From mass defect to the binding-energy curve — how nuclei hold together, decay, and power stars and reactors
🔬 Interactive 3D · A nucleus is a densely packed drop of protons and neutrons held by the strong force — almost all of an atom's mass in a volume a trillion times smaller than the atom.
Every atom hides almost all of its mass in a speck at its centre. Rutherford's scattering experiments showed that the positive charge and more than 99.9% of an atom's mass sit in a nucleus whose radius is smaller than the atom's by a factor of about ten thousand — so the nuclear volume is roughly $10^{-12}$ of the atomic volume. If an atom were blown up to the size of a classroom, its nucleus would be a pinhead. This chapter asks what that pinhead is made of, what holds it together against the fierce electrical repulsion of its packed protons, and why rearranging it releases a million times more energy than any chemical reaction. Those three questions — composition, binding, and transformation — organise everything that follows. 🔉⇢
The constituents are nucleons: protons and neutrons. A proton carries one unit of positive charge; the neutron, discovered by Chadwick in 1932, is very slightly heavier and electrically neutral. The atomic number $Z$ counts the protons, the neutron number $N$ counts the neutrons, and the mass number $A=Z+N$ counts the nucleons. A nuclide is written $^{A}_{Z}X$. Nuclides of one element that differ only in $N$ are isotopes; those with the same $A$ are isobars; those with the same $N$ are isotones. Because chemistry is decided by the electron cloud, and the electron count is fixed by $Z$, isotopes are chemically identical and share a place in the periodic table — the very word means 'same place'. Masses on this scale are quoted in the atomic mass unit $u$, defined as one-twelfth of the mass of a $^{12}\text{C}$ atom: $1u=1.660539\times10^{-27}\,\text{kg}$, equivalently $931.5\,\text{MeV}/c^{2}$. 🔉⇢
The size of a nucleus obeys a startlingly simple law. Electron- and $\alpha$-scattering find that a nucleus of mass number $A$ has radius $R=R_{0}A^{1/3}$ with $R_{0}=1.2\,\text{fm}$. Since volume scales as $R^{3}\propto A$, and mass scales as $A$ too, the density is the same for every nucleus — about $2.3\times10^{17}\,\text{kg\,m^{-3}}$, fourteen orders of magnitude denser than water. Nuclear matter behaves like an incompressible liquid drop of universal density; the matter in a neutron star is compressed to essentially this same value, which is why such a star is, in effect, one giant nucleus. 🔉⇢
The deepest idea in the chapter is that mass is a form of energy. Einstein's $E=mc^{2}$ means the mass of a bound nucleus is measurably less than the sum of the masses of its separated nucleons. That deficit, the mass defect $\Delta M=[Z m_{p}+(A-Z)m_{n}]-M$, times $c^{2}$, is the binding energy $E_{b}$: the energy you would have to pour in to pull the nucleus apart. Dividing by $A$ gives the binding energy per nucleon $E_{b}/A$, the single most useful number in nuclear physics. Plotted against $A$ it rises steeply for light nuclei, flattens to a broad plateau near $8.75\,\text{MeV}$ around iron ($A\approx56$), and falls gently to about $7.6\,\text{MeV}$ by uranium ($A=238$). Read that curve correctly and the whole chapter opens up: because iron sits at the top, both splitting a heavy nucleus (fission) and joining light nuclei (fusion) move their nucleons to more tightly bound states and therefore release energy. The near-constant plateau is itself evidence that the nuclear force is short-ranged and saturates — each nucleon binds only to its immediate neighbours. 🔉⇢
What holds the nucleus together cannot be gravity (far too weak) and cannot be the electrical force (which repels the protons). It is a distinct strong nuclear force: attractive and very strong beyond about $0.8\,\text{fm}$, sharply repulsive at shorter range, essentially the same between any pair of nucleons regardless of charge, and effectively zero beyond a few femtometres. Its short range and saturation explain why binding energy per nucleon stays flat once a nucleus is large enough that most nucleons are surrounded rather than on the surface. 🔉⇢
