\lambda_e=h/\sqrt{2mE}$.
EXECUTE $\dfrac{\lambda_e}{\lambda_{ph}}=\dfrac{h/\sqrt{2mE}}{hc/E}=\dfrac{E}{c\sqrt{2mE}}=\dfrac{1}{c}\sqrt{\dfrac{E}{2m}}=\dfrac{v_e}{2c}$, where $v_e$ is the electron's speed.
REFLECT Since $v_e\ll c$ for a few-eV electron, its de Broglie wavelength is far shorter than that of a photon of the same energy.

Source: JEE-pattern

WE16 · Momentum kick from an absorbed light pulse · JEE Advanced 🔉⇢

SITUATION A $100\,\text{ns}$ pulse of light of power $30\,\text{mW}$ is completely absorbed by a small object initially at rest.
TARGET Find the momentum delivered to the object.
STRATEGY Total energy $U=P\,t$; for full absorption the momentum is $\Delta p=U/c$.
EXECUTE $U=(30\times10^{-3})(100\times10^{-9})=3.0\times10^{-9}\,\text{J}$; $\Delta p=\dfrac{3.0\times10^{-9}}{3\times10^{8}}=1.0\times10^{-17}\,\text{kg m s}^{-1}$.
REFLECT A tiny but definite kick — the particle nature of light made mechanical.

Source: JEE-pattern

WE17 · Two frequencies on the same metal: eliminating the work function · hard 🔉⇢

SITUATION On one metal, light of frequency $\nu_1$ gives stopping potential $V_1$ and light of frequency $\nu_2$ gives $V_2$.
TARGET Show how to find Planck's constant without knowing the work function.
STRATEGY Write $eV_1=h\nu_1-\phi_0$ and $eV_2=h\nu_2-\phi_0$ and subtract.
EXECUTE $e(V_1-V_2)=h(\nu_1-\nu_2)\Rightarrow h=\dfrac{e(V_1-V_2)}{\nu_1-\nu_2}$.
REFLECT Subtracting cancels $\phi_0$: this is exactly the slope method Millikan used, in two-point form.

Source: JEE-pattern

WE18 · Work function well and the escape condition · medium 🔉⇢

SITUATION A metal surface has work function $4.5\,\text{eV}$. Light of wavelength $200\,\text{nm}$ is incident.
TARGET Find the maximum kinetic energy of the photoelectrons.
STRATEGY $K_{max}=hc/\lambda-\phi_0$; use $hc=1240\,\text{eV nm}$.
EXECUTE $E=\dfrac{1240}{200}=6.2\,\text{eV}$; $K_{max}=6.2-4.5=1.7\,\text{eV}=2.7\times10^{-19}\,\text{J}$.
REFLECT The $200\,\text{nm}$ ultraviolet is well above threshold, so electrons emerge with a healthy $1.7\,\text{eV}$.

Source: JEE-pattern

WE19 · Photoelectron bent into a circle by a magnetic field · JEE Advanced 🔉⇢

SITUATION Light of wavelength $200\,\text{nm}$ falls on a metal of work function $4.2\,\text{eV}$. The fastest photoelectrons enter a uniform magnetic field $B=5.0\times10^{-4}\,\text{T}$ perpendicular to their velocity.
TARGET Find the radius of the circular path of the fastest photoelectrons.
STRATEGY Get $K_{max}=hc/\lambda-\phi_0$, then $v_{max}=\sqrt{2K_{max}/m_e}$, then $r=\dfrac{m_ev_{max}}{eB}=\dfrac{\sqrt{2m_eK_{max}}}{eB}$.
EXECUTE $E=\dfrac{1240}{200}=6.2\,\text{eV}$; $K_{max}=6.2-4.2=2.0\,\text{eV}=3.2\times10^{-19}\,\text{J}$; $p=\sqrt{2m_eK_{max}}=\sqrt{2(9.11\times10^{-31})(3.2\times10^{-19})}=7.6\times10^{-25}\,\text{kg m s}^{-1}$; $r=\dfrac{p}{eB}=\dfrac{7.6\times10^{-25}}{(1.6\times10^{-19})(5.0\times10^{-4})}=9.5\times10^{-3}\,\text{m}\approx9.5\,\text{mm}.$
REFLECT A classic Advanced cross-topic problem: the photoelectric effect fixes the electron's energy, then magnetism bends it. The radius carries the momentum $p=\sqrt{2m_eK_{max}}$ directly.

Source: JEE-pattern

WE20 · Two wavelengths, one metal: find work function and Planck's constant · JEE Advanced 🔉⇢

SITUATION On the same metal, light of wavelength $\lambda_1=350\,\text{nm}$ gives stopping potential $V_1=1.45\,\text{V}$ and $\lambda_2=550\,\text{nm}$ gives $V_2=0.16\,\text{V}$.
TARGET Find Planck's constant and the work function of the metal.
STRATEGY Write $eV=hc/\lambda-\phi_0$ for each wavelength and subtract to eliminate $\phi_0$; then back-substitute for $\phi_0$.
EXECUTE $e(V_1-V_2)=hc\left(\dfrac{1}{\lambda_1}-\dfrac{1}{\lambda_2}\right)$. Numerically $\dfrac{1}{\lambda_1}-\dfrac{1}{\lambda_2}=\dfrac{1}{350}-\dfrac{1}{550}=1.04\times10^{-3}\,\text{nm}^{-1}$, and $V_1-V_2=1.29\,\text{V}$, giving $hc=\dfrac{1.29}{1.04\times10^{-3}}\approx1240\,\text{eV nm}$, i.e. $h=6.63\times10^{-34}\,\text{J s}$. Then $\phi_0=\dfrac{1240}{550}-0.16=2.09\,\text{eV}.$
REFLECT Subtracting the two equations cancels the work function and isolates $h$ — the two-point form of Millikan's slope method, a staple of Advanced numericals.

Source: JEE-pattern

WE21 · de Broglie wavelength of a thermal neutron · JEE Advanced 🔉⇢

SITUATION A neutron (mass $1.67\times10^{-27}\,\text{kg}$) is in thermal equilibrium at $T=300\,\text{K}$, with average kinetic energy $K=\tfrac32 k_BT$ ($k_B=1.38\times10^{-23}\,\text{J K}^{-1}$).
TARGET Estimate its de Broglie wavelength.
STRATEGY Compute $K=\tfrac32 k_BT$, then $\lambda=\dfrac{h}{\sqrt{2mK}}$.
EXECUTE $K=\tfrac32(1.38\times10^{-23})(300)=6.2\times10^{-21}\,\text{J}$; $\sqrt{2mK}=\sqrt{2(1.67\times10^{-27})(6.2\times10^{-21})}=4.5\times10^{-24}\,\text{kg m s}^{-1}$; $\lambda=\dfrac{6.63\times10^{-34}}{4.5\times10^{-24}}=1.5\times10^{-10}\,\text{m}=0.15\,\text{nm}.$
REFLECT A thermal neutron's wavelength is about an atomic spacing — which is exactly why neutron diffraction is a standard probe of crystal structure, just like electron and X-ray diffraction.

Source: JEE-pattern

WE22 · Radiation pressure of a reflected laser beam · hard 🔉⇢

SITUATION A laser delivers power $P=1.0\,\text{W}$ onto a perfectly reflecting mirror at normal incidence.
TARGET Find the force the beam exerts on the mirror.
STRATEGY For full reflection the momentum delivered per second is doubled: $F=\dfrac{2P}{c}$.
EXECUTE $F=\dfrac{2(1.0)}{3\times10^{8}}=6.7\times10^{-9}\,\text{N}.$
REFLECT A perfect reflector reverses each photon's momentum, so it receives twice the push of a perfect absorber ($F=P/c$). Tiny, but it is the principle of the solar sail.

Source: JEE-pattern

WE23 · Ratio of de Broglie wavelengths at equal kinetic energy · JEE Advanced 🔉⇢

SITUATION An electron, a proton and an alpha particle each have the same kinetic energy $K$.
TARGET Order their de Broglie wavelengths.
STRATEGY $\lambda=\dfrac{h}{\sqrt{2mK}}\propto\dfrac{1}{\sqrt{m}}$ at fixed $K$; the lightest has the longest wavelength.
EXECUTE Masses: $m_e\lt m_p\lt m_\alpha$ (electron $\approx9.11\times10^{-31}$, proton $\approx1.67\times10^{-27}$, alpha $\approx4m_p$). Hence $\lambda_e\gt\lambda_p\gt\lambda_\alpha$, with $\dfrac{\lambda_e}{\lambda_p}=\sqrt{\dfrac{m_p}{m_e}}\approx43$ and $\dfrac{\lambda_p}{\lambda_\alpha}=\sqrt{\dfrac{m_\alpha}{m_p}}=2.$
REFLECT At equal kinetic energy the wavelength ranking follows inverse-root-mass, so the electron's wave is by far the longest and the alpha's the shortest.

Source: JEE-pattern

On the concept tabs

These worked examples are taught in full alongside their interactive scene:

📐 Formula Sheet Printable · every formula cited

Electron emission and the work function

QuantityFormulaWhat it means / when to useSource
Work function (escape energy) 🔉⇢$\phi_0 = h\nu_0$minimum energy to free a surface electron; $\nu_0$ = threshold frequencyNCERT Class XII Ch 11
Electron volt 🔉⇢$1\,\text{eV} = 1.602\times10^{-19}\,\text{J}$energy gained across a $1\,\text{V}$ dropNCERT Class XII Ch 11
Specific charge of the electron 🔉⇢$\dfrac{e}{m} = 1.76\times10^{11}\,\text{C kg}^{-1}$Thomson; independent of the emitterNCERT Class XII Ch 11

Photoelectric effect

QuantityFormulaWhat it means / when to useSource
Maximum kinetic energy from stopping potential 🔉⇢$K_{max} = eV_0$$V_0$ = stopping (cut-off) potentialNCERT Class XII Ch 11
Einstein's photoelectric equation 🔉⇢$K_{max} = h\nu - \phi_0$one photon, one electronNCERT Class XII Ch 11
Threshold frequency 🔉⇢$\nu_0 = \dfrac{\phi_0}{h}$below $\nu_0$: no emission at any intensityNCERT Class XII Ch 11
Stopping potential vs frequency 🔉⇢$V_0 = \dfrac{h}{e}\,\nu - \dfrac{\phi_0}{e}$straight line, slope $h/e = 4.14\times10^{-15}\,\text{V s}$NCERT Class XII Ch 11
Threshold wavelength 🔉⇢$\lambda_0 = \dfrac{hc}{\phi_0}$emission if $\lambda \lt \lambda_0$NCERT Class XII Ch 11

The photon

QuantityFormulaWhat it means / when to useSource
Photon energy 🔉⇢$E = h\nu = \dfrac{hc}{\lambda}$$hc = 1240\,\text{eV nm}$NCERT Class XII Ch 11
Photon momentum 🔉⇢$p = \dfrac{h\nu}{c} = \dfrac{h}{\lambda}$$E = pc$ (massless)NCERT Class XII Ch 11
Photon flux of a beam 🔉⇢$N = \dfrac{P}{h\nu}$photons per second from power $P$NCERT Class XII Ch 11
Radiation force (full absorption) 🔉⇢$F = \dfrac{P}{c}$double it for a perfect reflectorNCERT Class XII Ch 11

Matter waves

QuantityFormulaWhat it means / when to useSource
de Broglie wavelength 🔉⇢$\lambda = \dfrac{h}{p} = \dfrac{h}{mv}$wave nature of a moving particleNCERT Class XII Ch 11
For an accelerated charge 🔉⇢$\lambda = \dfrac{h}{\sqrt{2mqV}}$from $qV = p^2/2m$NCERT Class XII Ch 11
Electron shortcut 🔉⇢$\lambda = \dfrac{1.227}{\sqrt{V}}\,\text{nm}$$V$ in volts, electron from restNCERT Class XII Ch 11
From kinetic energy 🔉⇢$\lambda = \dfrac{h}{\sqrt{2mK}}$equal $K$: lighter particle has longer $\lambda$NCERT Class XII Ch 11

Constants

QuantityFormulaWhat it means / when to useSource
Planck's constant 🔉⇢$h = 6.63\times10^{-34}\,\text{J s} = 4.14\times10^{-15}\,\text{eV s}$Planck's constantNCERT Class XII Ch 11
Electron mass / charge 🔉⇢$m_e = 9.11\times10^{-31}\,\text{kg},\ e = 1.6\times10^{-19}\,\text{C}$Electron mass / chargeNCERT Class XII Ch 11
hc (handy) 🔉⇢$hc = 1240\,\text{eV nm} = 1.99\times10^{-25}\,\text{J m}$hc (handy)NCERT Class XII Ch 11

📜 Previous-Year Questions Authentic NTA · 67 questions

Every PYQ traces to an official NTA/JAB paper (year + shift + paper). Answer keys match the official key. OCR-extracted questions are flagged for SME verification.

JEE Main 2021 · Paper 1 · August 27 Shift 1 · Q13 (official key (printed in paper)) Answer: (D) increases the number of photons incident and the K.E. of the ejected electrons remains unchanged

In a photoelectric experiment, increasing the intensity of incident light :

  • (A) increases the number of photons incident and also increases the K.E. of the ejected electrons
  • (B) increases the frequency of photons incident and increases the K.E. of the ejected electrons
  • (C) increases the frequency of photons incident and the K.E. of the ejected electrons remains unchanged
  • (D) increases the number of photons incident and the K.E. of the ejected electrons remains unchanged
JEE Main 2021 · Paper 1 · August 27 Shift 2 · Q13 (official key (printed in paper)) Answer: (B) 1.25 V

A monochromatic neon lamp with wavelength of 670.5 nm illuminates a photo-sensitive material which has a stopping voltage of 0.48 V. What will be the stopping voltage if the source light is changed with another source of wavelength of 474.6 nm?

  • (A) 0.96 V
  • (B) 1.25 V
  • (C) 0.24 V
  • (D) 1.5 V
JEE Main 2021 · Paper 1 · March 16 Shift 1 · Q18 (official key (printed in paper)) Answer: (D) Intensity

The stopping potential in the context of photoelectric effect depends on the following property of incident electromagnetic radiation :

  • (A) Phase
  • (B) Frequency
  • (C) Amnplitude
  • (D) Intensity
JEE Main 2021 · Paper 1 · August 26 Shift 2 · Q2 (official key (printed in paper)) Answer: (B) ${7 \over 9}$E

The de-Broglie wavelength of a particle having kinetic energy E is $\lambda$. How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to 75% of the initial value?

  • (A) ${1 \over 9}$E
  • (B) ${7 \over 9}$E
  • (C) E
  • (D) ${16 \over 9}$E
JEE Main 2021 · Paper 1 · August 31 Shift 2 · Q8 (official key (printed in paper)) Answer: (C) $\sqrt {{{{m_p}} \over {{m_e}}}}$

Consider two separate ideal gases of electrons and protons having same number of particles. The temperature of both the gases are same. The ratio of the uncertainty in determining the position of an electron to that of a proton is proportional to :-

  • (A) ${\left( {{{{m_p}} \over {{m_e}}}} \right)^{3/2}}$
  • (B) $\sqrt {{{{m_e}} \over {{m_p}}}}$
  • (C) $\sqrt {{{{m_p}} \over {{m_e}}}}$
  • (D) ${{{m_p}} \over {{m_e}}}$
JEE Main 2023 · Paper 1 · January 25 Shift 1 · Q1 (official key (printed in paper)) Answer: (D) $\frac{\lambda_0}{2}$

Electron beam used in an electron microscope, when accelerated by a voltage of 20 kV, has a de-Broglie wavelength of $\lambda_0$. IF the voltage is increased to 40 kV, then the de-Broglie wavelength associated with the electron beam would be :

  • (A) 3 $\lambda_0$
  • (B) 9 $\lambda_0$
  • (C) $\frac{\lambda_0}{\sqrt2}$
  • (D) $\frac{\lambda_0}{2}$
JEE Main 2023 · Paper 1 · January 31 Shift 2 · Q10 (official key (printed in paper)) Answer: (D) Both metals A and B will emit photo-electrons

If the two metals $\mathrm{A}$ and $\mathrm{B}$ are exposed to radiation of wavelength $350 \mathrm{~nm}$. The work functions of metals $\mathrm{A}$ and $\mathrm{B}$ are $4.8 \mathrm{eV}$ and $2.2 \mathrm{eV}$. Then choose the correct option.