Nuclei that lie off the line of stability transform themselves by radioactivity, discovered by Becquerel in 1896. There are three modes. In $\alpha$-decay the nucleus emits a helium nucleus $^{4}_{2}\text{He}$, so $Z\to Z-2$ and $A\to A-4$. In $\beta$-decay an electron (or positron) is emitted along with an (anti)neutrino: $\beta^{-}$ raises $Z$ by one at fixed $A$, $\beta^{+}$ lowers it — the neutrino is what balances energy, momentum and lepton number, and its introduction is what rescues the conservation laws that $\beta$-decay first seemed to break. In $\gamma$-decay the nucleus sheds excess energy as a high-energy photon with no change in $Z$ or $A$. Whichever the mode, the number of nuclei left undecayed falls exponentially: $N=N_{0}e^{-\lambda t}$. The decay constant $\lambda$ fixes the half-life $T_{1/2}=\ln2/\lambda=0.693/\lambda$ and the mean life $\tau=1/\lambda$, and the activity — the number of disintegrations per second — is $R=\lambda N$, measured in becquerel ($1\,\text{Bq}=1$ decay/s) or curie ($1\,\text{Ci}=3.7\times10^{10}\,\text{Bq}$). 🔉⇢
Finally the chapter turns to nuclear energy. Fission is a neutron-induced splitting of a heavy nucleus such as $^{235}_{92}\text{U}$ into two intermediate-mass fragments plus a few free neutrons, releasing about $200\,\text{MeV}$ per event; those spare neutrons can trigger further fissions, and whether the reaction dies out or grows is governed by a multiplication factor — the principle of both the reactor (controlled, $k=1$) and the bomb (uncontrolled). Fusion is the opposite move: light nuclei joined into a heavier one, the energy source of the Sun, whose proton–proton cycle nets $4\,^{1}_{1}\text{H}\to{}^{4}_{2}\text{He}$ with $26.7\,\text{MeV}$ released. Fusion demands enormous temperatures because the nuclei must be flung hard enough to beat their mutual Coulomb barrier before the short-range attraction can grab them. Every one of these processes is bookkept by a single quantity, the $Q$-value, $Q=(\sum m_{\text{initial}}-\sum m_{\text{final}})c^{2}$: positive means energy released (exothermic), negative means energy absorbed (endothermic). 🔉⇢
It helps to hold the whole chapter as one story about a single graph. The binding-energy-per-nucleon curve is the spine: its rise, plateau and gentle fall encode the composition (what nucleons are), the size law (why density is fixed), the nuclear force (why the plateau is flat), radioactivity (why nuclei off the stability line rearrange), and nuclear energy (why moving toward iron liberates energy). A student who can sketch that curve, mark iron at the top, and say in one breath 'above iron, split; below iron, fuse' already understands the chapter's central claim. Everything numerical then hangs off two conversions — mass to energy through $c^{2}$, and counts to rates through $\lambda$ — and off the conserved quantities: charge, nucleon number, and, once the neutrino is admitted, energy, momentum and lepton number. 🔉⇢
A few orders of magnitude are worth carrying in your head, because they turn abstract formulas into physical intuition. Nuclear sizes are femtometres ($10^{-15}\,\text{m}$); nuclear energies are millions of electron-volts, a million times the electron-volts of chemistry; fissioning one kilogram of uranium releases about $10^{14}\,\text{J}$ against the $10^{7}\,\text{J}$ from burning a kilogram of coal. The Sun's core sits at about $1.5\times10^{7}\,\text{K}$, well below the naive $3\times10^{9}\,\text{K}$ the Coulomb barrier seems to demand, which tells you fusion there rides on the fast tail of the energy distribution rather than the average. Keeping these scales straight is often enough to eliminate two wrong options before any arithmetic. Above all, guard the atoms–nuclei boundary: a question about hydrogen-like electron transitions or spectral lines belongs to the previous chapter no matter how many nuclear-sounding words it carries, while binding energy, decay, fission and fusion are the business of this one. 🔉⇢
For JEE this chapter is compact but reliably examined, contributing a few marks every year, and it rewards precise bookkeeping over memorised formulas. The recurring skills are: convert a mass defect in $u$ to an energy in MeV using $1u=931.5\,\text{MeV}/c^{2}$; read and use the binding-energy-per-nucleon curve to decide whether a process releases energy; track $Z$ and $A$ through a decay or a reaction; and handle the decay law without confusing the exponential form with the halving form $N=N_{0}(1/2)^{t/T_{1/2}}$. Get those four moves clean and almost every nuclei question in the paper becomes routine. 🔉⇢
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.
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The nucleus of an atom is the small, massive, positively charged central region built from protons and neutrons; the number of protons is the atomic number $Z$, the number of neutrons is $N$, and the total number of nucleons is the mass number $A = Z + N$.
The radius $R$ of a nucleus grows with mass number as $R = R_0 A^{1/3}$ with $R_0 \approx 1.2$ femtometres, so the volume is proportional to $A$ and the density of the nucleus is a constant, independent of $A$, nearly the same for all nuclei.
Einstein gave the famous mass-energy equivalence relation $E = mc^2$, which states that mass is a form of energy: a mass $m$ is equivalent to an energy $E$ equal to the mass multiplied by the square of the speed of light $c$.
The mass of a nucleus is always less than the total mass of the separate protons and neutrons that make it up; this difference is the mass defect $\Delta M$, and the energy equivalent $\Delta M c^2$ is the binding energy, the energy needed to separate the nucleus into its free nucleons.
The binding energy per nucleon is the total binding energy of a nucleus divided by its mass number, $E_{bn} = E_b / A$; we can think of binding energy per nucleon as the average energy needed to remove one nucleon from the nucleus, and its variation with mass number governs which nuclei can release energy.
The nuclear force is the strong attractive force that binds protons and neutrons together in the nucleus; it acts only over a very short range of a few femtometres, is much stronger than the electric force at that range, is nearly the same between any pair of nucleons, and falls rapidly to zero beyond its range.
Radioactivity is the process in which an unstable nucleus spontaneously emits radiation and changes into another nucleus; this is referred to as radioactive decay. Three types of radioactive decay occur in nature: alpha decay, beta decay and gamma decay.
The law of radioactive decay states that the number of nuclei that decay per unit time is proportional to the number of undecayed nuclei present, so the number remaining falls as $N = N_0 e^{-\lambda t}$; radioactivity is a nuclear phenomenon in which an unstable nucleus undergoes a decay, and $\lambda$ is the decay constant.
The half-life $T_{1/2}$ is the time in which half of the undecayed nuclei present decay, and the mean life $\tau$ is the average time a nucleus survives; they are related to the decay constant by $T_{1/2} = 0.693/\lambda$ and $\tau = 1/\lambda$, so that $T_{1/2} = 0.693\,\tau$.
The activity of a radioactive sample is the number of nuclei that decay in it per unit time; it equals the decay constant multiplied by the number of undecayed nuclei present, $R = \lambda N$, and is measured in becquerel, where one becquerel is one decay per second.
A most important neutron-induced nuclear reaction is fission, in which a heavy nucleus such as uranium absorbs a neutron and splits into two middle-sized fragments, releasing a few neutrons and a large amount of energy, about $200$ million electron volt per fission.
In nuclear fusion two light nuclei join to form a heavier nucleus; when two light nuclei fuse to form a larger nucleus, energy is released because the product is more tightly bound, but the nuclei must first overcome their electric repulsion, which requires very high temperature.
The Q value of a nuclear reaction $A + b \to C + d$ is the energy released, equal to the difference between the total rest energy of the reactants and that of the products, $Q = [(m_A + m_b) - (m_C + m_d)]c^2$; a positive $Q$ means energy is released and a negative $Q$ means energy must be supplied.
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A nucleus is Z protons and N neutrons packed together. The proton count Z fixes the element (and the charge +Ze); adding neutrons changes A but not the element — that is what isotopes are.