  • (A) Metal B will not emit photo-electrons
  • (B) Both metals $\mathrm{A}$ and $\mathrm{B}$ will not emit photo-electrons
  • (C) Metal A will not emit photo-electrons
  • (D) Both metals A and B will emit photo-electrons
JEE Main 2023 · Paper 1 · February 1 Shift 1 · Q12 (official key (printed in paper)) Answer: (B) 2 : 1

A proton moving with one tenth of velocity of light has a certain de Broglie wavelength of $\lambda$. An alpha particle having certain kinetic energy has the same de-Brogle wavelength $\lambda$. The ratio of kinetic energy of proton and that of alpha particle is:

  • (A) 1 : 4
  • (B) 2 : 1
  • (C) 4 : 1
  • (D) 1 : 2
JEE Main 2023 · Paper 1 · January 31 Shift 1 · Q12 (official key (printed in paper)) Answer: (D) A is correct but R is not correct

Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R Assertion A : The beam of electrons show wave nature and exhibit interference and diffraction. Reason R : Davisson Germer Experimentally verified the wave nature of electrons. In the light of the above statements, choose the most appropriate answer from the options given below :

  • (A) A is not correct but R is correct.
  • (B) Both A and R are correct and R is the correct explanation of A
  • (C) Both A and R are correct but R is Not the correct explanation of A
  • (D) A is correct but R is not correct
JEE Main 2023 · Paper 1 · January 29 Shift 1 · Q18 (official key (printed in paper)) Answer: (D) B and C only

The threshold wavelength for photoelectric emission from a material is 5500 $\overset{\circ}{A}$. Photoelectrons will be emitted, when this material is illuminated with monochromatic radiation from a A. 75 W infra-red lamp B. 10 W infra-red lamp C. 75 W ultra-violet lamp D. 10 W ultra-violet lamp Choose the correct answer from the options given below :

  • (A) C only
  • (B) A and D only
  • (C) C and D only
  • (D) B and C only
JEE Main 2023 · Paper 1 · January 25 Shift 2 · Q19 (official key (printed in paper)) Answer: (D) Statement I is incorrect but Statement II is correct

Given below are two statements : Statement I : Stopping potential in photoelectric effect does not depend on the power of the light source. Statement II : For a given metal, the maximum kinetic energy of the photoelectron depends on the wavelength of the incident light. In the light of above statements, choose the most appropriate answer from the options given below

  • (A) Both Statement I and Statement II are incorrect
  • (B) Statement I is correct but Statement II is incorrect
  • (C) Both Statement I and Statement II are correct
  • (D) Statement I is incorrect but Statement II is correct
JEE Main 2023 · Paper 1 · January 31 Shift 1 · Q2 (official key (printed in paper)) Answer: (C) Micro waves

If a source of electromagnetic radiation having power $15 \mathrm{~kW}$ produces $10^{16}$ photons per second, the radiation belongs to a part of spectrum is. (Take Planck constant $h=6 \times 10^{-34} \mathrm{Js}$ )

  • (A) Gamma rays
  • (B) Radio waves
  • (C) Micro waves
  • (D) Ultraviolet rays
JEE Main 2023 · Paper 1 · January 24 Shift 1 · Q3 (official key (printed in paper)) Answer: (C) B, C only

From the photoelectric effect experiment, following observations are made. Identify which of these are correct. A. The stopping potential depends only on the work function of the metal. B. The saturation current increases as the intensity of incident light increases. C. The maximum kinetic energy of a photo electron depends on the intensity of the incident light. D. Photoelectric effect can be explained using wave theory of light. Choose the correct answer from the options given below :

  • (A) A, B, D only
  • (B) A, C, D only
  • (C) B, C only
  • (D) B only
JEE Main 2023 · Paper 1 · January 24 Shift 2 · Q3 (official key (printed in paper)) Answer: (B) ${\lambda _\alpha } > {\lambda _p} > {\lambda _e}$

An $\alpha$-particle, a proton and an electron have the same kinetic energy. Which one of the following is correct in case of their de-Broglie wavelength:

  • (A) ${\lambda _\alpha } > {\lambda _p} < {\lambda _e}$
  • (B) ${\lambda _\alpha } > {\lambda _p} > {\lambda _e}$
  • (C) ${\lambda _\alpha } = {\lambda _p} = {\lambda _e}$
  • (D) ${\lambda _\alpha } < {\lambda _p} < {\lambda _e}$
JEE Main 2023 · Paper 1 · January 30 Shift 1 · Q7 (official key (printed in paper)) Answer: (B) $0.5 \times 10^{-17} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$

A small object at rest, absorbs a light pulse of power $20 \mathrm{~mW}$ and duration $300 \mathrm{~ns}$. Assuming speed of light as $3 \times 10^{8} \mathrm{~m} / \mathrm{s}$, the momentum of the object becomes equal to :

  • (A) $1 \times 10^{-17} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$
  • (B) $0.5 \times 10^{-17} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$
  • (C) $3 \times 10^{-17} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$
  • (D) $2 \times 10^{-17} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$
IIT-JEE 2009 · Paper 1 · Q56 (official key) Answer: A

When a particle is restricted to move along $x$-axis between $x = 0$ and $x = a$, where $a$ is of nanometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends $x = 0$ and $x = a$. The wavelength of this standing wave is related to the linear momentum $p$ of the particle according to the de Broglie relation. The energy of the particle of mass $m$ is related to its linear momentum as $E = \dfrac{p^2}{2m}$. Thus, the energy of the particle can be denoted by a quantum number '$n$' taking values $1, 2, 3, \ldots$ ($n = 1$, called the ground state) corresponding to the number of loops in the standing wave. Use the model described above to answer the following question for a particle moving in the line $x = 0$ to $x = a$. Take $h = 6.6\times10^{-34}$ J s and $e = 1.6\times10^{-19}$ C. The allowed energy for the particle for a particular value of $n$ is proportional to

  • (A) $a^{-2}$
  • (B) $a^{-3/2}$
  • (C) $a^{-1}$
  • (D) $a^{2}$
IIT-JEE 2009 · Paper 1 · Q57 (official key) Answer: B

When a particle is restricted to move along $x$-axis between $x = 0$ and $x = a$, where $a$ is of nanometer dimension, its energy can take only certain specific values. The allowed energies correspond to standing waves with nodes at $x = 0$ and $x = a$; the wavelength is related to the linear momentum $p$ by the de Broglie relation, and $E = \dfrac{p^2}{2m}$. The energy is denoted by a quantum number $n = 1, 2, 3, \ldots$ ($n = 1$ is the ground state) corresponding to the number of loops in the standing wave. Take $h = 6.6\times10^{-34}$ J s and $e = 1.6\times10^{-19}$ C. If the mass of the particle is $m = 1.0\times10^{-30}$ kg and $a = 6.6$ nm, the energy of the particle in its ground state is closest to

  • (A) 0.8 meV
  • (B) 8 meV
  • (C) 80 meV
  • (D) 800 meV
IIT-JEE 2009 · Paper 1 · Q58 (official key) Answer: D

When a particle is restricted to move along $x$-axis between $x = 0$ and $x = a$, where $a$ is of nanometer dimension, its energy can take only certain specific values. The allowed energies correspond to standing waves with nodes at $x = 0$ and $x = a$; the wavelength is related to the linear momentum $p$ by the de Broglie relation, and $E = \dfrac{p^2}{2m}$. The energy is denoted by a quantum number $n = 1, 2, 3, \ldots$ ($n = 1$ is the ground state) corresponding to the number of loops in the standing wave. The speed of the particle, that can take discrete values, is proportional to

  • (A) $n^{-3/2}$
  • (B) $n^{-1}$
  • (C) $n^{1/2}$
  • (D) $n$
IIT-JEE 2010 · Paper 1 · Q76 (official key) Answer: 3

An $\alpha$-particle and a proton are accelerated from rest by a potential difference of 100 V. After this, their de Broglie wavelengths are $\lambda_\alpha$ and $\lambda_p$ respectively. The ratio $\dfrac{\lambda_p}{\lambda_\alpha}$, to the nearest integer, is

IIT-JEE 2011 · Paper 2 · Q38 (official key) Answer: 7

A silver sphere of radius 1 cm and work function 4.7 eV is suspended from an insulating thread in free-space. It is under continuous illumination of 200 nm wavelength light. As photoelectrons are emitted, the sphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from the sphere is $A\times10^{Z}$ (where $1<A<10$). The value of $Z$ is

JEE Advanced 2013 · Paper 1 · Q16 (official key) Answer: 1

The work functions of Silver and Sodium are $4.6$ and $2.3$ eV, respectively. The ratio of the slope of the stopping potential versus frequency plot for Silver to that of Sodium is

JEE Advanced 2013 · Paper 1 · Q6 (official key) Answer: B

A pulse of light of duration $100$ ns is absorbed completely by a small object initially at rest. Power of the pulse is $30$ mW and the speed of light is $3\times10^{8}\ \text{m s}^{-1}$. The final momentum of the object is

  • (A) $0.3\times10^{-17}\ \text{kg m s}^{-1}$
  • (B) $1.0\times10^{-17}\ \text{kg m s}^{-1}$
  • (C) $3.0\times10^{-17}\ \text{kg m s}^{-1}$
  • (D) $9.0\times10^{-17}\ \text{kg m s}^{-1}$
JEE Advanced 2014 · Paper 2 · Q7 (official key) Answer: A

A metal surface is illuminated by light of two different wavelengths $248$ nm and $310$ nm. The maximum speeds of the photoelectrons corresponding to these wavelengths are $u_1$ and $u_2$, respectively. If the ratio $u_1 : u_2 = 2 : 1$ and $hc = 1240$ eV nm, the work function of the metal is nearly

  • (A) $3.7$ eV
  • (B) $3.2$ eV
  • (C) $2.8$ eV
  • (D) $2.5$ eV
JEE Advanced 2016 · Paper 1 · Q1 (official key) Answer: B

In a historical experiment to determine Planck's constant, a metal surface was irradiated with light of different wavelengths. The emitted photoelectron energies were measured by applying a stopping potential. The relevant data for the wavelength ($\lambda$) of incident light and the corresponding stopping potential ($V_0$) are given below: $\lambda$ ($\mu$m) : $0.3$, $0.4$, $0.5$ $V_0$ (Volt) : $2.0$, $1.0$, $0.4$ Given that $c = 3 \times 10^8$ m s$^{-1}$ and $e = 1.6 \times 10^{-19}$ C, Planck's constant (in units of J s) found from such an experiment is

  • (A) $6.0 \times 10^{-34}$
  • (B) $6.4 \times 10^{-34}$
  • (C) $6.6 \times 10^{-34}$
  • (D) $6.8 \times 10^{-34}$
JEE Advanced 2016 · Paper 2 · Q8 (official key) Answer: C

Light of wavelength $\lambda_{\text{ph}}$ falls on a cathode plate inside a vacuum tube as shown in the figure. The work function of the cathode surface is $\phi$ and the anode is a wire mesh of conducting material kept at a distance $d$ from the cathode. A potential difference $V$ is maintained between the electrodes. If the minimum de Broglie wavelength of the electrons passing through the anode is $\lambda_e$, which of the following statement(s) is(are) true?

  • (A) $\lambda_e$ decreases with increase in $\phi$ and $\lambda_{\text{ph}}$
  • (B) $\lambda_e$ is approximately halved, if $d$ is doubled
  • (C) For large potential difference ($V \gg \phi/e$), $\lambda_e$ is approximately halved if $V$ is made four times
  • (D) $\lambda_e$ increases at the same rate as $\lambda_{\text{ph}}$ for $\lambda_{\text{ph}} < hc/\phi$
JEE Advanced 2017 · Paper 2 · Q3 (official key) Answer: D

A photoelectric material having work-function $\phi_0$ is illuminated with light of wavelength $\lambda$ $\left(\lambda < \dfrac{hc}{\phi_0}\right)$. The fastest photoelectron has a de Broglie wavelength $\lambda_d$. A change in wavelength of the incident light by $\Delta\lambda$ results in a change $\Delta\lambda_d$ in $\lambda_d$. Then the ratio $\Delta\lambda_d/\Delta\lambda$ is proportional to

  • (A) $\lambda_d/\lambda$
  • (B) $\lambda_d^2/\lambda^2$
  • (C) $\lambda_d^3/\lambda$
  • (D) $\lambda_d^3/\lambda^2$
JEE Advanced 2018 · Paper 2 · Q13 (official key) Answer: 24.00

In a photoelectric experiment a parallel beam of monochromatic light with power of $200\ \mathrm{W}$ is incident on a perfectly absorbing cathode of work function $6.25\ \mathrm{eV}$. The frequency of light is just above the threshold frequency so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is 100%. A potential difference of $500\ \mathrm{V}$ is applied between the cathode and the anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force $F = n \times 10^{-4}\ \mathrm{N}$ due to the impact of the electrons. The value of $n$ is __________. Mass of the electron $m_e = 9 \times 10^{-31}\ \mathrm{kg}$ and $1.0\ \mathrm{eV} = 1.6 \times 10^{-19}\ \mathrm{J}$.

JEE Advanced 2019 · Paper 2 · Q12 (official key) Answer: 1

A perfectly reflecting mirror of mass $M$ mounted on a spring constitutes a spring-mass system of angular frequency $\Omega$ such that $\frac{4\pi M\Omega}{h} = 10^{24}\ \mathrm{m^{-2}}$ with $h$ as Planck's constant. $N$ photons of wavelength $\lambda = 8\pi \times 10^{-6}$ m strike the mirror simultaneously at normal incidence such that the mirror gets displaced by 1 $\mu$m. If the value of $N$ is $x \times 10^{12}$, then the value of $x$ is ____. [Consider the spring as massless]

JEE Advanced 2021 · Paper 2 · Q19 (official key) Answer: 6

In a photoemission experiment, the maximum kinetic energies of photoelectrons from metals $P$, $Q$ and $R$ are $E_P$, $E_Q$ and $E_R$, respectively, and they are related by $E_P = 2E_Q = 2E_R$. In this experiment, the same source of monochromatic light is used for metals $P$ and $Q$ while a different source of monochromatic light is used for the metal $R$. The work functions for metals $P$, $Q$ and $R$ are 4.0 eV, 4.5 eV and 5.5 eV, respectively. The energy of the incident photon used for metal $R$, in eV, is ___.

JEE Advanced 2022 · Paper 2 · Q16 (official key) Answer: A

When light of a given wavelength is incident on a metallic surface, the minimum potential needed to stop the emitted photoelectrons is $6.0\ V$. This potential drops to $0.6\ V$ if another source with wavelength four times that of the first one and intensity half of the first one is used. What are the wavelength of the first source and the work function of the metal, respectively? [Take $\dfrac{hc}{e} = 1.24 \times 10^{-6}\ J\ m\ C^{-1}$.]

  • (A) $1.72 \times 10^{-7}\ m$, $1.20\ eV$
  • (B) $1.72 \times 10^{-7}\ m$, $5.60\ eV$
  • (C) $3.78 \times 10^{-7}\ m$, $5.60\ eV$
  • (D) $3.78 \times 10^{-7}\ m$, $1.20\ eV$
JEE Main 2026 · Paper 1 · April 5 Shift 1 · Q35 (official key) Answer: C

An electron of mass $m$ is moving in an electric field $\vec{E}=-2 E_{\mathrm{o}} \hat{i}\left(E_{\mathrm{o}}=\right.$ constant $\left.>0\right)$, with an initial velocity $\vec{V}=v_{\mathrm{o}} \hat{i} \left(v_{\mathrm{o}}=\right.$ constant $\left.>0\right)$. If $\lambda_{\mathrm{o}}=\frac{h}{4 m v_{\mathrm{o}}}$, its de Broglie wavelength at time $t$ is $\_\_\_\_$ . ( $e=$ charge of electron)

  • (A) $\frac{4 \lambda_{\mathrm{o}}}{\left[1-\frac{E_{\mathrm{o}} e}{2 m} \frac{t}{v_{\mathrm{o}}}\right]}$
  • (B) $\frac{4 \lambda_{\mathrm{o}}}{\left[1+\frac{E_{\mathrm{o}} e}{2 m} \frac{t}{v_{\mathrm{o}}}\right]}$
  • (C) $\frac{4 \lambda_{\mathrm{o}}}{\left[1+\frac{2 E_{\mathrm{o}} e}{m} \frac{t}{v_{\mathrm{o}}}\right]}$
  • (D) $\frac{4 \lambda_{\mathrm{o}}}{\left[1-\frac{2 E_{\mathrm{o}} e}{m} \frac{t}{v_{\mathrm{o}}}\right]}$
JEE Main 2026 · Paper 1 · April 8 Shift 2 · Q38 (official key) Answer: D

A monochromatic source of light operating at 15 kW emits $2.5 \times 10^{22}$ photons $/ \mathrm{s}$. The region of an electromagnetic spectrum to which the emitted electromagnetic radiation belongs to $\_\_\_\_$. (Take $h=6.6 \times 10^{-34} \mathrm{~J} . \mathrm{s}$ and $c=3 \times 10^8 \mathrm{~m} / \mathrm{s}$ ).

  • (A) Microwave
  • (B) Infrared
  • (C) Visible
  • (D) Ultraviolet
JEE Main 2026 · Paper 1 · April 5 Shift 1 · Q41 (official key) Answer: C

Light source having wavelength 331 nm is used to generate photo-electrons whose stopping potential is 0.2 V . The work function of the used metal in the experiment is $\alpha \times 10^{-19} \mathrm{~J}$. The value of $\alpha$ is $\_\_\_\_$ . $\left(\mathrm{h}=6.62 \times 10^{-34} \mathrm{~J} \mathrm{~s}, \mathrm{e}=1.6 \times 10^{-19} \mathrm{C} \text { and } \mathrm{c}=3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)$

  • (A) 3.68
  • (B) 4.68
  • (C) 5.68
  • (D) 2.68
JEE Main 2026 · Paper 1 · April 8 Shift 2 · Q43 (official key) Answer: D

$K_1$ and $K_2$ be the maximum kinetic energies of photoelectrons emitted from a surface of a given material for the light of wavelength $\lambda_1$ and $\lambda_2$, respectively. If $\lambda_1=2 \lambda_2$ then the work function of material is given by :

  • (A) $K_2+2 K_1$
  • (B) $2 K_2-K_1$
  • (C) $K_1-2 K_2$
  • (D) $K_2-2 K_1$
JEE Main 2020 · Paper 1 · January 7 Shift 2 · Q13 (published compilation) Answer: B⚑ verify

An electron (of mass m) and a photon have the same energy E in the range of a few eV. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is (c = speed of light in vaccuum)

  • (A) ${1 \over c}{\left( {{{2E} \over m}} \right)^{{1 \over 2}}}$
  • (B) ${1 \over c}{\left( {{E \over {2m}}} \right)^{{1 \over 2}}}$
  • (C) ${\left( {{E \over {2m}}} \right)^{{1 \over 2}}}$
  • (D) $c{\left( {2mE} \right)^{{1 \over 2}}}$
JEE Main 2020 · Paper 1 · January 9 Shift 2 · Q18 (published compilation) Answer: C⚑ verify

An electron of mass m and magnitude of charge |e| initially at rest gets accelerated by a constant electric field E. The rate of change of de-Broglie wavelength of this electron at time t ignoring relativistic effects is :

  • (A) ${{ - h} \over {\left| e \right|Et}}$
  • (B) ${{ - h} \over {\left| e \right|E\sqrt t }}$
  • (C) ${{ - h} \over {\left| e \right|E{t^2}}}$
  • (D) ${{\left| e \right|Et} \over h}$
JEE Main 2020 · Paper 1 · September 5 Shift 2 · Q22 (published compilation) Answer: 2⚑ verify

The surface of a metal is illuminated alternately with photons of energies $E_{1}$ = 4 eV and $E_{2}$ = 2.5 eV respectively. The ratio of maximum speeds of the photoelectrons emitted in the two cases is 2. The work function of the metal in (eV) is _____.

JEE Main 2020 · Paper 1 · September 3 Shift 2 · Q3 (published compilation) Answer: A⚑ verify

Two sources of light emit X-rays of wavelength 1 nm and visible light of wavelength 500 nm, respectively. Both the sources emit light of the same power 200 W. The ratio of the number density of photons of X-rays to the number density of photons of the visible light of the given wavelengths is :

  • (A) ${1 \over {500}}$
  • (B) 500
  • (C) 250
  • (D) ${1 \over {250}}$
JEE Main 2020 · Paper 1 · January 8 Shift 1 · Q8 (published compilation) Answer: B⚑ verify

When photon of energy 4.0 eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy $T_{A}$ eV end de-Broglie wavelength $\lambda _A$. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50 eV is $T_{B}$ = ($T_{A}$ – 1.5) eV. If the de-Broglie wavelength of these photoelectrons $\lambda _B$ = 2$\lambda _A$, then the work function of metal B is :

  • (A) 1.5eV
  • (B) 4eV
  • (C) 2eV
  • (D) 3eV
JEE Main 2020 · Paper 1 · September 3 Shift 1 · Q9 (published compilation) Answer: C⚑ verify

When the wavelength of radiation falling on a metal is changed from 500 nm to 200 nm, the maximum kinetic energy of the photoelectrons becomes three times larger. The work function of the metal is close to :

  • (A) 1.02 eV
  • (B) 0.81 eV
  • (C) 0.61 eV
  • (D) 0.52 eV
JEE Main 2021 · Paper 1 · February 25 Shift 2 · Q11 (published compilation) Answer: B⚑ verify

An electron of mass $m_{e}$ and a proton of mass $m_{p}$ = 1836 $m_{e}$ are moving with the same speed. The ratio of their de Broglie wavelength ${{{}^\lambda electron} \over {{}^\lambda proton}}$ will be :

  • (A) 1
  • (B) 1836
  • (C) ${1 \over {1836}}$
  • (D) 918
JEE Main 2021 · Paper 1 · February 25 Shift 2 · Q7 (published compilation) Answer: D⚑ verify

The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength 491 nm is 0.710 V. When the incident wavelength is changed to a new value, the stopping potential is 1.43 V. The new wavelength is :

  • (A) 400 nm
  • (B) 329 nm
  • (C) 309 nm
  • (D) 382 nm
JEE Main 2021 · Paper 1 · February 24 Shift 1 · Q9 (published compilation) Answer: D⚑ verify

Given below are two statements : Statement I : Two photons having equal linear momenta have equal wavelengths. Statement II : If the wavelength of photon is decreased, then the momentum and energy of a photon will also decrease. In the light of the above statements, choose the correct answer from the options given below.

  • (A) Statement I is false but Statement II is true
  • (B) Both Statement I and Statement II are false
  • (C) Both Statement I and Statement II are true
  • (D) Statement I is true but Statement II is false
JEE Main 2022 · Paper 1 · July 29 Shift 2 · Q10 (published compilation) Answer: B⚑ verify

An $\alpha$ particle and a proton are accelerated from rest through the same potential difference. The ratio of linear momenta acquired by above two particles will be:

  • (A) $\sqrt2$ : 1
  • (B) 2$\sqrt2$ : 1
  • (C) 4$\sqrt2$ : 1
  • (D) 8 : 1
JEE Main 2022 · Paper 1 · June 27 Shift 1 · Q13 (published compilation) Answer: B⚑ verify

An $\alpha$ particle and a carbon 12 atom has same kinetic energy K. The ratio of their de-Broglie wavelengths $({\lambda _\alpha }:{\lambda _{C12}})$ is :

  • (A) $1:\sqrt 3$
  • (B) $\sqrt 3 :1$
  • (C) $3:1$
  • (D) $2:\sqrt 3$
JEE Main 2022 · Paper 1 · June 25 Shift 1 · Q15 (published compilation) Answer: A⚑ verify

Given below are two statements : Statement I : Davisson-Germer experiment establishes the wave nature of electrons. Statement II : If electrons have wave nature, they can interfere and show diffraction. In the light of the above statements choose the correct answer from the option given below :

  • (A) Both Statement I and Statement II are true.
  • (B) Both Statement I and Statement II are false.
  • (C) Statement I is true but Statement II is false.
  • (D) Statement I is false but Statement II is true.
JEE Main 2022 · Paper 1 · July 29 Shift 1 · Q16 (published compilation) Answer: B⚑ verify

The kinetic energy of emitted electron is E when the light incident on the metal has wavelength $\lambda$. To double the kinetic energy, the incident light must have wavelength:

  • (A) $\frac{\mathrm{hc}}{\mathrm{E} \lambda-\mathrm{hc}}$
  • (B) $\frac{\mathrm{hc} \lambda}{\mathrm{E} \lambda+\mathrm{hc}}$
  • (C) $\frac{\mathrm{h} \lambda}{\mathrm{E} \lambda+\mathrm{hc}}$
  • (D) $\frac{\text { hc } \lambda}{\mathrm{E} \lambda-\mathrm{hc}}$
JEE Main 2022 · Paper 1 · July 25 Shift 1 · Q17 (published compilation) Answer: C⚑ verify

A metal exposed to light of wavelength $800 \mathrm{~nm}$ and and emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of photo-electron doubles when light of wavelength $500 \mathrm{~nm}$ is used. The workfunction of the metal is : (Take hc $=1230 \,\mathrm{eV}-\mathrm{nm}$ ).

  • (A) 1.537 eV
  • (B) 2.46 eV
  • (C) 0.615 eV
  • (D) 1.23 eV
JEE Main 2022 · Paper 1 · July 27 Shift 2 · Q17 (published compilation) Answer: B⚑ verify

With reference to the observations in photo-electric effect, identify the correct statements from below : (A) The square of maximum velocity of photoelectrons varies linearly with frequency of incident light. (B) The value of saturation current increases on moving the source of light away from the metal surface. (C) The maximum kinetic energy of photo-electrons decreases on decreasing the power of LED (light emitting diode) source of light. (D) The immediate emission of photo-electrons out of metal surface can not be explained by particle nature of light/electromagnetic waves. (E) Existence of threshold wavelength can not be explained by wave nature of light/ electromagnetic waves. Choose the correct answer from the options given below :

  • (A) and (B) only
  • (A) and (E) only
  • (C) and (E) only
  • (D) and (E) only
JEE Main 2022 · Paper 1 · July 28 Shift 2 · Q17 (published compilation) Answer: C⚑ verify

Two streams of photons, possessing energies equal to five and ten times the work function of metal are incident on the metal surface successively. The ratio of maximum velocities of the photoelectron emitted, in the two cases respectively, will be

  • (A) 1 : 2
  • (B) 1 : 3
  • (C) 2 : 3
  • (D) 3 : 2
JEE Main 2022 · Paper 1 · June 26 Shift 1 · Q17 (published compilation) Answer: B⚑ verify

An electron with speed v and a photon with speed c have the same de-Broglie wavelength. If the kinetic energy and momentum of electron are $E_{e}$ and $p_{e}$ and that of photon are $E_{ph}$ and $p_{ph}$ respectively. Which of the following is correct?

  • (A) ${{{E_e}} \over {{E_{ph}}}} = {{2c} \over v}$
  • (B) ${{{E_e}} \over {{E_{ph}}}} = {v \over {2c}}$
  • (C) ${{{p_e}} \over {{p_{ph}}}} = {{2c} \over v}$
  • (D) ${{{p_e}} \over {{p_{ph}}}} = {v \over {2c}}$
JEE Main 2022 · Paper 1 · June 26 Shift 2 · Q17 (published compilation) Answer: A⚑ verify

A metal surface is illuminated by a radiation of wavelength 4500 $\overset{\circ}{A}$. The ejected photo-electron enters a constant magnetic field of 2 mT making an angle of 90$^\circ$ with the magnetic field. If it starts revolving in a circular path of radius 2 mm, the work function of the metal is approximately :

  • (A) 1.36 eV
  • (B) 1.69 eV
  • (C) 2.78 eV
  • (D) 2.23 eV
JEE Main 2022 · Paper 1 · July 26 Shift 1 · Q4 (published compilation) Answer: B⚑ verify

A parallel beam of light of wavelength $900 \mathrm{~nm}$ and intensity $100 \,\mathrm{Wm}^{-2}$ is incident on a surface perpendicular to the beam. The number of photons crossing $1 \mathrm{~cm}^{2}$ area perpendicular to the beam in one second is :

  • (A) $3 \times 10^{16}$
  • (B) $4.5 \times 10^{16}$
  • (C) $4.5 \times 10^{17}$
  • (D) $4.5 \times 10^{20}$
JEE Main 2022 · Paper 1 · July 25 Shift 2 · Q7 (published compilation) Answer: D⚑ verify

The ratio of wavelengths of proton and deuteron accelerated by potential $V_{p}$ and $V_{d}$ is 1 : $\sqrt2$. Then the ratio of $V_{p}$ to $V_{d}$ will be :

  • (A) 1 : 1
  • (B) $\sqrt2$ : 1
  • (C) 2 : 1
  • (D) 4 : 1
JEE Main 2022 · Paper 1 · June 29 Shift 1 · Q7 (published compilation) Answer: B⚑ verify

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R : Assertion A : The photoelectric effect does not takes place, if the energy of the incident radiation is less than the work function of a metal. Reason R : Kinetic energy of the photoelectrons is zero, if the energy of the incident radiation is equal to the work function of a metal. In the light of the above statements, choose the most appropriate answer from the options given below.

  • (A) Both A and R are correct and R is the correct explanation of A.
  • (B) Both A and R are correct but R is not the correct explanation of A.
  • (C) A is correct but R is not correct.
  • (D) A is not correct but R is correct.
JEE Main 2023 · Paper 1 · April 11 Shift 2 · Q38 (published compilation) Answer: A⚑ verify

The ratio of the de-Broglie wavelengths of proton and electron having same Kinetic energy : (Assume $m_{p}=m_{e} \times 1849$ )

  • (A) 1:43
  • (B) 1:62
  • (C) 2:43
  • (D) 1:30
JEE Main 2023 · Paper 1 · April 8 Shift 1 · Q40 (published compilation) Answer: A⚑ verify

Proton $(\mathrm{P})$ and electron (e) will have same de-Broglie wavelength when the ratio of their momentum is (assume, $\mathrm{m}_{\mathrm{p}}=1849 \mathrm{~m}_{\mathrm{e}}$ ):

  • (A) 1 : 1
  • (B) 1 : 43
  • (C) 1 : 1849
  • (D) 43 : 1
JEE Main 2023 · Paper 1 · April 13 Shift 2 · Q42 (published compilation) Answer: B⚑ verify

Given below are two statements: Statement I : Out of microwaves, infrared rays and ultraviolet rays, ultraviolet rays are the most effective for the emission of electrons from a metallic surface. Statement II : Above the threshold frequency, the maximum kinetic energy of photoelectrons is inversely proportional to the frequency of the incident light. In the light of above statements, choose the correct answer form the options given below

  • (A) Both Statement I and Statement II are true
  • (B) Statement I is true but statement II is false
  • (C) Statement I is false but statement II is true
  • (D) Both Statement I and Statement II are false
JEE Main 2023 · Paper 1 · April 6 Shift 2 · Q49 (published compilation) Answer: C⚑ verify

The work functions of Aluminium and Gold are $4.1 ~\mathrm{eV}$ and and $5.1 ~\mathrm{eV}$ respectively. The ratio of the slope of the stopping potential versus frequency plot for Gold to that of Aluminium is

  • (A) 1.5
  • (B) 1.24
  • (C) 1
  • (D) 2
JEE Main 2023 · Paper 1 · April 8 Shift 2 · Q50 (published compilation) Answer: D⚑ verify

In photo electric effect A. The photocurrent is proportional to the intensity of the incident radiation B. Maximum Kinetic energy with which photoelectrons are emitted depends on the intensity of incident light. C. Max. K.E with which photoelectrons are emitted depends on the frequency of incident light. D. The emission of photoelectrons require a minimum threshold intensity of incident radiation. E. Max. K.E of the photoelectrons is independent of the frequency of the incident light. Choose the correct answer from the options given below:

  • (A) A and E only
  • (B) A and B only
  • (C) B and C only
  • (D) A and C only
JEE Main 2023 · Paper 1 · April 11 Shift 1 · Q54 (published compilation) Answer: 3⚑ verify

A monochromatic light is incident on a hydrogen sample in ground state. Hydrogen atoms absorb a fraction of light and subsequently emit radiation of six different wavelengths. The frequency of incident light is $x \times 10^{15} \mathrm{~Hz}$. The value of $x$ is ____________. (Given h $=4.25 \times 10^{-15} ~\mathrm{eVs}$ )

JEE Main 2023 · Paper 1 · April 13 Shift 2 · Q59 (published compilation) Answer: 4125⚑ verify

An atom absorbs a photon of wavelength $500 \mathrm{~nm}$ and emits another photon of wavelength $600 \mathrm{~nm}$. The net energy absorbed by the atom in this process is $n \times 10^{-4} ~\mathrm{eV}$. The value of n is __________. [Assume the atom to be stationary during the absorption and emission process] (Take $\mathrm{h}=6.6 \times 10^{-34} ~\mathrm{Js}$ and $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ )

JEE Main 2026 · Paper 1 · April 6 Shift 2 · Q47 (published compilation) Answer: 4⚑ verify

The de Broglie wavelength for an electron accelerated through the potential difference of $V_1$ volt is $\lambda_1$. When the potential difference is changed to $V_2$ volt, the associated de Broglie wavelength is increased by $50 \%$. If $\left(V_1 / V_2\right)=(9 / \alpha)$, then the value of $\alpha$ is $\_\_\_\_$.

JEE Main 2024 · Paper 1 · January 27 Shift 2 · Q1 (published compilation) Answer: C⚑ verify

The threshold frequency of a metal with work function $6.63 \mathrm{~eV}$ is :

  • (A) $16 \times 10^{15} \mathrm{~Hz}$
  • (B) $16 \times 10^{12} \mathrm{~Hz}$
  • (C) $1.6 \times 10^{15} \mathrm{~Hz}$
  • (D) $1.6 \times 10^{12} \mathrm{~Hz}$
JEE Main 2026 · Paper 1 · April 2 Shift 1 · Q42 (published compilation) Answer: B⚑ verify

For a certain metal, when monochromatic light of wavelength $\lambda$ is incident, the stopping potential for photoelectrons is $3V_0$. When the same metal is illuminated by light of wavelength $2\lambda$, then the stopping potential becomes $V_0$. The threshold wavelength for photoelectric emission for the given metal is $\alpha \lambda$. The value of $\alpha$ is ______.

  • (A) 1
  • (B) 4
  • (C) 2
  • (D) 3
JEE Main 2026 · Paper 1 · April 5 Shift 2 · Q43 (published compilation) Answer: C⚑ verify

An electron is travelling with a velocity $v$ in free space and when it enters a medium, its velocity is reduced by $20 \%$. The de Broglie wavelength of electron in the medium is $\alpha \lambda_0$, where $\lambda_0$ is its de Broglie wavelength in free space. The value of $\alpha$ is $\_\_\_\_$ .

  • (A) 1.20
  • (B) 1.0
  • (C) 1.25
  • (D) 0.75
JEE Main 2026 · Paper 1 · April 4 Shift 2 · Q44 (published compilation) Answer: A⚑ verify

The de Broglie wavelength associated with an electron accelerated through a potential difference V is $\lambda_{\mathrm{e}}$ and the de Broglie wavelength associated with a proton accelerated through the same potential difference is $\lambda_{\mathrm{p}}$. If their corresponding masses are $m_{\mathrm{e}}$ and $m_{\mathrm{p}}$, respectively, then the ratio of their de Broglie wavelengths $\left(\frac{\lambda_e}{\lambda_p}\right)$ is $\_\_\_\_$ .