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Every atom is electrically neutral and consists of a very small central nucleus surrounded by electrons. The nucleus contains nearly the whole mass of the atom, while the electrons in the outer region contribute almost nothing to the mass. Experiments by Rutherford, Geiger and Marsden on the scattering of alpha particles by thin foils revealed that the positive charge and nearly all the mass of the atom reside in this small central nucleus. The size of the nucleus is smaller than the size of the atom by a large factor, so the atom is mostly empty space with a dense core. 🔉⇢
The nucleus is built from two kinds of particles. The proton carries a positive charge equal in magnitude to the charge of the electron, and the neutron is neutral, carrying no charge. Because a proton or a neutron inside the nucleus is called a nucleon, the nucleus is a bound system of nucleons. The number of protons in the nucleus is the atomic number, denoted $Z$. Since the atom is neutral, the number of electrons outside the nucleus equals the number of protons inside it, so $Z$ also determines the chemical behaviour of the element and its place in the periodic table. 🔉⇢
The number of neutrons in the nucleus is the neutron number, denoted $N$. The total number of nucleons, protons together with neutrons, is the mass number, denoted $A$. These numbers obey the simple relation $A = Z + N$, so the neutron number follows as $N = A - Z$. A nucleus of a particular element is written with its chemical symbol $X$, with the mass number as a superscript and the atomic number as a subscript, in the form $\,{}^{A}_{Z}X$. For example, the nucleus of gold is written $\,{}^{197}_{79}Au$, which contains $79$ protons and $197 - 79 = 118$ neutrons. 🔉⇢
A species of nucleus with a definite value of $Z$ and a definite value of $N$ is called a nuclide. Nuclides are grouped in useful ways. Nuclei with the same atomic number $Z$ but different mass number $A$ are called isotopes; they contain the same number of protons but different numbers of neutrons. Because $Z$ is the same, isotopes are nuclei of the same chemical element and occupy the same place in the periodic table. Hydrogen, for example, exists as three isotopes: ordinary hydrogen $\,{}^{1}_{1}H$ with one proton and no neutron, deuterium $\,{}^{2}_{1}H$ with one proton and one neutron, and tritium $\,{}^{3}_{1}H$ with one proton and two neutrons. 🔉⇢
Nuclei with the same mass number $A$ but different atomic number $Z$ are called isobars. Since $A$ is the same but $Z$ differs, isobars are nuclei of different elements that contain the same total number of nucleons. Nuclei with the same neutron number $N$ but different atomic number $Z$ are called isotones. These three groupings, isotopes, isobars and isotones, are simply different ways of holding one of the numbers $Z$, $A$ or $N$ constant while the others differ, and they help in studying how nuclear properties depend on the composition of the nucleus. 🔉⇢
The masses of atoms and of nuclei are so small that expressing them in kilogram is inconvenient. Instead they are measured in the atomic mass unit, denoted $u$, which is defined as one-twelfth of the mass of one atom of the carbon isotope $\,{}^{12}_{6}C$. In this unit the mass of a proton is close to $1.00727\ u$ and the mass of a neutron is close to $1.00866\ u$, both nearly equal to one atomic mass unit. The mass of an electron is far smaller, close to $0.00055\ u$, which is why the electrons contribute so little to the mass of the atom and nearly all the mass is concentrated in the nucleus. 🔉⇢
The masses of nuclei can be measured accurately with a mass spectrometer. In such a device charged atoms are separated according to their mass, and the measurement reveals that most elements found in nature are a mixture of several isotopes. The relative amount of each isotope present in a natural sample is called its abundance. The atomic mass of an element that appears in the periodic table is therefore an average value, obtained by weighting the mass of each isotope by its abundance in the natural mixture. 🔉⇢
Chlorine provides a clear example of this weighted average. Chlorine found in nature consists mainly of two isotopes, $\,{}^{35}_{17}Cl$ with mass close to $34.98\ u$ and $\,{}^{37}_{17}Cl$ with mass close to $36.96\ u$, present in the approximate proportions $75.4$ per cent and $24.6$ per cent. Because the two isotopes are present in these amounts, the average atomic mass of chlorine is close to $35.5\ u$, a value that lies between the masses of the two isotopes and closer to the more abundant one. The average is not the mass of any single chlorine nucleus but a property of the natural mixture. 🔉⇢
The proton was identified as a fundamental constituent of the nucleus early in the study of nuclear structure, since the lightest nucleus, that of ordinary hydrogen, is a single proton. The neutron, however, was harder to detect because it carries no charge and therefore does not readily produce the effects that charged particles produce. The neutron was discovered by James Chadwick, who studied the particles emitted when beryllium was bombarded by alpha particles. He showed that the emitted radiation consisted of neutral particles with mass close to that of the proton, and for this discovery he was awarded the Nobel Prize. 🔉⇢
Before the discovery of the neutron it was difficult to explain the composition of the nucleus. The mass number $A$ of a nucleus is greater than its atomic number $Z$ for every element beyond hydrogen, so the nucleus must contain something in addition to protons that adds mass without adding charge. The neutron supplied exactly this. With protons and neutrons as the two constituents, a nucleus of mass number $A$ and atomic number $Z$ is understood to contain $Z$ protons and $A - Z$ neutrons, and this simple picture explains the masses and charges of all nuclei. 🔉⇢
It is important to distinguish clearly between the atomic number and the mass number, because they play different roles. The atomic number $Z$ fixes the identity of the element and equals the number of protons and also the number of electrons in the neutral atom. The mass number $A$ fixes the total number of nucleons and therefore determines, to a good approximation, the mass of the nucleus in atomic mass units. Two nuclei can share the same $A$ yet belong to different elements, and two nuclei can belong to the same element yet have different $A$; only when both $Z$ and $N$ are given is the nuclide completely specified. 🔉⇢
The relative masses of the nucleons and the electron have an important consequence for the mass of the atom. Since each nucleon has a mass close to one atomic mass unit while the electron mass is smaller by a factor of nearly two thousand, the mass of an atom of mass number $A$ is close to $A$ atomic mass units. This is why the mass number is called by that name: it is very nearly the mass of the atom expressed in atomic mass units. The small difference between the actual mass and the whole number $A$ carries important information about the binding of the nucleus, which is studied separately. 🔉⇢
The composition of the nucleus also underlies the notation used throughout nuclear physics. When a nuclear process is written down, the mass number and the atomic number are conserved: the total number of nucleons before a process equals the total number after, and the total charge is likewise conserved. Keeping careful account of $Z$ and $A$ on both sides of a nuclear process is the first step in understanding how one nucleus transforms into another, and it rests entirely on the simple counting of protons and neutrons that defines the composition of the nucleus. 🔉⇢
In summary, the atom is a neutral system with a tiny massive nucleus at its centre and light electrons around it. The nucleus is built from $Z$ protons and $N$ neutrons, together called nucleons, with total number $A = Z + N$. The atomic number $Z$ names the element, the mass number $A$ counts the nucleons, and the neutron number $N = A - Z$ completes the description. Isotopes share $Z$, isobars share $A$, and isotones share $N$. Masses are measured in the atomic mass unit, natural elements are mixtures of isotopes, and the average atomic mass is the abundance-weighted mean of the isotope masses. 🔉⇢
The idea that the nucleus is a bound system of protons and neutrons developed gradually from experiment. The scattering of alpha particles by thin foils, performed by Geiger and Marsden and explained by Rutherford, first established that the atom has a small massive central nucleus carrying the whole positive charge. The measured deflections of the alpha particles could only be explained if the positive charge and nearly all the mass were concentrated in a region far smaller than the atom, and this region is the nucleus. The number of protons in this nucleus was then identified with the atomic number that fixes the chemical element. 🔉⇢
The role of the atomic number as the number of protons is confirmed by the regular way the elements are arranged in the periodic table. As the atomic number increases by one from one element to the next, one more proton is present in the nucleus and one more electron is present in the neutral atom, and the chemical properties change in a regular manner. The number of protons in the nucleus therefore determines both the charge of the nucleus and the number of electrons that surround it, and hence the whole chemical behaviour of the atom. 🔉⇢
The three isotopes of hydrogen bring out the meaning of the neutron number clearly. Ordinary hydrogen has a single proton and no neutron, so its mass number is one. Deuterium has one proton and one neutron, giving mass number two, and its nucleus is called the deuteron. Tritium has one proton and two neutrons, giving mass number three. All three are hydrogen because all three have a single proton and hence atomic number one, yet they differ in mass because they contain different numbers of neutrons. This is the essential feature of isotopes. 🔉⇢
Isobars and isotones can be understood by the same kind of counting. Two nuclei are isobars when they have equal mass number but different atomic number, so they contain the same total number of nucleons shared differently between protons and neutrons. Two nuclei are isotones when they contain the same number of neutrons but different numbers of protons, so they belong to different elements. Studying nuclei related in these ways helps reveal how the stability and other properties of a nucleus depend separately on the number of protons and the number of neutrons it contains. 🔉⇢