  • (A) $\text { } \sqrt{\frac{m_p}{m_e}}$
  • (B) $\sqrt{\frac{m_e}{m_p}}$
  • (C) $\frac{m_p}{m_e}$
  • (D) $\left(\frac{m_p}{m_e}\right)^2$

🎯 Question Bank 113 MCQs · graded

Distribution — advanced: 14 · easy: 36 · hard: 16 · medium: 47. Every question carries a source trace; each ends in an SME-verify solution.

Q1 The minimum energy required by an electron to just escape from a metal surface is called the easy
Step solution + source
The work function $\phi_0$ is the least energy needed to free an electron from the metal surface, as the surface pulls an escaping electron back. It is a property of the metal and its surface, measured in eV, and is different from nuclear binding or ionisation of an isolated atom. 🔉⇢

Source: NCERT §11.2

Q2 Which process supplies the escape energy in photoelectric emission? easy
Step solution + source
In photoelectric emission the escape energy comes from an absorbed light quantum $h\nu$. Thermionic emission uses heat and field emission uses a strong field ($\sim 10^8\,\text{V/m}$); those are the other two routes but are not the photoelectric one. 🔉⇢

Source: NCERT §11.2

Q3 One electron volt equals easy
Step solution + source
$1\,\text{eV}$ is the energy gained by an electron accelerated through $1\,\text{V}$: $W=eV=(1.6\times10^{-19}\,\text{C})(1\,\text{V})=1.6\times10^{-19}\,\text{J}$. 🔉⇢

Source: NCERT §11.2

Q4 The work function of a metal depends on easy
Step solution + source
The work function $\phi_0$ is set by the metal and its surface condition. It does not depend on the incident light's intensity or on the external voltage; those change the current or stopping potential, not $\phi_0$. 🔉⇢

Source: NCERT §11.2

Q5 Thermionic emission, field emission and photoelectric emission all achieve the same end, which is to medium
Step solution + source
Each route supplies at least $\phi_0$ to a free electron so it can cross the surface barrier: heat (thermionic), a strong field (field emission) or a photon (photoelectric). The electron's charge is unchanged and no atoms are ionised. 🔉⇢

Source: NCERT §11.2

Q6 The work function of caesium is $2.14\,\text{eV}$. In joules this is closest to medium
Step solution + source
$\phi_0=2.14\,\text{eV}\times1.6\times10^{-19}\,\text{J/eV}=3.42\times10^{-19}\,\text{J}$. 🔉⇢

Source: NCERT-derived

Q7 Alkali metals such as sodium, potassium and caesium are photosensitive even to visible light because they have medium
Step solution + source
Visible photons carry only a few eV, so only metals with a small $\phi_0$ (the alkali metals) can be triggered by them. High-$\phi_0$ metals like zinc need ultraviolet light of larger $h\nu$. 🔉⇢

Source: NCERT §11.3

Q8 A metal has work function $4.5\,\text{eV}$. The longest wavelength of light that can just eject an electron is about medium
Step solution + source
Threshold wavelength $\lambda_0=hc/\phi_0=1240\,\text{eV nm}/4.5\,\text{eV}\approx276\,\text{nm}$, in the ultraviolet. Longer wavelengths carry too little energy per photon to free an electron. 🔉⇢

Source: NCERT-derived

Q9 In field emission electrons are pulled out of a metal by medium
Step solution + source
Field emission uses an intense electric field (about $10^8\,\text{V m}^{-1}$, as at a spark-plug tip) to lower and thin the surface barrier so electrons tunnel/are pulled out; no heating or light is needed. 🔉⇢

Source: NCERT §11.2

Q10 Two metals X and Y have work functions $\phi_X=2\,\text{eV}$ and $\phi_Y=5\,\text{eV}$. For the same incident frequency above both thresholds, the emitted electrons from X have hard
Step solution + source
$K_{max}=h\nu-\phi_0$. For the same $\nu$, the metal with the smaller work function loses less to escape, so X ($\phi=2\,\text{eV}$) gives $K_{max}$ larger by $\phi_Y-\phi_X=3\,\text{eV}$ than Y. 🔉⇢

Source: NCERT-derived

Q11 The free electrons in a metal are 'free' in the sense that they hard
Step solution + source
Inside the metal the conduction electrons move in an almost constant potential, so they are free to roam; but the surface presents a barrier (the work function) that needs extra energy to cross. They still obey Pauli's exclusion principle, which shapes their energy distribution. 🔉⇢

Source: NCERT (Points to Ponder)

Q12 The photoelectric effect was first observed in 1887 by easy
Step solution + source
Hertz noticed, during his electromagnetic-wave spark experiments, that ultraviolet light on the emitter enhanced the spark. This chance observation of light-assisted electron escape was the discovery of the photoelectric effect. Einstein later explained it in 1905 with the quantum relation $K_{max}=h\nu-\phi_0$. 🔉⇢

Source: NCERT §11.3

Q13 Electrons emitted from a metal surface when light of suitable frequency falls on it are called easy
Step solution + source
Light-ejected electrons are named photoelectrons and the resulting current is the photocurrent. They are ordinary electrons; the prefix simply records that light produced them; the resulting photocurrent obeys $i\propto I$ at fixed frequency. 🔉⇢

Source: NCERT §11.3

Q14 Hallwachs found that a clean zinc plate connected to an electroscope, when illuminated with ultraviolet light, if initially negatively charged, easy
Step solution + source
UV light ejects electrons from the zinc, so a negatively charged plate loses charge, an uncharged plate turns positive, and a positive plate becomes more positive. All three point to negative particles (electrons) leaving under illumination, requiring photon energy $h\nu\ge\phi_0$. 🔉⇢

Source: NCERT §11.3

Q15 Lenard observed that photocurrent flows in the tube easy
Step solution + source
Lenard saw the current appear the instant UV struck the emitter and vanish when the light stopped, showing the light directly liberates the electrons that carry the current across the evacuated tube, with onset time $\sim10^{-9}\,\text{s}$. 🔉⇢

Source: NCERT §11.3

Q16 In the experimental tube for studying the photoelectric effect, the plate C that emits electrons is the easy
Step solution + source
Monochromatic light falls on the photosensitive plate C, the emitter (cathode); the freed electrons are gathered by plate A, the collector (anode), completing the circuit; each freed electron needs energy $\ge\phi_0$ to escape C. 🔉⇢

Source: NCERT §11.4

Q17 Below the threshold frequency, increasing the intensity of the incident light medium
Step solution + source
There is a cut-off (threshold) frequency $\nu_0$ below which no emission occurs, however intense or prolonged the light. Intensity multiplies photons but not their individual energy $h\nu$, so if $h\nu\lt\phi_0$ nothing is emitted. 🔉⇢

Source: NCERT §11.3

Q18 Metals like zinc, cadmium and magnesium show photoelectric effect only with medium
Step solution + source
These higher work-function metals respond only to short-wavelength ultraviolet light, whose photons carry enough energy ($h\nu\ge\phi_0$). Visible or longer wavelengths have too little energy per photon. 🔉⇢

Source: NCERT §11.3

Q19 A key experimental feature Hallwachs and Lenard established was that the emission depends on the light's medium
Step solution + source
They found a threshold frequency $\nu_0$ characteristic of each emitter; below it no electrons come out at any intensity. This frequency dependence, not mere brightness, is the signature that the wave picture later failed to explain. 🔉⇢

Source: NCERT §11.3

Q20 The photosensitive plate in the experimental tube is enclosed in an evacuated envelope so that medium
Step solution + source
A vacuum lets the photoelectrons travel freely from emitter C to collector A without colliding with air molecules, so the measured photocurrent faithfully reflects the emission process, for which $K_{max}=h\nu-\phi_0$. 🔉⇢

Source: NCERT §11.4

Q21 Which observation about the TIME of emission was hardest for the wave theory to accept? hard
Step solution + source
Experiments show emission starts within about $10^{-9}\,\text{s}$ of illumination, however faint the beam. The wave picture predicted a long build-up time for a single electron to soak up enough energy, so this instantaneity was a decisive clue for the photon model. 🔉⇢

Source: NCERT §11.4

Q22 A quartz window (rather than ordinary glass) is used on the tube because quartz hard
Step solution + source
Ordinary glass absorbs ultraviolet, but the higher work-function emitters need UV photons. A transparent quartz window lets UV through to the photosensitive plate so emission can occur once $h\nu\gt\phi_0$. 🔉⇢

Source: NCERT §11.4

Q23 At a fixed frequency and accelerating voltage, the photoelectric current is found to be easy
Step solution + source
More intense light of a given frequency means more photons per second, hence more photoelectrons per second, so the photocurrent rises linearly with intensity: $i\propto I$. 🔉⇢

Source: NCERT §11.4

Q24 The maximum, levelled-off value of photocurrent reached at large positive collector voltage is called the easy
Step solution + source
Once the collector is positive enough to gather every emitted electron, the current stops rising: this plateau is the saturation current. It corresponds to all photoelectrons reaching the collector, so $i_{sat}\propto I$. 🔉⇢

Source: NCERT §11.4

Q25 The minimum negative (retarding) collector voltage that reduces the photocurrent to zero is the easy
Step solution + source
The stopping potential $V_0$ is the retarding voltage at which even the most energetic photoelectrons are turned back, so the current is exactly zero: $eV_0=K_{max}$. 🔉⇢

Source: NCERT §11.4

Q26 The maximum kinetic energy of photoelectrons is related to the stopping potential $V_0$ by easy
Step solution + source
At the stopping potential the retarding work $eV_0$ just removes the largest electron kinetic energy: $K_{max}=eV_0=\tfrac12 m v_{max}^2$. 🔉⇢

Source: NCERT §11.4

Q27 Keeping frequency fixed but increasing the intensity from $I_1$ to $I_2$ ($I_2\gt I_1$), the stopping potential medium
Step solution + source
Intensity sets the NUMBER of electrons (so saturation current $\propto I$) but not their maximum energy, which depends only on frequency. Hence $V_0$ is unchanged but the plateau current rises. 🔉⇢

Source: NCERT §11.4

Q28 If the stopping potential for a certain metal and frequency is $1.5\,\text{V}$, the maximum kinetic energy of the photoelectrons is medium
Step solution + source
$K_{max}=eV_0$. With $V_0=1.5\,\text{V}$, $K_{max}=1.5\,\text{eV}=1.5\times1.6\times10^{-19} =2.4\times10^{-19}\,\text{J}$. In electron-volts it is simply $1.5\,\text{eV}$. 🔉⇢

Source: NCERT-derived

Q29 The maximum speed of a photoelectron with $K_{max}=1.5\,\text{eV}$ is closest to ($m_e=9.1\times10^{-31}\,\text{kg}$) advanced
Step solution + source
$\tfrac12 m v^2=1.5\times1.6\times10^{-19}=2.4\times10^{-19}\,\text{J}$. So $v=\sqrt{2\times2.4\times10^{-19}/9.1\times10^{-31}}=\sqrt{5.27\times10^{11}}\approx7.3\times10^{5}\,\text{m/s}$. 🔉⇢

Source: NCERT-derived

Q30 On a graph of photocurrent versus collector plate potential, the current continues to flow even at a small NEGATIVE potential because medium
Step solution + source
Photoelectrons leave with a spread of kinetic energies. Until the retarding voltage reaches $V_0$, the faster electrons still overcome $eV$ and arrive, so a shrinking current persists into negative voltages. 🔉⇢

Source: NCERT §11.4

Q31 For the same frequency, curves of photocurrent versus voltage for three intensities $I_1\lt I_2\lt I_3$ share the medium
Step solution + source
Since intensity changes only the number of electrons, all three curves cut the voltage axis at the same $V_0$ (same $K_{max}$) but saturate at currents in the ratio $I_1:I_2:I_3$. 🔉⇢

Source: NCERT §11.4

Q32 Saturation current corresponds physically to the situation where medium
Step solution + source
At saturation the collector is positive enough to sweep up all emitted electrons; increasing the voltage further cannot raise the current because there are no more electrons to collect, so $i_{sat}\propto I$ and is independent of $V$ beyond this point. 🔉⇢

Source: NCERT §11.4

Q33 A photocell gives a saturation current of $8\,\mu\text{A}$ under a certain beam. If the intensity is halved (frequency unchanged), the new saturation current is advanced
Step solution + source
Saturation current is proportional to intensity, $i_{sat}\propto I$. Halving $I$ halves the number of photoelectrons per second, so $i_{sat}=8/2=4\,\mu\text{A}$. The stopping potential would be unchanged. 🔉⇢

Source: NCERT-derived

Q34 The number of photoelectrons emitted per second from the plate C is directly proportional to hard
Step solution + source
Intensity equals photons per unit area per second; each above-threshold photon can free one electron, so the emission rate (and hence saturation current) scales with intensity, while the electrons' energy is set by frequency through $K_{max}=h\nu-\phi_0$. 🔉⇢

Source: NCERT §11.4

Q35 For a given metal, the stopping potential $V_0$ varies with the frequency $\nu$ of incident light as a easy
Step solution + source
From $eV_0=h\nu-\phi_0$, $V_0=(h/e)\nu-\phi_0/e$: a straight line of slope $h/e$ and intercept $-\phi_0/e$, cutting the frequency axis at the threshold $\nu_0$. 🔉⇢

Source: NCERT §11.4

Q36 The threshold frequency $\nu_0$ of a metal is the frequency at which the stopping potential is easy
Step solution + source
At $\nu=\nu_0$ the photon energy just equals $\phi_0$, so $K_{max}=0$ and no retarding voltage is needed: $V_0=0$. Below $\nu_0$ there is no emission at all. 🔉⇢

Source: NCERT §11.4

Q37 Increasing the frequency of incident light (above threshold) at fixed intensity makes the stopping potential easy
Step solution + source
Higher frequency means more energetic photons, so $K_{max}=h\nu-\phi_0$ rises and a larger retarding voltage is needed to stop the electrons: $V_0$ increases. 🔉⇢

Source: NCERT §11.4

Q38 The threshold frequency is related to the work function by easy
Step solution + source
Emission just begins when $h\nu_0=\phi_0$, giving $\nu_0=\phi_0/h$. A larger work function means a higher threshold frequency. 🔉⇢

Source: NCERT §11.6

Q39 The work function of caesium is $2.14\,\text{eV}$. Its threshold frequency is about medium
Step solution + source
$\nu_0=\phi_0/h=(2.14\times1.6\times10^{-19})/(6.63\times10^{-34})=3.42\times10^{-19}/6.63\times10^{-34} \approx5.16\times10^{14}\,\text{Hz}$. 🔉⇢

Source: NCERT §11.6 (Example 11.2)

Q40 The threshold wavelength for a metal of work function $2.0\,\text{eV}$ is about medium
Step solution + source
$\lambda_0=hc/\phi_0=1240\,\text{eV nm}/2.0\,\text{eV}=620\,\text{nm}$, in the red region of the visible spectrum, consistent with alkali metals responding to visible light. 🔉⇢

Source: NCERT-derived

Q41 For a metal of threshold frequency $3.3\times10^{14}\,\text{Hz}$, light of $8.2\times10^{14}\,\text{Hz}$ gives a stopping potential of about ($h/e=4.14\times10^{-15}\,\text{V s}$) hard
Step solution + source
$eV_0=h(\nu-\nu_0)$, so $V_0=(h/e)(\nu-\nu_0)=4.14\times10^{-15}\times(8.2-3.3)\times10^{14} =4.14\times10^{-15}\times4.9\times10^{14}\approx2.0\,\text{V}$. 🔉⇢

Source: NCERT §11.6 (Exercise 11.6)

Q42 Two metals A and B have threshold frequencies $\nu_A\lt\nu_B$. For the same incident frequency above both thresholds, the larger stopping potential is observed for hard
Step solution + source
$eV_0=h(\nu-\nu_0)$. For fixed $\nu$, the smaller $\nu_0$ gives the larger $(\nu-\nu_0)$ and hence the larger $V_0$. Since $\nu_A\lt\nu_B$, metal A has the larger stopping potential. 🔉⇢

Source: NCERT-derived

Q43 The stopping potential is independent of the intensity of light because medium
Step solution + source
Each electron absorbs one photon of energy $h\nu$; raising intensity adds more photons but not more energy per photon, so $K_{max}$ and therefore $V_0=K_{max}/e$ are unchanged by intensity. 🔉⇢

Source: NCERT §11.4

Q44 If light of frequency below $\nu_0$ is used, the stopping potential graph medium
Step solution + source
Below the threshold there are no photoelectrons at all (no matter the intensity), so there is no current and the idea of a stopping potential does not apply since $h\nu\lt\phi_0$. 🔉⇢

Source: NCERT §11.4

Q45 Light of frequency $7.21\times10^{14}\,\text{Hz}$ ejects electrons of maximum speed $6.0\times10^{5}\,\text{m/s}$. The threshold frequency is about advanced
Step solution + source
$K_{max}=\tfrac12 m v^2=\tfrac12(9.11\times10^{-31})(6.0\times10^5)^2=1.64\times10^{-19}\,\text{J}$. Then $\nu_0=\nu-K_{max}/h=7.21\times10^{14}-1.64\times10^{-19}/6.63\times10^{-34} =7.21\times10^{14}-2.47\times10^{14}\approx4.7\times10^{14}\,\text{Hz}$. 🔉⇢

Source: NCERT §11 (Exercise 11.8)

Q46 For a photosensitive surface, the plot of $V_0$ against $\nu$ for two different metals gives two lines that are advanced
Step solution + source
The slope is $h/e$, a universal constant, so all metals give parallel lines; the intercept on the frequency axis is $\nu_0=\phi_0/h$, which differs from metal to metal. 🔉⇢