The choice of the carbon isotope $\,{}^{12}_{6}C$ to define the atomic mass unit is a matter of convention, but it fixes the unit very precisely. By defining the mass of one carbon $\,{}^{12}_{6}C$ atom to be exactly twelve atomic mass units, the masses of all other atoms and nuclei are measured relative to it. On this scale the proton and the neutron each have a mass close to one atomic mass unit, so a nucleus of mass number $A$ has a mass close to $A$ atomic mass units, and the small departures from whole numbers carry the information about nuclear binding. 🔉⇢
The discovery of the neutron by Chadwick completed the picture of nuclear composition. When beryllium was bombarded by alpha particles, a penetrating neutral radiation was emitted, and Chadwick showed that this radiation consisted of neutral particles of mass close to that of the proton. Because the neutron is neutral, it is not repelled by the positive charge of a nucleus and can approach and enter a nucleus readily, which is why neutrons are so effective in producing nuclear reactions. The neutral neutron and the positive proton together account for the mass and charge of every nucleus. 🔉⇢
The conservation of the atomic number and the mass number governs how one nucleus changes into another. In any nuclear process the total number of nucleons is the same before and after, so the mass numbers balance, and the total charge is the same, so the atomic numbers balance. These two balances are simply statements that protons and neutrons are neither created nor destroyed in ordinary nuclear processes but are only rearranged, and they follow directly from the definition of the atomic number and the mass number in terms of the counting of protons and neutrons. 🔉⇢
Because the electrons are so light and lie far outside the nucleus, they play almost no part in determining the mass or the size of the nucleus, yet they determine the chemistry of the atom. This separation of roles, with the heavy nucleus fixing the mass and the light electrons fixing the chemical behaviour, is a central feature of atomic structure. The number of protons ties the two together, since it fixes both the charge of the nucleus and the number of electrons, so that the atomic number is the single most important number describing an atom. 🔉⇢
Double the nucleons and the radius barely grows — R rises only as A^(1/3). Because volume ∝ A, every nucleus packs matter to the same colossal density, ~2.3×10¹⁷ kg/m³, independent of which nucleus it is.
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The size of a nucleus is measured by scattering fast charged particles at it and studying how they are deflected. In the classic experiment, a beam of alpha particles is directed at a thin foil, and the way the particles are scattered reveals how the positive charge and the mass of the target atoms are distributed. Rutherford analysed the scattering performed by Geiger and Marsden and concluded that the positive charge is concentrated in a very small central nucleus. Later experiments with faster particles measured the actual radius of the nucleus for many elements. 🔉⇢
These measurements show that a nucleus is roughly spherical and that its radius depends on the number of nucleons it contains. The radius $R$ of a nucleus of mass number $A$ is found to follow the simple relation $R = R_0 A^{1/3}$, where $R_0$ is a constant with the value close to $1.2$ femtometres. A femtometre, sometimes called a fermi, is $10^{-15}$ metre, so nuclear radii are of the order of a few femtometres. This is smaller than the size of the atom by a factor of about ten thousand to one hundred thousand. 🔉⇢
The cube-root dependence of the radius on the mass number has a direct meaning. If the radius grows as the cube root of $A$, then the volume, which grows as the cube of the radius, must grow in direct proportion to $A$ itself. A nucleus is approximately a sphere of radius $R$, so its volume is proportional to $R^3$, and since $R^3 = R_0^3 A$, the volume is proportional to $A$. In other words, each added nucleon contributes very nearly the same amount of volume to the nucleus, whatever the size of the nucleus. 🔉⇢
This result leads to one of the most striking features of nuclear matter. The mass of a nucleus is proportional to $A$, since each nucleon has nearly the same mass, and the volume is also proportional to $A$. The density, which is mass divided by volume, therefore does not depend on $A$ at all. Consequently the density of nucleus is a constant, independent of A, for all nuclei. A small nucleus and a large nucleus, although they differ greatly in the number of nucleons they contain, have nearly the same density in their interior. 🔉⇢
The constancy of nuclear density can be made quantitative. Taking a nucleon mass close to $1.67 \times 10^{-27}$ kilogram and a value of $R_0$ close to $1.2$ femtometres, the density of nuclear matter comes out to be of the order of $10^{17}$ kilogram per cubic metre. This is an enormous value, far greater than the density of ordinary matter such as water or metals, whose densities are only a few thousand kilogram per cubic metre. The high nuclear density reflects the fact that the nucleons are packed closely together in the small volume of the nucleus. 🔉⇢