Source: NCERT §11.6

Q47 A metal is illuminated by light whose frequency is exactly the threshold frequency. The photoelectrons are emitted with medium
Step solution + source
At $\nu=\nu_0$, $K_{max}=h\nu_0-\phi_0=0$. The photon energy is just enough to free the electron with no energy to spare, so the electrons barely escape. 🔉⇢

Source: NCERT §11.6

Q48 According to the classical wave theory, the maximum kinetic energy of photoelectrons should increase with the easy
Step solution + source
In the wave picture a brighter beam carries larger field amplitudes and delivers more energy to each electron, so $K_{max}$ ought to grow with intensity. Experiment flatly contradicts this. 🔉⇢

Source: NCERT §11.5

Q49 The observed independence of $K_{max}$ from intensity is a problem for the wave theory because the wave theory predicts easy
Step solution + source
The wave model ties energy delivery to amplitude (intensity). Since experiment shows $K_{max}$ depends on frequency and not intensity, the wave prediction of intensity-dependent energy fails. 🔉⇢

Source: NCERT §11.5

Q50 The wave theory cannot explain the existence of a threshold frequency because it predicts that medium
Step solution + source
Classically, a sufficiently intense beam of ANY frequency, given enough time, should feed an electron the escape energy. So the wave theory forbids a sharp cut-off frequency, yet a sharp threshold $\nu_0=\phi_0/h$ is always observed. 🔉⇢

Source: NCERT §11.5

Q51 The wave theory predicts a measurable TIME LAG before emission for weak light because medium
Step solution + source
With energy spread continuously over the wavefront and shared among many electrons, calculations give hours for one electron to accumulate $\phi_0$ from a dim beam. Observation shows emission within $\sim10^{-9}\,\text{s}$, contradicting the wave estimate. 🔉⇢

Source: NCERT §11.5

Q52 Which set of photoelectric observations does the classical wave theory FAIL to explain? medium
Step solution + source
The wave theory handles interference/diffraction well but fails on three photoelectric facts: a cut-off frequency, $K_{max}$ set by frequency (not intensity), and the near-instant onset of emission. 🔉⇢

Source: NCERT §11.5

Q53 In the wave picture the energy of the electromagnetic field is medium
Step solution + source
Classically the field's energy is spread continuously across the region the wave occupies and grows with amplitude squared (intensity). This continuous-energy assumption is exactly what fails for the photoelectric effect, where energy comes in quanta $E=h\nu$. 🔉⇢

Source: NCERT §11.5

Q54 The greater the intensity of radiation in the wave picture, the greater is the hard
Step solution + source
Intensity in the wave model is proportional to the square of the field amplitude. Frequency and wavelength are independent of intensity, and 'photon energy' is not a wave-theory concept at all; there intensity $\propto E_0^2$. 🔉⇢

Source: NCERT §11.5

Q55 Why does the wave theory succeed for interference but fail for the photoelectric effect? hard
Step solution + source
Light shows a dual nature: propagation phenomena (interference, diffraction) reveal its wave character, while energy-momentum exchange with matter (photoelectric, Compton) reveals its particle character. Each model works only in its own domain; energy exchange follows $E=h\nu$. 🔉⇢

Source: NCERT §11.5

Q56 A very dim ultraviolet source above threshold is switched on. The wave theory and experiment disagree most sharply about advanced
Step solution + source
Both agree emission occurs (above threshold), but the wave theory predicts a long delay for dim light while experiment finds emission within about $10^{-9}\,\text{s}$. The onset time is the decisive point of disagreement. 🔉⇢

Source: NCERT §11.5

Q57 The resolution of the wave theory's failures came from Einstein's assumption that light energy is advanced
Step solution + source
Einstein proposed that radiation is made of quanta of energy $h\nu$ and that one electron absorbs one quantum. This single-photon, single-electron picture immediately explains the threshold, the frequency-set $K_{max}$, and the instantaneity. 🔉⇢

Source: NCERT §11.6

Q58 Einstein's photoelectric equation is easy
Step solution + source
An electron absorbs one quantum $h\nu$; after paying the work function $\phi_0$ to escape, the surplus is its maximum kinetic energy: $K_{max}=h\nu-\phi_0$. 🔉⇢

Source: NCERT §11.6

Q59 In Einstein's equation, the term $\phi_0$ represents easy
Step solution + source
$\phi_0$ is the work function, the least energy an electron must be given to cross the surface barrier. The photon spends $\phi_0$ on escape and the rest becomes kinetic energy. 🔉⇢

Source: NCERT §11.6

Q60 Einstein's photoelectric equation is essentially a statement of easy
Step solution + source
$h\nu=\phi_0+K_{max}$ balances the energy in (one photon) against energy spent escaping plus the kinetic energy carried away, i.e. energy conservation for one absorption event. 🔉⇢

Source: NCERT §11.6

Q61 Photoelectric emission by Einstein's picture is possible only if easy
Step solution + source
$K_{max}=h\nu-\phi_0$ must be non-negative, so a photon can free an electron only when its energy exceeds the work function, $h\nu\gt\phi_0$ (equivalently $\nu\gt\nu_0$). 🔉⇢

Source: NCERT §11.6

Q62 Light of energy $3.0\,\text{eV}$ falls on a metal of work function $2.0\,\text{eV}$. The maximum kinetic energy of the photoelectrons is easy
Step solution + source
$K_{max}=h\nu-\phi_0=3.0-2.0=1.0\,\text{eV}$. The photon spends $2.0\,\text{eV}$ freeing the electron and the remaining $1.0\,\text{eV}$ appears as kinetic energy. 🔉⇢

Source: NCERT-derived

Q63 The work function of caesium is $2.14\,\text{eV}$. Light of frequency $6.0\times10^{14}\,\text{Hz}$ gives $K_{max}$ of about medium
Step solution + source
Photon energy $h\nu=6.63\times10^{-34}\times6.0\times10^{14}=3.98\times10^{-19}\,\text{J}=2.48\,\text{eV}$. Then $K_{max}=2.48-2.14\approx0.34\,\text{eV}$. 🔉⇢

Source: NCERT §11 (Exercise 11.2)

Q64 Light of wavelength $400\,\text{nm}$ strikes a metal of work function $2.0\,\text{eV}$. The maximum kinetic energy is about medium
Step solution + source
Photon energy $=hc/\lambda=1240/400=3.1\,\text{eV}$. So $K_{max}=3.1-2.0=1.1\,\text{eV}$. 🔉⇢

Source: NCERT-derived

Q65 Doubling the FREQUENCY of the incident light (keeping it above threshold) will medium
Step solution + source
$K_{max}=h\nu-\phi_0$. Doubling $\nu$ doubles $h\nu$ but $\phi_0$ is fixed, so the new $K'_{max}=2h\nu-\phi_0=2(h\nu-\phi_0)+\phi_0=2K_{max}+\phi_0\gt2K_{max}$. 🔉⇢

Source: NCERT-derived

Q66 According to Einstein's picture, increasing the intensity of light of a fixed frequency (above threshold) increases medium
Step solution + source
Intensity equals photons per second; more photons free more electrons, raising the current, but each electron still gains $h\nu-\phi_0$, so $K_{max}$ is unchanged. 🔉⇢

Source: NCERT §11.6

Q67 Which graph is a straight line with slope equal to Planck's constant $h$? medium
Step solution + source
$K_{max}=h\nu-\phi_0$ is linear in $\nu$ with slope $h$ and intercept $-\phi_0$. Plotting $K_{max}$ against $\nu$ therefore yields $h$ from the slope. 🔉⇢

Source: NCERT §11.6

Q68 A photon of energy $5\,\text{eV}$ ejects an electron from a surface with $K_{max}=3\,\text{eV}$. The work function is easy
Step solution + source
$\phi_0=h\nu-K_{max}=5-3=2\,\text{eV}$. The photon spends $2\,\text{eV}$ on escape and gives the electron $3\,\text{eV}$ of kinetic energy. 🔉⇢

Source: NCERT-derived

Q69 Light of wavelength $330\,\text{nm}$ is incident on a metal of work function $4.2\,\text{eV}$. Photoemission hard
Step solution + source
Photon energy $=1240/330=3.76\,\text{eV}\lt4.2\,\text{eV}=\phi_0$. Since $h\nu\lt\phi_0$, no electrons are emitted, regardless of how intense the light is. 🔉⇢

Source: NCERT §11 (Exercise 11.7)

Q70 The stopping potential form of Einstein's equation is medium
Step solution + source
Since $K_{max}=eV_0$, substituting into $K_{max}=h\nu-\phi_0$ gives $eV_0=h\nu-\phi_0$, valid for $\nu\ge\nu_0$. 🔉⇢

Source: NCERT §11.6

Q71 Two photons of energies $4\,\text{eV}$ and $6\,\text{eV}$ eject electrons from the same metal ($\phi_0=2\,\text{eV}$). The ratio of their maximum kinetic energies is advanced
Step solution + source
$K_1=4-2=2\,\text{eV}$ and $K_2=6-2=4\,\text{eV}$, so $K_1:K_2=2:4=1:2$. Note the ratio is not $4:6$ because the fixed work function is subtracted from each. 🔉⇢

Source: NCERT-derived

Q72 For a metal, when the wavelength of incident light is decreased (still above threshold), the maximum kinetic energy of photoelectrons medium
Step solution + source
$K_{max}=hc/\lambda-\phi_0$. Smaller $\lambda$ means larger photon energy $hc/\lambda$, so $K_{max}$ increases. Shorter wavelength (higher frequency) yields more energetic electrons. 🔉⇢

Source: NCERT-derived

Q73 A surface of work function $2.5\,\text{eV}$ is lit by $6.2\,\text{eV}$ photons. If instead $\phi_0$ were $1.5\,\text{eV}$ for the same photons, $K_{max}$ would advanced
Step solution + source
$K_{max}=h\nu-\phi_0$. Lowering $\phi_0$ from $2.5$ to $1.5\,\text{eV}$ removes $1.0\,\text{eV}$ less from the same photon energy, so $K_{max}$ rises by exactly $1.0\,\text{eV}$ (from $3.7$ to $4.7\,\text{eV}$). 🔉⇢

Source: NCERT-derived

Q74 Millikan determined Planck's constant $h$ from photoelectric experiments by measuring the easy
Step solution + source
Since $V_0=(h/e)\nu-\phi_0/e$, the $V_0$-vs-$\nu$ graph is a straight line of slope $h/e$. Measuring that slope and multiplying by the known $e$ gives $h$. 🔉⇢

Source: NCERT §11.6

Q75 The slope of the stopping-potential versus frequency graph equals easy
Step solution + source
$V_0=(h/e)\nu-\phi_0/e$; comparing with $y=mx+c$ gives slope $m=h/e$, a universal constant independent of the metal. 🔉⇢

Source: NCERT §11.6

Q76 Because the slope $h/e$ is the same for every metal, the $V_0$-vs-$\nu$ lines for different metals are easy
Step solution + source
All metals share the slope $h/e$; only the intercept (set by $\phi_0$) differs. Equal slopes with different intercepts means the lines are parallel. 🔉⇢

Source: NCERT §11.6

Q77 If the slope of the $V_0$-vs-$\nu$ line is $4.12\times10^{-15}\,\text{V s}$, then Planck's constant is about medium
Step solution + source
slope $=h/e$, so $h=\text{slope}\times e=4.12\times10^{-15}\times1.6\times10^{-19} =6.6\times10^{-34}\,\text{J s}$. 🔉⇢

Source: NCERT §11 (Exercise 11.5)

Q78 Millikan originally set out to test Einstein's equation intending to disprove it, but his precise measurements instead medium
Step solution + source
Over 1906–1916 Millikan verified the linear $V_0$-vs-$\nu$ relation for several alkali metals, obtaining an $h$ consistent with Planck's blackbody value and confirming Einstein's photon picture. 🔉⇢

Source: NCERT §11.6

Q79 The intercept of the $V_0$-vs-$\nu$ line on the stopping-potential axis (at $\nu=0$) is medium
Step solution + source
Setting $\nu=0$ in $V_0=(h/e)\nu-\phi_0/e$ gives $V_0=-\phi_0/e$. Physically the line is only real for $\nu\ge\nu_0$, but its extrapolated intercept encodes the work function. 🔉⇢

Source: NCERT §11.6

Q80 From a $V_0$-vs-$\nu$ line, the work function is obtained from medium
Step solution + source
The line crosses $V_0=0$ at $\nu=\nu_0$ (the threshold). Then $\phi_0=h\nu_0$, so the frequency intercept gives the work function once $h$ is known from the slope. 🔉⇢

Source: NCERT §11.6

Q81 The ratio of the slopes of the $V_0$-vs-$\nu$ lines for silver and sodium is hard
Step solution + source
The slope is $h/e$ for every metal, so the ratio is exactly $1:1$. Silver and sodium differ only in their intercepts (work functions), not in slope. 🔉⇢

Source: NCERT §11.6

Q82 Two metals give $V_0$-vs-$\nu$ lines with frequency intercepts $\nu_{0,1}=5\times10^{14}\,\text{Hz}$ and $\nu_{0,2}=10\times10^{14}\,\text{Hz}$. The ratio of their work functions $\phi_1:\phi_2$ is hard
Step solution + source
$\phi_0=h\nu_0$, so $\phi_1:\phi_2=\nu_{0,1}:\nu_{0,2}=5:10=1:2$. The metal with the higher threshold frequency has the larger work function. 🔉⇢

Source: NCERT-derived

Q83 In SI units, verifying $h$ from the photoelectric slope requires knowing medium
Step solution + source
The measured quantity is slope $=h/e$. To extract $h$ one multiplies by the independently known elementary charge $e=1.6\times10^{-19}\,\text{C}$. 🔉⇢

Source: NCERT §11.6

Q84 Two students measure the $V_0$-vs-$\nu$ slope for the same metal but at different intensities. Their slopes should be advanced
Step solution + source
Intensity changes only the saturation current, not $K_{max}$ or $V_0$ at a given frequency. Hence the $V_0$-vs-$\nu$ slope stays $h/e$ regardless of intensity, and both students get the same value. 🔉⇢

Source: NCERT §11.6

Q85 The energy of a photon of frequency $\nu$ is easy
Step solution + source
Each photon carries a quantum of energy $E=h\nu=hc/\lambda$; higher frequency (shorter wavelength) means a more energetic photon. 🔉⇢

Source: NCERT §11.7

Q86 The momentum of a photon of wavelength $\lambda$ is easy
Step solution + source
A photon carries momentum $p=h\nu/c=h/\lambda$. Though massless, it has definite energy and momentum, the hallmarks of a particle. 🔉⇢

Source: NCERT §11.7

Q87 Photons are easy
Step solution + source
Photons carry no charge, so external electric and magnetic fields do not deflect them. This distinguishes them from the charged photoelectrons; a photon still carries momentum $p=h/\lambda$. 🔉⇢

Source: NCERT §11.7

Q88 The energy of a photon of frequency $6.0\times10^{14}\,\text{Hz}$ is about medium
Step solution + source
$E=h\nu=6.63\times10^{-34}\times6.0\times10^{14}=3.98\times10^{-19}\approx4.0\times10^{-19}\,\text{J}$ (about $2.5\,\text{eV}$). 🔉⇢

Source: NCERT §11 (Example 11.1)

Q89 A laser emits $2.0\times10^{-3}\,\text{W}$ at $6.0\times10^{14}\,\text{Hz}$. The number of photons emitted per second is about medium
Step solution + source
Each photon has $E=3.98\times10^{-19}\,\text{J}$. Number per second $N=P/E =2.0\times10^{-3}/3.98\times10^{-19}\approx5.0\times10^{15}$ photons per second. 🔉⇢

Source: NCERT §11 (Example 11.1)

Q90 The momentum of a photon of wavelength $500\,\text{nm}$ is about medium
Step solution + source
$p=h/\lambda=6.63\times10^{-34}/5.0\times10^{-7}=1.33\times10^{-27}\,\text{kg m/s}$. 🔉⇢

Source: NCERT-derived

Q91 If the wavelength of a photon is halved, its momentum medium
Step solution + source
$p=h/\lambda$, so $p\propto1/\lambda$. Halving $\lambda$ doubles the momentum (and also doubles the energy $E=hc/\lambda$). 🔉⇢

Source: NCERT §11.7

Q92 Increasing the intensity of light of a given wavelength increases medium
Step solution + source
Intensity is the number of photons crossing unit area per unit time. Each photon still has energy $h\nu$ and momentum $h/\lambda$; only their number rises with intensity. 🔉⇢

Source: NCERT §11.7

Q93 The relation between a photon's energy $E$ and its momentum $p$ is medium
Step solution + source
$E=h\nu$ and $p=h\nu/c$, so $E=pc$. This is the massless-particle energy–momentum relation, obtained by setting rest mass to zero. 🔉⇢

Source: NCERT §11.7

Q94 A $100\,\text{ns}$ pulse of power $30\,\text{mW}$ is fully absorbed by a small object at rest. The momentum delivered is about advanced
Step solution + source
Energy in the pulse $U=Pt=30\times10^{-3}\times100\times10^{-9}=3.0\times10^{-9}\,\text{J}$. For full absorption momentum $p=U/c=3.0\times10^{-9}/3\times10^{8}=1.0\times10^{-17}\,\text{kg m/s}$. 🔉⇢

Source: NCERT-derived (JEE-pattern)

Q95 A photon and an electron have the same wavelength. Compared with the electron, the photon has hard
Step solution + source
For BOTH, $p=h/\lambda$. Equal wavelengths therefore mean equal momenta, even though their energies differ (photon $E=pc$; electron $E=p^2/2m$). 🔉⇢