The fact that the density is the same for all nuclei tells us something important about how nucleons are arranged. If the nucleons were packed more tightly in large nuclei than in small ones, the density would increase with $A$; if they were packed more loosely, it would decrease. The observed constancy means that the nucleons keep a nearly fixed spacing from one another, so that adding more nucleons simply makes the nucleus larger without changing how densely the interior is filled. This behaviour resembles the way a drop of liquid grows when more molecules are added, keeping the same density. 🔉⇢
The comparison with a liquid drop is useful. In a drop of liquid the molecules are close together and nearly touch, and adding more molecules enlarges the drop while the density stays the same. Nuclear matter behaves in a similar way, with the nucleons playing the role of the molecules. This picture, in which the nucleus resembles a drop of very dense liquid, helps explain not only the constant density but also several other nuclear properties, and it follows directly from the relation $R = R_0 A^{1/3}$. 🔉⇢
It is worth appreciating how small the nucleus is compared with the atom. The radius of a typical atom is of the order of $10^{-10}$ metre, while the radius of a nucleus is of the order of a few times $10^{-15}$ metre. The nucleus is therefore smaller than the atom by a factor of about ten thousand to one hundred thousand in radius, and by the cube of that factor in volume. If the atom were enlarged to the size of a large room, the nucleus would be no bigger than a pinhead at its centre, yet that pinhead would contain nearly all the mass. 🔉⇢
Because nearly all the mass of the atom is in the nucleus and the nucleus occupies such a tiny fraction of the volume of the atom, the density of the nucleus is enormously greater than the average density of the atom. The average density of ordinary matter is set by the atoms as a whole, which are mostly empty space, whereas the nuclear density is set by the tightly packed core. This contrast, of a factor of the order of $10^{14}$ between nuclear density and ordinary density, is one of the most remarkable results of the study of nuclear size. 🔉⇢
The relation $R = R_0 A^{1/3}$ also lets us compare the sizes of different nuclei directly. Since the radius grows only as the cube root of the mass number, a nucleus with eight times as many nucleons has a radius only twice as large, because the cube root of eight is two. A heavy nucleus with a mass number close to $216$ has a radius about six times that of a light nucleus with mass number close to one, since the cube root of $216$ is six. Thus even the heaviest nuclei are only a few times larger in radius than the lightest. 🔉⇢
The value of the constant $R_0$ is obtained from the scattering measurements themselves. By measuring the radius $R$ for nuclei of known mass number $A$ and plotting $R$ against $A^{1/3}$, the data fall close to a straight line whose slope gives $R_0$. The measurements for many different nuclei agree well with a single value of $R_0$ near $1.2$ femtometres, which is strong evidence that the simple relation $R = R_0 A^{1/3}$ describes real nuclei and that the density is indeed constant across the whole range of nuclei. 🔉⇢
The astronomical importance of nuclear density is seen in certain stars. When a massive star collapses at the end of its life, the matter can be compressed until its density approaches that of nuclear matter, and a very dense object is formed in which the material is packed almost as tightly as inside a nucleus. Such objects show, on a giant scale, the same enormous density that the relation $R = R_0 A^{1/3}$ predicts for the interior of a single nucleus, connecting the smallest and the largest objects in nature. 🔉⇢
In summary, the radius of a nucleus is given by $R = R_0 A^{1/3}$, with $R_0$ close to $1.2$ femtometres. The volume is therefore proportional to the mass number $A$, and since the mass is also proportional to $A$, the density is the same for all nuclei, with a value of the order of $10^{17}$ kilogram per cubic metre. This constant, very high density means the nucleons are packed closely together with a fixed spacing, and the nucleus behaves much like a small, extremely dense drop of liquid. 🔉⇢
The measurement of nuclear size rests on the scattering of fast charged particles, and the interpretation of these experiments gives the radius directly. When particles of high enough energy are directed at a nucleus, the way they are scattered depends on how the charge and matter of the nucleus are distributed over its volume. By studying the scattering for nuclei of many different mass numbers, the radius of each nucleus is determined, and the collected results are described by the single relation $R = R_0 A^{1/3}$ with $R_0$ close to $1.2$ femtometres. 🔉⇢
The cube-root law can be understood by thinking of the nucleus as a collection of nucleons each occupying a fixed small volume. If every nucleon takes up the same volume and the nucleons are packed together with a fixed spacing, then the total volume of the nucleus is simply the volume of one nucleon multiplied by the number of nucleons, so the volume is proportional to $A$. A sphere whose volume is proportional to $A$ has a radius proportional to the cube root of $A$, which is exactly the observed relation. 🔉⇢