Source: NCERT §11.7/11.8

Q96 In a photon–electron collision (e.g. Compton scattering), what is conserved? hard
Step solution + source
Such collisions conserve total energy and total momentum. The number of photons need not be conserved, as a photon can be absorbed or a new one created; the scattered photon's wavelength changes, with $E=pc$ for each photon. 🔉⇢

Source: NCERT §11.7

Q97 Two sources emit equal power: one X-rays of $\lambda=1\,\text{nm}$, the other visible light of $\lambda=500\,\text{nm}$. The visible source emits photons at a rate that is advanced
Step solution + source
Photon energy $\propto1/\lambda$, so a visible photon has $1/500$ the energy of an X-ray photon. For equal power $P=N E$, the rate $N\propto1/E\propto\lambda$, so the visible source emits $500\times$ as many photons per second. 🔉⇢

Source: NCERT-derived (JEE-pattern)

Q98 The de Broglie wavelength of a particle of momentum $p$ is easy
Step solution + source
de Broglie proposed matter waves with $\lambda=h/p=h/mv$. The left side is a wave attribute and the right a particle attribute, linked by Planck's constant. 🔉⇢

Source: NCERT §11.8

Q99 The de Broglie wavelength is significant (measurable) only for easy
Step solution + source
Because $\lambda=h/mv$ and $h$ is tiny, only very small masses/momenta give a wavelength large enough to matter (comparable to atomic spacings). Everyday objects have utterly negligible wavelengths. 🔉⇢

Source: NCERT §11.8

Q100 The de Broglie wavelength of a moving particle depends on its easy
Step solution + source
$\lambda=h/p$ involves only the momentum. It is independent of the particle's charge or nature; a proton and an electron of equal momentum have the same de Broglie wavelength. 🔉⇢

Source: NCERT §11.8

Q101 For a de Broglie wave, which velocity is physically meaningful and equals the particle's velocity? medium
Step solution + source
The phase velocity of a matter wave has no direct physical significance, but the group velocity of the wave packet is meaningful and equals the particle speed, consistent with $\lambda=h/p$. 🔉⇢

Source: NCERT (Points to Ponder)

Q102 A ball of mass $0.12\,\text{kg}$ moves at $20\,\text{m/s}$. Its de Broglie wavelength is about medium
Step solution + source
$p=mv=0.12\times20=2.4\,\text{kg m/s}$; $\lambda=h/p=6.63\times10^{-34}/2.4 =2.76\times10^{-34}\,\text{m}$, far too small to detect, which is why the ball shows no wave nature. 🔉⇢

Source: NCERT §11.8

Q103 An electron moving at $5.4\times10^{6}\,\text{m/s}$ has a de Broglie wavelength of about medium
Step solution + source
$p=mv=9.11\times10^{-31}\times5.4\times10^{6}=4.92\times10^{-24}\,\text{kg m/s}$; $\lambda=h/p=6.63\times10^{-34}/4.92\times10^{-24}=1.35\times10^{-10}\,\text{m}=0.135\,\text{nm}$, comparable to X-ray wavelengths. 🔉⇢

Source: NCERT §11 (Example 11.3)

Q104 For a particle of kinetic energy $K$ and mass $m$, the de Broglie wavelength is medium
Step solution + source
Non-relativistically $K=p^2/2m$, so $p=\sqrt{2mK}$ and $\lambda=h/p=h/\sqrt{2mK}$. Higher energy means shorter wavelength. 🔉⇢

Source: NCERT §11.8

Q105 An electron accelerated through a potential difference $V$ has de Broglie wavelength (in nm) approximately medium
Step solution + source
$K=eV$, so $p=\sqrt{2meV}$ and $\lambda=h/\sqrt{2meV}=1.227/\sqrt{V}\,\text{nm}$ (with $V$ in volts). For $V=100\,\text{V}$, $\lambda\approx0.123\,\text{nm}$. 🔉⇢

Source: NCERT §11.8

Q106 An electron accelerated through $100\,\text{V}$ has a de Broglie wavelength of about hard
Step solution + source
$\lambda=1.227/\sqrt{V}\,\text{nm}=1.227/\sqrt{100}=1.227/10=0.1227\approx0.123\,\text{nm}$, of the order of atomic-plane spacings, so such electrons diffract from crystals. 🔉⇢

Source: NCERT-derived

Q107 A proton and an electron are accelerated through the same potential difference. The ratio of their de Broglie wavelengths $\lambda_e/\lambda_p$ is advanced
Step solution + source
$\lambda=h/\sqrt{2mqV}$; with equal $q$ and $V$, $\lambda\propto1/\sqrt{m}$. So $\lambda_e/\lambda_p=\sqrt{m_p/m_e}\approx\sqrt{1836}\approx43$: the lighter electron has the longer wavelength. 🔉⇢

Source: NCERT-derived

Q108 An electron and a photon have the same de Broglie wavelength. The ratio of the electron's momentum to the photon's momentum is advanced
Step solution + source
For both, $p=h/\lambda$. Equal wavelengths give equal momenta, so the ratio is exactly $1$, even though their energies differ ($E_{ph}=pc$ vs $E_e=p^2/2m$). 🔉⇢

Source: NCERT §11.8

Q109 The de Broglie wavelength of a photon (treated by $\lambda=h/p$) equals hard
Step solution + source
For a photon $p=h\nu/c=h/\lambda_{EM}$, so $h/p=\lambda_{EM}$. The de Broglie formula reproduces the photon's own electromagnetic wavelength, a consistency check on $\lambda=h/p$. 🔉⇢

Source: NCERT §11.8

Q110 Who first hypothesised that moving material particles have an associated wavelength? easy
Step solution + source
In 1924 Louis de Broglie proposed matter waves, reasoning by symmetry with light's dual nature. Davisson and Germer later confirmed it experimentally with electron diffraction, matching $\lambda=h/p$. 🔉⇢

Source: NCERT §11.8

Q111 The Davisson–Germer experiment established the medium
Step solution + source
Davisson and Germer scattered low-energy electrons from a nickel crystal and observed a diffraction maximum, confirming that electrons behave as waves with the de Broglie wavelength $\lambda=h/p$. 🔉⇢

Source: NCERT §11.8 (matter waves)

Q112 An electron (mass $m$) and a photon have the same energy $E$ (a few eV). The ratio of the electron's de Broglie wavelength to the photon's wavelength is advanced
Step solution + source
Photon: $\lambda_{ph}=hc/E$. Electron: $\lambda_e=h/\sqrt{2mE}$. Ratio $\lambda_e/\lambda_{ph}=(h/\sqrt{2mE})(E/hc)=\sqrt{E}/(\sqrt{2m}\,c)=\sqrt{E/2mc^2}$, which is much less than 1 for a few-eV energy. 🔉⇢

Source: NCERT-derived (JEE-pattern)

Q113 As the speed of a particle increases, its de Broglie wavelength easy
Step solution + source
$\lambda=h/mv$, so $\lambda\propto1/v$. Faster particles (larger momentum) have shorter de Broglie wavelengths; this is why highly energetic electrons resolve fine crystal structure. 🔉⇢

Source: NCERT §11.8

⏱️ Mock Test 30 Q · 60 min · +4 correct, −1 wrong, 0 unattempted

Rules: No calculator beyond basic arithmetic. Take $h=6.63\times10^{-34}\,\text{J s}$, $e=1.6\times10^{-19}\,\text{C}$ and $hc=1240\,\text{eV nm}$. Convert the work function to joules or work throughout in electron volts. Negative marking rewards accuracy over guessing.

🎬 Video Lectures clip-indexed · NPTEL / IIT / MIT

MIT OCW / Walter Lewin, 3Blue1Brown, Khan, Veritasium. Each video is broken into clips — click a clip to play just that segment.

Stopping Potential in Photoelectric effect- IIT JEE & NEET | Vineet Khatri Sir | ATP STAR Kota 🔉⇢
IIT JEE Prep by iQuanta

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📚 Teaches: Tier-1 educational channel (IIT JEE Prep by iQuanta); found via yt-dlp search 'stopping potential threshold frequency photoelectric effect physics', oEmbed-verified live.

📑 Clips (2)
  • 0:00–1:51What stopping potential isDefines stopping potential as the reverse voltage applied to stop photoelectrons moving from cathode to anode, halting their kinetic energy.stopping potential
  • 1:51–3:40Relating stopping potential to Einstein's equationDerives that kinetic energy equals eV, then substitutes into the photoelectric equation to solve for the stopping potential.stopping potential
Photoelectric Effect Explained in Simple Words for Beginners 🔉⇢
Science ABC

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📚 Teaches: Tier-1 educational channel (Science ABC); found via yt-dlp search 'thermionic photoelectric field emission work function physics', oEmbed-verified live.

📑 Clips (3)
  • 0:00–1:51Photoelectric effect via toll-booth analogyPhotons eject electrons only if frequency meets the threshold; a toll-booth analogy explains why intensity alone is not enough.photoelectric effect
  • 1:51–3:41Work function and threshold frequencyWork function is the minimum energy to free an electron; it equals Planck's constant times the threshold frequency.work function
  • 3:41–4:18Real-world uses of the photoelectric effectLists applications: solar power, photodetectors, cameras, light meters, X-ray imaging, automatic doors, and barcode scanners.applications
Hallwachs & Lenard's Observation - Dual Nature of Radiation & Matter | Class 12 Physics | CBSE/JE... 🔉⇢
Magnet Brains

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📚 Teaches: Tier-1 educational channel (Magnet Brains); found via yt-dlp search 'photoelectric effect Hertz Hallwachs Lenard experiment physics', oEmbed-verified live.

📑 Clips (3)
  • 1:52–3:44Lenard's observation of photocurrentIn an evacuated tube with two plates, UV radiation on the emitter plate produces a current detected by the meter.Lenard's observation
  • 3:44–5:36Hallwachs: charged zinc plate loses chargeUV light on a negatively charged zinc plate reduces its negative charge; a neutral plate becomes positively charged.Hallwachs observation
  • 7:29–9:21Electron emission explains the effectAfter the 1897 discovery of the electron, results are explained: suitable-frequency radiation ejects electrons from the metal.electron emission
45.Chapter - 10 | Dual Nature Of Radiation & Matter | Lenard's Experiment | Physics Baba 2.O 🔉⇢
Rankplus

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📚 Teaches: Tier-1 educational channel (Rankplus); found via yt-dlp search 'photoelectric effect Hertz Hallwachs Lenard experiment physics', oEmbed-verified live.

📑 Clips (3)
  • 3:41–5:32Intensity increases the photocurrentExplains intensity as photons per unit area per second; more intensity means more electrons, so photocurrent rises with intensity.intensity
  • 7:22–9:13Saturation current at high positive voltageRaising positive collector voltage collects more electrons until all are gathered and current levels off at the saturation current.saturation current
  • 12:54–16:36Stopping potential depends on frequencyStopping potential is set by electron kinetic energy, so it rises with frequency but is independent of light intensity.stopping potential
Failure Of Wave Theory Of Light To Explain Photoelectric Effect - CLASS 12 Physics 🔉⇢
Physics by Suman Dhull

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📚 Teaches: Tier-1 educational channel (Physics by Suman Dhull); found via yt-dlp search 'why wave theory fails photoelectric effect physics', oEmbed-verified live.

📑 Clips (2)
  • 0:42–2:32Light energy spread over the metal surfaceDescribes the classical picture where light energy is distributed uniformly over the metal surface and shared among its free electrons.wave picture
  • 2:32–5:50Energy and kinetic energy of electronsRough captions touching on how more energy relates to maximum kinetic energy and how many photoelectrons are produced.kinetic energy
Photoelectric Effect, Work Function, Threshold Frequency, Wavelength, Speed & Kinetic Energy, Electr 🔉⇢
The Organic Chemistry Tutor

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📚 Teaches: Tier-1 educational channel (The Organic Chemistry Tutor); found via yt-dlp search 'Einstein photoelectric equation The Organic Chemistry Tutor', oEmbed-verified live.

📑 Clips (0)

Full lecture — no clip index.

determination of Planck's constant using Photoelectric effect 🔉⇢
Pranjali Sharma

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📚 Teaches: Tier-1 educational channel (Pranjali Sharma); found via yt-dlp search 'determination of Planck's constant photoelectric experiment physics', oEmbed-verified live.

📑 Clips (2)
  • 0:01–1:51Photocell setup for Planck's constantDescribes the photo cell apparatus and procedure: mount a filter, set current display and intensity, apply negative voltage.photocell experiment
  • 1:51–3:28Stopping voltage graph, slope h/eStopping voltage per filter fits Vs=h*mu/e minus work function; the straight-line graph has slope h/e. Notes precautions.stopping potential
Plank's Constant 🔉⇢
Autonomous Academy

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📚 Teaches: Tier-1 educational channel (Autonomous Academy); found via yt-dlp search 'determination of Planck's constant photoelectric experiment physics', oEmbed-verified live.

📑 Clips (0)

Full lecture — no clip index.

How does the photon have momentum without mass? 🔉⇢
Quasicrystals' Log-Lin metric

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📚 Teaches: Tier-1 educational channel (Quasicrystals' Log-Lin metric); found via yt-dlp search 'photon momentum E=pc physics explained', oEmbed-verified live.

📑 Clips (3)
  • 0:02–1:52How a massless photon has momentumThough a photon has zero rest mass, relativity gives E = pc, so its momentum equals energy over c and is inversely related to wavelength.photon momentum
  • 1:52–3:44Phase vs group velocity of a photonThe wave packet's phase and group velocities are equal for light, and with zero rest mass the photon energy stays proportional to momentum.group velocity
  • 3:44–4:19Rest mass vs dynamic massFor massive particles rest mass ties to phase velocity and relativistic dynamic mass to group velocity; energy change means momentum change.relativistic mass
De Broglie Hypothesis | De Broglie Wavelength 🔉⇢
Najam Academy

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📚 Teaches: Tier-1 educational channel (Najam Academy); found via yt-dlp search 'de Broglie wavelength matter waves Khan Academy', oEmbed-verified live.

📑 Clips (3)
  • 0:00–1:53Wave-particle nature of lightLight behaves as a wave (diffraction, interference) and as a particle (photoelectric effect, photons); intro to wave terms.wave-particle duality
  • 3:45–5:38de Broglie hypothesis and matter wavesde Broglie (1924) proposes any particle like an electron also behaves as a wave, with c=lambda*f and momentum p=h/lambda.de Broglie hypothesis
  • 7:29–9:06Calculating de Broglie wavelengthWorked example: lambda=h/mv for a 60 g ball at 80 m/s gives about 1.38e-34 m, too small to detect.de Broglie wavelength
De Broglie Hypothesis — Matter Waves Explained (+ Examples) 🔉⇢
For the Love of Physics

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📚 Teaches: Tier-1 educational channel (For the Love of Physics); found via yt-dlp search 'de Broglie hypothesis matter waves physics', oEmbed-verified live.

📑 Clips (3)
  • 0:02–1:53Photon picture and dual nature of lightPhotoelectric and black-body effects show light as photons of energy E=h*nu, while interference/diffraction show wave nature.dual nature of light
  • 5:36–7:27Matter waves: lambda = h/pde Broglie says matter has a wave with wavelength lambda=h/p; wave and particle properties are inversely correlated.matter waves
  • 16:47–18:37Bohr orbits as electron standing wavesRearranging mvr=nh/2pi gives circumference=n*lambda, so a stable orbit fits an integer number of de Broglie wavelengths.Bohr quantization
Davisson Germer Experiment 🔉⇢
Educational Videos

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📚 Teaches: Tier-1 educational channel (Educational Videos); found via yt-dlp search 'Davisson Germer experiment electron diffraction physics', oEmbed-verified live.

📑 Clips (2)
  • 0:02–1:54de Broglie hypothesis and Davisson-Germer setupIntroduces the wave-particle nature of electrons and the Davisson-Germer apparatus: electron gun, nickel crystal target, and detector.de Broglie waves
  • 1:54–3:52Electron diffraction peak confirms wave natureScattered electrons show an intensity peak from constructive interference, maximal near 54 volts, verifying the wave nature of matter.electron diffraction
Photoelectric Effect | Photoelectric Effect | The Most Interesting Experiment in Physics | Physic... 🔉⇢
VisualLearning-हिन्दी

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📚 Teaches: Tier-1 educational channel (VisualLearning-हिन्दी); found via yt-dlp search 'प्रकाश विद्युत प्रभाव photoelectric effect Hindi physics NCERT', oEmbed-verified live.

📑 Clips (3)
  • 0:00–1:51Photoelectric effect and its history (Hindi)Introduces the photoelectric effect discovered by Hertz, explained by Lenard and Einstein, showing light has a particle (photon) nature.photoelectric effect
  • 1:51–3:43Setup, saturation current, stopping potential (Hindi)Describes the vacuum-tube experiment and how photocurrent rises with intensity to a saturation current, and reverse voltage gives…saturation current
  • 3:43–4:08Effect of radiation frequency (Hindi)Shows that at equal intensity but different frequencies, both the photocurrent and stopping potential take different values.frequency effect
डी ब्रोग्ली समीकरण | Di brogli samikaran | De broglie equation | 12th physics | by monu sir 🔉⇢
MK Sir Inspiration

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📚 Teaches: Tier-1 educational channel (MK Sir Inspiration); found via yt-dlp search 'दे ब्रोग्ली तरंगदैर्ध्य de Broglie wavelength Hindi physics', oEmbed-verified live.

📑 Clips (3)
  • 0:00–1:51Intro to de Broglie matter waves (Hindi)Explains that any moving particle, such as an electron, behaves like a wave, and this associated wave is called the de Broglie wave.de Broglie waves
  • 1:51–3:43Deriving wavelength from E=hf and E=mc2 (Hindi)Sets photon energy hf equal to mc-squared to relate mass, frequency and the speed of light while building the de Broglie relation.wavelength derivation
  • 5:37–7:29Wavelength via momentum and kinetic energy (Hindi)Expresses de Broglie wavelength as h over momentum, h over mv, and h over root(2mK) when kinetic energy is given.de Broglie wavelength
Photoelectric effect | Electronic structure of atoms | Chemistry | Khan Academy 🔉⇢
Khan Academy Organic Chemistry

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📚 Teaches: Tier-1 educational channel (Khan Academy Organic Chemistry); found via yt-dlp search 'electron emission work function metal Khan Academy', oEmbed-verified live.

📑 Clips (3)
  • 0:01–1:52Photoelectric effect as photon collisionA photon strikes a bound electron and, if energetic enough, frees it as a photoelectron with kinetic energy 1/2 mv^2.photoelectric effect
  • 3:43–5:36Photon energy E = hc/lambdaUses c=lambda*nu to get E=hc/lambda; for 525 nm light the photon energy is about 3.78e-19 J.photon energy
  • 9:17–10:24Longer wavelength gives no emissionAt 625 nm the photon energy 3.2e-19 J is below the work function, so no photoelectron is produced regardless of intensity.threshold wavelength
Which metals will show photoelectric emission | Dual nature of light | Physics | Khan Academy 🔉⇢
Khan Academy India - English

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📚 Teaches: Tier-1 educational channel (Khan Academy India - English); found via yt-dlp search 'electron emission work function metal Khan Academy', oEmbed-verified live.

📑 Clips (3)
  • 0:00–1:50Einstein's photoelectric equationPhoton energy hf frees a trapped electron: part is the work function, the rest becomes maximum kinetic energy K_max.photoelectric equation
  • 3:40–5:32Finding photon energy from wavelengthDerives frequency=velocity/wavelength, then computes photon energy E=hc/lambda for 430 nm light.photon energy
  • 7:23–9:10eV conversion and which metals emitConverts photon energy to 2.89 eV, then compares with each work function; three metals show photoelectric emission.work function
Photoelectric effect explanation using quantum theory | Dual nature of light | Khan Academy 🔉⇢
Khan Academy India - English

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📚 Teaches: Tier-1 educational channel (Khan Academy India - English); found via yt-dlp search 'photoelectric effect explained Khan Academy', oEmbed-verified live.

📑 Clips (3)
  • 0:00–1:50Why the wave model of light failsZinc shows photoeffect for dim UV but not bright visible light; the wave model wrongly predicts intensity, not frequency, should matter.wave model
  • 3:42–7:25Photon energy hf and the ping-pong analogyA single low-energy photon can't free an electron no matter how many arrive; photon energy equals hf, so frequency decides the effect.photon energy
  • 7:25–10:17Quantum model explains intensity, frequency, timingMore intensity means more electrons not more KE; higher frequency raises KE; instant photon absorption makes emission instantaneous.photoelectric laws
Experimental setup & saturation current: photoelectric effect | Dual nature of light | Khan Academy 🔉⇢
Khan Academy India - English

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📚 Teaches: Tier-1 educational channel (Khan Academy India - English); found via yt-dlp search 'photoelectric effect intensity saturation current physics', oEmbed-verified live.

📑 Clips (3)
  • 0:00–1:50Why build a photoelectron-counting experimentZinc emits electrons under UV light; a second metal is added as a collector to count how many electrons come out per second.emitter and collector
  • 3:40–7:21Applying voltage to collect all electronsA battery makes the collector positive to attract all electrons; raising the voltage until current stops rising confirms full collection.collecting voltage
  • 7:21–11:05Saturation current and the photocurrent graphSaturation current equals photoelectrons per second; predicts the photocurrent-vs-voltage graph and notes the vacuum requirement.saturation current
All photoelectric effect graphs: Effect of intensity/frequency | Dual nature of light | Khan Academy 🔉⇢
Khan Academy India - English

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📚 Teaches: Tier-1 educational channel (Khan Academy India - English); found via yt-dlp search 'photoelectric effect intensity saturation current physics', oEmbed-verified live.

📑 Clips (3)
  • 0:00–1:50Recap: saturation current and stopping potentialReviews the photocurrent-vs-voltage graph: saturation current shows photoelectrons per second; stopping potential gives max kinetic energy.photocurrent graph
  • 5:30–7:20Higher intensity raises the saturation currentMore intensity means more photons per second and more electrons per second, so the saturation current on the graph becomes larger.intensity effect
  • 9:11–10:37Stopping voltage vs frequency from Einstein's equationUses Einstein's equation to show stopping voltage rises linearly with frequency but stays zero below the threshold frequency.threshold frequency
Stopping potential & maximum kinetic energy | Dual nature of light | Physics | Khan Academy 🔉⇢
Khan Academy India - English

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📚 Teaches: Tier-1 educational channel (Khan Academy India - English); found via yt-dlp search 'stopping potential threshold frequency photoelectric effect physics', oEmbed-verified live.

📑 Clips (3)
  • 0:00–1:51Saturation current recap: counting electronsRecaps how a collector battery drives photocurrent to a maximum, the saturation current, which indicates electrons emitted per second.saturation current
  • 3:43–7:25Reversing voltage to find the stopping potentialApplying negative collector voltage repels electrons; raising it until current hits zero gives the stopping voltage, e.g. 3 volts.stopping potential
  • 7:25–11:05Max kinetic energy from stopping voltageUses potential difference meaning to show fastest electron's max KE equals stopping voltage in electron volts (3 eV = 4.8e-19 J).kinetic energy
Stopping potential vs frequency graph | Dual nature of light | Physics | Khan Academy 🔉⇢
Khan Academy India - English

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📚 Teaches: Tier-1 educational channel (Khan Academy India - English); found via yt-dlp search 'stopping potential photoelectric Khan Academy', oEmbed-verified live.

📑 Clips (3)
  • 0:00–1:50Why Millikan thought Einstein's equation was wrongIntroduces Millikan's skepticism and simplifies Einstein's equation to relate frequency of light to stopping potential in electron volts.Einstein's equation
  • 3:40–5:30Slope h/e is a universal constantWrites stopping voltage vs frequency as y=mx+c; the slope h/e is the same for every material, which is what Millikan found strange.slope h/e
  • 7:21–9:34Millikan's experiments confirm EinsteinAfter ten years Millikan found the same slope for all materials, proving Einstein right and letting h be measured experimentally.experimental proof
Photo electric effect & failure of wave theory | Dual nature of light | Physics | Khan Academy 🔉⇢
Khan Academy India - English

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📚 Teaches: Tier-1 educational channel (Khan Academy India - English); found via yt-dlp search 'why wave theory fails photoelectric effect physics', oEmbed-verified live.

📑 Clips (3)
  • 0:01–1:51Hertz discovers the photoelectric effectShining light on metals ejects electrons; this photoelectric effect, found by Hertz, forced physicists to rethink the nature of light.photoelectric effect
  • 5:32–7:23Frequency controls energy; threshold frequencyRaising frequency raises electron kinetic energy, and below a minimum threshold frequency no electrons are emitted no matter how bright.threshold frequency
  • 9:15–11:07Emission is instantaneous, not delayedWave theory predicted a time delay to gather energy, but electrons are emitted instantly on illumination regardless of brightness.time delay
Einsteins photoelectric equation & work function | Dual nature of light | Physics | Khan Academy 🔉⇢
Khan Academy India - English

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📚 Teaches: Tier-1 educational channel (Khan Academy India - English); found via yt-dlp search 'Einstein photoelectric equation Khan Academy', oEmbed-verified live.

📑 Clips (3)
  • 0:00–1:52What work function meansExplains work function as the minimum energy to pull an electron from a metal, using gold at about 5 eV and defining the electron volt.work function
  • 3:43–5:34Why electrons emerge with a range of energiesShows electrons can lose energy in collisions or be more tightly bound, so kinetic energies vary rather than being a single value.energy spectrum
  • 7:25–9:48Deriving Einstein's photoelectric equationUses a cesium example to show maximum kinetic energy equals photon energy minus work function, giving Einstein's equation and why K is max.Einstein's equation
Photon Momentum | Quantum physics | Physics | Khan Academy 🔉⇢
Khan Academy

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📑 Clips (2)
  • 0:00–1:51Light has momentum but no massExplains why massless light still carries momentum, so p = mv does not apply to photons and special relativity is needed.photon momentum
  • 1:51–2:56Photon momentum formula p = h/lambdaGives the photon momentum as Planck's constant over wavelength; tiny per photon but enough to push a solar sail.photon momentum
Photon Energy | Physical Processes | MCAT | Khan Academy 🔉⇢
khanacademymedicine

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📚 Teaches: Tier-1 educational channel (khanacademymedicine); found via yt-dlp search 'photon energy and momentum physics Khan Academy', oEmbed-verified live.

📑 Clips (3)
  • 0:01–1:51Why light is both a wave and a particleLight shows diffraction and interference like a wave, yet deposits energy only in discrete quantum chunks, revealing particle-like behavior.wave-particle duality
  • 3:41–5:33Photon energy formula E = hfEnergy of a single photon equals Planck's constant times frequency; Planck's constant is tiny (~6.626e-34 J s), so quantization is hard…photon energy
  • 7:23–9:14Energy absorbed in discrete stepsA detector absorbs whole photons at a time, giving a step-like build-up that looks smooth on a macroscopic scale but is quantized up close.energy quantization
De Broglie wavelength | Physics | Khan Academy 🔉⇢
Khan Academy Physics

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📚 Teaches: Tier-1 educational channel (Khan Academy Physics); found via yt-dlp search 'de Broglie wavelength matter waves Khan Academy', oEmbed-verified live.

📑 Clips (3)
  • 0:00–1:50Light as wave and particleThe photoelectric effect needed light delivering energy in discrete packets (hf), while the double-slit pattern showed wave behavior.wave-particle duality
  • 1:50–3:40De Broglie proposes matter wavesIn 1924 de Broglie suggested that electrons, thought to be particles, may also have a wavelength and behave like waves.de Broglie hypothesis
  • 7:21–9:11Davisson-Germer confirms electron wavesElectrons fired through an atomic-scale double slit produced a diffraction pattern, proving matter particles have a wavelength h/p.electron diffraction

💬 Doubt Solving Ask-any-doubt RAG + FAQ

🤖 Ask any doubt RAG · NCERT-grounded, cited

Answers are grounded in tier-1 material (NCERT / JEE PYQ) and cited — the AI will say so if a doubt isn't in the indexed syllabus.

Frequently-asked doubts

Does increasing the intensity of the light increase the maximum kinetic energy of the photoelectrons?
No. Intensity fixes how MANY photons arrive per second, so it fixes the saturation current (the number of electrons emitted per second). The maximum kinetic energy is fixed by the frequency through $K_{max}=h\nu-\phi_0$ and is completely independent of intensity. On the current-voltage graph, raising intensity lifts the saturation plateau but leaves the stopping potential exactly where it was.
Why does a very bright red light fail to eject electrons when a dim blue light succeeds?
Because each photon acts on a single electron, and a red photon carries less energy than a blue one. If a red photon's energy $hc/\lambda$ is below the work function, no single red photon can free an electron, and adding more red photons (more intensity) just adds more too-weak photons. A blue photon carries more energy; if it exceeds $\phi_0$, even a few of them (dim light) cause emission.
What exactly does the stopping potential measure?
It measures the maximum kinetic energy of the emitted electrons: $K_{max}=eV_0$. When the collector is made just negative enough that even the fastest electron is turned back, all of that electron's kinetic energy has been spent climbing the potential $eV_0$. So a voltmeter reading $V_0$ is a direct, convenient measurement of the electron's maximum energy in electron volts.
Is the slope of the stopping-potential-versus-frequency graph the same for all metals?
Yes. The slope is $h/e$, a ratio of two fundamental constants, so it is universal — every metal gives a line of the same slope. What differs between metals is the intercept: the threshold frequency $\nu_0=\phi_0/h$, and equivalently the work function. This is why Millikan could measure $h$ from the slope without ever needing the work function.
Why don't we see the wave nature of a moving cricket ball?
Because its de Broglie wavelength is absurdly small. A $0.12\,\text{kg}$ ball at $20\,\text{m/s}$ has $\lambda=h/mv\approx2.8\times10^{-34}\,\text{m}$, some twenty orders of magnitude smaller than an atom. No aperture or obstacle is anywhere near that small, so no diffraction or interference can occur. Only very light particles like electrons have wavelengths (fractions of a nanometre) large enough to diffract off crystals.
For the same kinetic energy, which has the longer de Broglie wavelength, an electron or a proton?
The electron. Since $\lambda=h/\sqrt{2mK}$, for a fixed kinetic energy $K$ the wavelength is inversely proportional to $\sqrt{m}$, so the lighter particle has the longer wavelength. Watch the condition, though: for the same SPEED the ratio flips ($\lambda\propto1/m$ at fixed $v$), and for the same MOMENTUM the wavelengths are equal.
Why do three photocurrent-versus-voltage curves at the same frequency but different intensities meet at the same stopping potential?
Because the stopping potential is fixed by the maximum kinetic energy $K_{max}=h\nu-\phi_0$, which depends only on the frequency and the metal — neither of which changed. Raising the intensity adds more photons per second, so more electrons are emitted per second and the saturation plateau on the positive-voltage side rises. But the fastest electron still has the same energy, so the same retarding voltage $V_0$ turns it back. All three curves therefore share one point on the negative-voltage axis and fan out only on the positive side.
Does the photoelectric effect happen instantly, and why does that matter?
Yes — provided the frequency is above threshold, emission begins within about $10^{-9}\,\text{s}$ of the light arriving, even for very dim light. This matters because the wave picture predicts the opposite: if a weak wave spread its energy over many electrons, a single electron would need hours to accumulate enough energy to escape. The instantaneous response is direct evidence that energy arrives in a single concentrated packet absorbed by one electron, not spread thinly over the wavefront.
Is a photon's momentum $p=mc$ with some effective mass?
No. A photon has zero rest mass and always travels at $c$; you cannot write $p=mc$ for it. Its momentum comes from $p=E/c$ together with $E=h\nu$, giving $p=h\nu/c=h/\lambda$. It is true that a photon carries momentum and can push on a surface (radiation pressure), but that momentum is a property of its energy and wavelength, not of any rest mass. Reserve $p=mv$ and $p=\sqrt{2mK}$ for massive particles.
How can I decide quickly whether a given light will eject electrons from a metal?
Compare the photon energy with the work function. Convert the wavelength to a photon energy with $E=hc/\lambda=1240/\lambda(\text{nm})\,\text{eV}$, then check whether $E\ge\phi_0$. If the photon energy is at least the work function, emission occurs (with $K_{max}=E-\phi_0$); if it is below, no emission occurs no matter how intense or prolonged the light. Equivalently, compare the wavelength with the threshold $\lambda_0=hc/\phi_0$: emission needs $\lambda\le\lambda_0$.
What is the difference between the work function and the threshold frequency?
They are two ways of stating the same escape condition. The work function $\phi_0$ is the minimum energy an electron needs to leave the metal; the threshold frequency $\nu_0=\phi_0/h$ is the smallest frequency whose photons carry that energy. A metal with a large work function has a high threshold frequency (needs bluer light), and vice versa. On the stopping-potential graph the work function sets the intercept while the universal slope $h/e$ is the same for every metal.
For the same accelerating voltage, which gets the shorter de Broglie wavelength, an electron or a proton?
The proton. Accelerated through the same $V$, both gain the same kinetic energy $eV$, so $\lambda=h/\sqrt{2mqV}$ with the same charge magnitude means $\lambda\propto1/\sqrt{m}$. The heavier proton has the larger momentum and hence the shorter wavelength — the ratio is $\lambda_e/\lambda_p=\sqrt{m_p/m_e}\approx43$. Note this is the same rule as 'same kinetic energy', because equal voltage gives equal kinetic energy for equal charge.
If I double the intensity of the light, does the stopping potential change?
No. The stopping potential depends only on the maximum kinetic energy of the electrons, $eV_0=K_{max}=h\nu-\phi_0$, which is set entirely by the frequency and the metal — not by how many photons arrive. Doubling the intensity doubles the number of photons and therefore doubles the saturation current, but each photon still carries the same energy $h\nu$, so the fastest electrons come off with exactly the same energy as before. The stopping potential stays put; only the height of the current plateau rises. Confusing 'more electrons' with 'faster electrons' is the single most common error on this topic.
Why does the photocurrent saturate at high collector voltage instead of rising forever?
Once the collector is positive enough, it sweeps up every electron the light manages to emit, so the current is limited by the emission rate, not by the collecting voltage. Making the voltage still larger cannot collect electrons that were never emitted, so the current levels off at a plateau — the saturation current. That plateau height is fixed by the intensity: more photons per second mean more electrons per second and a higher plateau. The plateau therefore measures how many electrons are freed, while the cut-off on the other side measures how energetic they are.
Can a single photon eject two electrons if it has more than twice the work function?
In the ordinary photoelectric effect, no: the process is one photon absorbed by one electron, and any energy above the work function becomes that one electron's kinetic energy, $K_{max}=h\nu-\phi_0$. A photon of energy $3\phi_0$ does not split into two escapes; it gives one electron a large kinetic energy. Two-electron emission from a single photon is an exotic higher-order effect, utterly negligible at ordinary intensities, and is never what an exam question intends. Treat photoemission as strictly one-photon-one-electron unless told otherwise.

Trap-answer taxonomy

Trap: Intensity vs frequency confusion

Treating a brighter light as if it made electrons faster, or a higher frequency as if it made more of them.

Fix: Say it every time: intensity sets the number of electrons (saturation current); frequency sets the energy of each (stopping potential, $K_{max}$).

Trap: Dropping the work function

Writing $K_{max}=h\nu$ instead of $K_{max}=h\nu-\phi_0$.

Fix: Always subtract $\phi_0$, and make sure both terms are in the same unit — convert eV to joules or use $hc=1240\,\text{eV nm}$.

Trap: Below-threshold intuition

Believing enough intensity or a long enough exposure will eventually cause emission below $\nu_0$.

Fix: Below the threshold frequency no single photon has enough energy, and photons act one at a time — no intensity or time helps. It is strictly all-or-nothing in frequency.

Trap: Photon vs matter momentum

Using $\lambda=h/mv$ for a photon (which has no rest mass) or $p=E/c$ for a slow electron.

Fix: Photon: $p=E/c=h/\lambda$. Massive particle: $p=\sqrt{2mK}$, $\lambda=h/\sqrt{2mK}$. Keep the two families of relations separate.

Trap: Unit slips in de Broglie problems

Mixing grams with kilograms, or forgetting to take the square root of the accelerating voltage.

Fix: Work in SI (kg, m/s, joules), or use the shortcut $\lambda=1.227/\sqrt{V}\,\text{nm}$ for an electron accelerated from rest through $V$ volts.

Trap: Misreading the stopping-potential graph

Reading the saturation-current height off the stopping-potential axis, or thinking a steeper $V_0$-vs-$\nu$ line means a different metal.

Fix: The stopping potential lives on the negative-voltage axis and measures energy; the saturation current is the positive-voltage plateau and measures electron number. And the $V_0$-vs-$\nu$ slope is $h/e$ for EVERY metal — only the intercept (threshold frequency) changes.

Trap: Confusing 'same energy', 'same speed' and 'same momentum'

Applying $\lambda\propto1/\sqrt{m}$ blindly when the problem actually fixes speed or momentum.

Fix: Same momentum $\Rightarrow$ equal $\lambda$. Same kinetic energy $\Rightarrow$ $\lambda\propto1/\sqrt{m}$ (lighter is longer). Same speed $\Rightarrow$ $\lambda\propto1/m$ (lighter is longer, more strongly). Read which quantity is held fixed before choosing the rule.

Trap: Forgetting the isolated-conductor charging effect

Assuming an isolated illuminated metal keeps emitting electrons indefinitely.

Fix: As electrons leave, the isolated conductor becomes positive and pulls later electrons back. Emission stops when its potential rises to $V=K_{max}/e$; the size of the conductor only sets how much charge that potential represents.

🚪 Dive Deeper Mystery room · 45 discoveries

Discovered 0 / 45

JEE Advanced archive — DISCOVER → ATTEMPT → REVEAL

A particle is confined to move along the $x$-axis between $x=0$ and $x=a$ (of nanometre size). Treating the confined particle as a standing de Broglie wave with nodes at both walls, find the allowed energies.

Attempt, then reveal full solution
Standing waves with nodes at $x=0,a$ require $a=n\dfrac{\lambda}{2}$, so $\lambda_n=\dfrac{2a}{n}$ ($n=1,2,\dots$). The de Broglie relation gives $p_n=\dfrac{h}{\lambda_n}=\dfrac{nh}{2a}$. With $E=\dfrac{p^2}{2m}$, $E_n=\dfrac{n^2h^2}{8ma^2}$. The energy is quantised purely because only whole numbers of half-wavelengths fit the box — this is the particle-in-a-box result, straight from de Broglie.

de Broglie standing-wave condition (JEE Advanced style)

A silver sphere of radius $1\,\text{cm}$ and work function $4.7\,\text{eV}$ hangs isolated in space and is illuminated continuously by monochromatic light of wavelength $200\,\text{nm}$. Explain qualitatively why the photoelectric emission stops after a while, and what fixes the final potential of the sphere.

Attempt, then reveal full solution
Each $200\,\text{nm}$ photon carries $hc/\lambda=6.2\,\text{eV}\gt 4.7\,\text{eV}$, so emission starts with $K_{max}=6.2-4.7=1.5\,\text{eV}$. But every electron that leaves makes the isolated sphere more positive, raising its potential $V$. An emitted electron must now also climb this potential, so effectively the required energy grows by $eV$. Emission ceases when $eV=K_{max}=1.5\,\text{eV}$, i.e. when the sphere's potential reaches $+1.5\,\text{V}$; the radius sets how much charge that potential corresponds to, not the stopping condition.

JEE Advanced style (photoelectric charging of an isolated conductor)

An electron and a photon each have a wavelength of $0.5\,\text{nm}$. Compare their momenta and their energies.

Attempt, then reveal full solution
Momentum depends only on wavelength: $p=h/\lambda$ is the SAME for both, $p=\dfrac{6.63\times10^{-34}}{0.5\times10^{-9}}=1.33\times10^{-24}\,\text{kg m s}^{-1}$. Energies differ: the photon has $E_{ph}=pc=(1.33\times10^{-24})(3\times10^{8})=4.0\times10^{-16}\,\text{J}\approx2.5\,\text{keV}$, while the electron has $E_e=\dfrac{p^2}{2m}=\dfrac{(1.33\times10^{-24})^2}{2(9.11\times10^{-31})}=9.7\times10^{-19}\,\text{J}\approx6\,\text{eV}$. Same wavelength, same momentum, but the photon has hundreds of times the energy.

Equal-wavelength comparison (JEE Advanced style)

Light of wavelength $400\,\text{nm}$ falls on a metal of work function $2.0\,\text{eV}$. Find the maximum kinetic energy of the photoelectrons and the stopping potential. Take $hc=1240\,\text{eV nm}$.

Attempt, then reveal full solution
The photon energy is $E=hc/\lambda=1240/400=3.1\,\text{eV}$. The maximum kinetic energy is $K_{max}=E-\phi_0=3.1-2.0=1.1\,\text{eV}$. Since $eV_0=K_{max}$, the stopping potential is $V_0=1.1\,\text{V}$. Note the stopping potential in volts is numerically equal to $K_{max}$ in electron-volts — a shortcut that saves a step whenever you work in these units.

Standard photoelectric numerical (JEE Main style)

When light of wavelength $\lambda_1=350\,\text{nm}$ illuminates a metal the stopping potential is $1.0\,\text{V}$; for $\lambda_2=250\,\text{nm}$ it is $2.4\,\text{V}$. From these data find Planck's constant. Take $c=3\times10^{8}\,\text{m s}^{-1}$ and $e=1.6\times10^{-19}\,\text{C}$.

Attempt, then reveal full solution
Write $eV_0=hc/\lambda-\phi_0$ for both. Subtracting removes $\phi_0$: $e(V_{02}-V_{01})=hc\left(\dfrac{1}{\lambda_2}-\dfrac{1}{\lambda_1}\right)$. The left side is $1.6\times10^{-19}\times1.4=2.24\times10^{-19}\,\text{J}$. The bracket is $(1/250-1/350)\times10^{9}=1.143\times10^{6}\,\text{m}^{-1}$, so $hc=2.24\times10^{-19}/1.143\times10^{6}=1.96\times10^{-25}\,\text{J m}$ and $h=hc/c=6.5\times10^{-34}\,\text{J s}$, close to the accepted value. Using two wavelengths eliminates the unknown work function entirely.

Two-wavelength determination of h (JEE Advanced style)

A metal has threshold wavelength $600\,\text{nm}$. Light of $400\,\text{nm}$ is incident. Find the maximum speed of the emitted electrons. Take $hc=1240\,\text{eV nm}$, $m_e=9.1\times10^{-31}\,\text{kg}$.

Attempt, then reveal full solution
The work function is $\phi_0=hc/\lambda_0=1240/600=2.07\,\text{eV}$. The photon energy is $1240/400=3.10\,\text{eV}$, so $K_{max}=3.10-2.07=1.03\,\text{eV}=1.65\times10^{-19}\,\text{J}$. Then $v_{max}=\sqrt{2K_{max}/m_e}=\sqrt{2\times1.65\times10^{-19}/9.1\times10^{-31}}\approx6.0\times10^{5}\,\text{m s}^{-1}$. The threshold wavelength is a convenient way to package the work function, since $\phi_0=hc/\lambda_0$.

Threshold wavelength to electron speed (JEE Main style)

An electron and a proton are accelerated from rest through the same potential difference $V$. Find the ratio of their de Broglie wavelengths.

Attempt, then reveal full solution
For a charge $q$ and mass $m$ through $V$, $\lambda=h/\sqrt{2mqV}$. Both carry the same charge magnitude, so $\lambda\propto1/\sqrt{m}$. Hence $\dfrac{\lambda_e}{\lambda_p}=\sqrt{\dfrac{m_p}{m_e}}=\sqrt{1836}\approx43$. The much lighter electron has the far longer wavelength. The key is that equal accelerating voltage means equal kinetic energy, so wavelength scales as the inverse square root of mass.

Electron vs proton at equal voltage (JEE Advanced style)

A monochromatic source emits $1.5\,\text{W}$ of light at wavelength $500\,\text{nm}$. How many photons does it emit per second? Take $hc=1240\,\text{eV nm}$, $1\,\text{eV}=1.6\times10^{-19}\,\text{J}$.

Attempt, then reveal full solution
Each photon has energy $E=hc/\lambda=1240/500=2.48\,\text{eV}=3.97\times10^{-19}\,\text{J}$. The number per second is the power divided by the energy per photon: $N=P/E=1.5/3.97\times10^{-19}\approx3.8\times10^{18}$ photons per second. The enormous count is why a macroscopic beam looks perfectly continuous even though its energy arrives in discrete quanta.

Photon flux from power (JEE Main style)

A parallel beam of light of intensity $I$ falls normally on a perfectly reflecting surface of area $A$. Derive the force the light exerts on the surface.

Attempt, then reveal full solution
In time $t$ the energy hitting the surface is $IAt$, carried by photons of total momentum $p=IAt/c$ (since a photon's momentum is its energy over $c$). On perfect reflection each photon reverses momentum, so the momentum delivered to the surface is $2IAt/c$. The force is the momentum per unit time, $F=2IA/c$, and the radiation pressure is $F/A=2I/c$. For a perfect absorber the factor of two disappears, giving $I/c$.

Radiation pressure derivation (JEE Advanced style)

A photoelectron of maximum kinetic energy $K_{max}=2.0\,\text{eV}$ enters a uniform magnetic field $B=1.0\times10^{-4}\,\text{T}$ perpendicular to its velocity. Find the radius of its circular path. Take $m_e=9.1\times10^{-31}\,\text{kg}$, $e=1.6\times10^{-19}\,\text{C}$.

Attempt, then reveal full solution
First the momentum: $K=2.0\,\text{eV}=3.2\times10^{-19}\,\text{J}$, so $p=\sqrt{2m_eK}=\sqrt{2\times9.1\times10^{-31}\times3.2\times10^{-19}}\approx7.6\times10^{-25}\,\text{kg m s}^{-1}$. In a magnetic field the radius is $r=p/(eB)=7.6\times10^{-25}/(1.6\times10^{-19}\times1.0\times10^{-4})\approx4.8\,\text{cm}$. This links the photoelectric effect to magnetic deflection: the photon sets the momentum, the field bends the path.

Photoelectron in a magnetic field (JEE Advanced style)

Show that a free electron cannot absorb a photon completely (that is, absorb all its energy and momentum and remain a free electron).

Attempt, then reveal full solution
Suppose it could. Conservation of momentum gives $h/\lambda=m_ev$ and conservation of energy (non-relativistically) gives $hc/\lambda=\tfrac12 m_ev^2$. Dividing the energy equation by the momentum equation gives $c=\tfrac12 v$, i.e. $v=2c$, which is impossible. So a free electron cannot fully absorb a photon; a third body (the metal lattice, or a nucleus) must be present to take up the balance of momentum. This is why the photoelectric effect needs bound electrons in a solid, while Compton scattering — partial transfer — is allowed for free electrons.

Kinematic impossibility of full absorption (JEE Advanced style)

X-rays of wavelength $0.10\,\text{nm}$ are Compton-scattered off electrons. Find the wavelength shift for photons scattered through $90^\circ$. Take the Compton wavelength $h/m_ec=2.43\times10^{-12}\,\text{m}$.

Attempt, then reveal full solution
The Compton formula is $\Delta\lambda=\dfrac{h}{m_ec}(1-\cos\theta)$. At $\theta=90^\circ$, $\cos\theta=0$, so $\Delta\lambda=h/m_ec=2.43\times10^{-12}\,\text{m}=0.00243\,\text{nm}$. The scattered wavelength is $0.10243\,\text{nm}$. The shift is independent of the incident wavelength and of the target material, depending only on the scattering angle — a hallmark of the photon picture.

Compton shift at 90 degrees (JEE Advanced style)

Estimate the minimum uncertainty in the speed of an electron known to be located within an atom of size $1\times10^{-10}\,\text{m}$. Take $h=6.63\times10^{-34}\,\text{J s}$, $m_e=9.1\times10^{-31}\,\text{kg}$.

Attempt, then reveal full solution
The uncertainty principle gives $\Delta p\gtrsim h/(4\pi\Delta x)$. With $\Delta x=1\times10^{-10}\,\text{m}$, $\Delta p\approx6.63\times10^{-34}/(4\pi\times10^{-10})\approx5.3\times10^{-25}\,\text{kg m s}^{-1}$. The speed uncertainty is $\Delta v=\Delta p/m_e\approx5.3\times10^{-25}/9.1\times10^{-31}\approx5.8\times10^{5}\,\text{m s}^{-1}$. That an electron confined to atomic size must have a speed spread of hundreds of kilometres per second is a direct consequence of its wave nature.

Uncertainty estimate from confinement (JEE Advanced style)

An electron and an alpha particle have the same kinetic energy. Find the ratio of their de Broglie wavelengths. Take the alpha mass as $7300\,m_e$.

Attempt, then reveal full solution
At equal kinetic energy, $\lambda=h/\sqrt{2mK}$, so $\lambda\propto1/\sqrt{m}$. Hence $\dfrac{\lambda_e}{\lambda_\alpha}=\sqrt{\dfrac{m_\alpha}{m_e}}=\sqrt{7300}\approx85$. The electron's wavelength is about eighty-five times the alpha particle's. The recurring lesson: at equal energy the lighter particle always has the longer wavelength, scaling as the inverse square root of mass.

Equal-energy wavelength ratio (JEE Advanced style)

Monochromatic light just above threshold is incident on a caesium surface ($\phi_0=2.14\,\text{eV}$). If the wavelength is $500\,\text{nm}$, by how much does the stopping potential change when the wavelength is reduced to $400\,\text{nm}$? Take $hc=1240\,\text{eV nm}$.

Attempt, then reveal full solution
Stopping potential in volts equals $K_{max}$ in eV. At $500\,\text{nm}$: $K_{max}=1240/500-2.14=2.48-2.14=0.34\,\text{eV}$, so $V_0=0.34\,\text{V}$. At $400\,\text{nm}$: $K_{max}=1240/400-2.14=3.10-2.14=0.96\,\text{eV}$, so $V_0=0.96\,\text{V}$. The change is $0.96-0.34=0.62\,\text{V}$. Notice the work function cancels in the difference: $\Delta(eV_0)=hc(1/\lambda_2-1/\lambda_1)$, so the shift depends only on the two wavelengths.

Change in stopping potential with wavelength (JEE Main style)

📊 Rank Predictor JoSAA/MCC-calibrated

Disclaimer: These bands are approximate and illustrative, built from publicly reported JoSAA 2023-24 closing-rank trends. Actual ranks depend on the number of candidates, paper difficulty and normalisation in a given year, and vary by category and shift. Use them for orientation, not as a guarantee.
What this does: Dual Nature of Radiation and Matter is a reliable high-yield chapter for JEE Main, almost always good for one or two questions, most often on Einstein's equation with a stopping potential, on the de Broglie wavelength of an accelerated electron, or on photon energy, momentum and flux. In JEE Advanced it usually appears folded into a multi-concept problem — a photoelectron then bent in a magnetic field, or a de Broglie wavelength set by an accelerating voltage. The bands below map an approximate overall JEE Main percentile to a JoSAA closing-rank range, to help you gauge where a given performance sits. They are indicative only.
How to read it: enter your score on a full chapter mock below. The tool maps it — via historical JEE marks→percentile→JoSAA closing-rank data — to the percentile and All-India-Rank band a student at that level typically lands in. It is a calibration signal for THIS chapter's mastery, not a full-exam rank.
Chapter-mock scorePercentile bandProjected AIR band
99.5+ percentile99.5+$\lt 1500$
99.0-99.5 percentile99.0-99.5$1500-4000$
98.0-99.0 percentile98.0-99.0$4000-9000$
95.0-98.0 percentile95.0-98.0$9000-25000$
90.0-95.0 percentile90.0-95.0$25000-55000$
80.0-90.0 percentile80.0-90.0$55000-120000$
$\lt 80$ percentile$\lt 80$$\gt 120000$

JoSAA 2023-24 closing-rank trends (indicative)

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