The constancy of nuclear density is a direct consequence and can be checked with numbers. The mass of the nucleus is close to $A$ times the mass of a single nucleon, and the volume is $\frac{4}{3}\pi R_0^3 A$. When the mass is divided by the volume to obtain the density, the mass number $A$ cancels between numerator and denominator, leaving a density that does not contain $A$ at all. The interior of a small nucleus and the interior of a large nucleus therefore have the same density, a value of the order of $10^{17}$ kilogram per cubic metre. 🔉⇢
This enormous density is difficult to grasp by comparison with ordinary matter. Water has a density of one thousand kilogram per cubic metre, and even the densest ordinary metals reach only about twenty thousand kilogram per cubic metre, yet nuclear matter is denser than these by a factor of the order of $10^{13}$. The reason is that in ordinary matter the atoms are mostly empty space, with tiny nuclei separated by distances far larger than their own size, whereas in nuclear matter the nucleons themselves are packed closely together with almost no empty space between them. 🔉⇢
The same relation lets one compare the radii of different nuclei quickly. Because the radius depends only on the cube root of the mass number, large changes in the number of nucleons produce only modest changes in the radius. A nucleus with mass number sixteen has a radius close to twice that of a nucleus with mass number two, since the cube root of eight is two. Even a very heavy nucleus with a mass number in the hundreds has a radius only a few times larger than that of the lightest nuclei, so all nuclei are of the same order of size, a few femtometres. 🔉⇢
The picture of the nucleus as a drop of very dense liquid follows naturally from the constant density. In a liquid the molecules are in contact and the density is fixed, so adding more molecules enlarges the drop while keeping the density the same, exactly as adding nucleons enlarges the nucleus at fixed density. This resemblance is more than an analogy, for it suggests that the nucleons interact mainly with their nearest neighbours, just as molecules in a liquid do, and this idea of short-range interaction between neighbouring nucleons explains several nuclear properties. 🔉⇢
The value of the constant $R_0$ is obtained by fitting the measured radii to the relation $R = R_0 A^{1/3}$. When the measured radius of each nucleus is compared with the cube root of its mass number, the points follow a straight line whose slope is $R_0$, and the same value near $1.2$ femtometres describes nuclei across the whole range of mass numbers. The success of a single value of $R_0$ in describing so many different nuclei is strong evidence that the simple relation is correct and that the density is truly constant. 🔉⇢
The extremely high density of nuclear matter has a counterpart on the astronomical scale. When a very massive star collapses at the end of its life, gravity can compress its matter until the density approaches that inside a nucleus, forming an extremely dense object. In such an object the matter is packed almost as tightly as the nucleons inside a single nucleus, so that a body of astronomical size behaves in some ways like an enormous piece of nuclear matter. This connects the smallest scale, the interior of a nucleus, with some of the densest objects known in the universe. 🔉⇢
The relation between size and mass number can be verified by comparing many nuclei at once. When the measured radius of each nucleus is set against the cube root of its mass number, the different nuclei fall together on a single straight line, which shows that one and the same constant $R_0$ describes them all. Nuclei ranging from the lightest to the heaviest agree with the relation $R = R_0 A^{1/3}$, and this agreement across the whole range of nuclei is the experimental foundation for treating the density of nuclear matter as a single fixed constant. 🔉⇢
It is instructive to state clearly how far the nucleus is from filling the atom. The radius of the atom is set by the electrons and is of the order of $10^{-10}$ metre, while the radius of the nucleus is only a few femtometres, that is a few times $10^{-15}$ metre. The nucleus is thus smaller than the atom by a factor of about ten thousand to one hundred thousand, and its volume is smaller by the cube of this factor. Nearly all the mass is packed into this tiny central volume, which is why the nuclear density is so enormously greater than the density of ordinary matter. 🔉⇢
The result that the density of the nucleus is a constant is one of the most important conclusions of the study of nuclear size, because it tells us how the nucleons are arranged inside the nucleus. A constant density means the nucleons keep a fixed spacing and interact mainly with their nearest neighbours, so that adding a nucleon enlarges the nucleus without changing how densely the interior is filled. This behaviour, together with the relation $R = R_0 A^{1/3}$, is the starting point for the picture of the nucleus as a small, very dense drop of nuclear matter. 🔉⇢
Mass is frozen energy. Convert a mass m and out comes E = mc² — and at nuclear scales the exchange rate is fixed: one atomic mass unit is worth 931.5 MeV.
Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.
One of the deepest results of the theory of relativity is that mass and energy are not separate and independent quantities but are equivalent, two aspects of the same physical quantity. Einstein gave the famous mass-energy equivalence relation in the form $E = mc^2$, where $E$ is the energy equivalent of a mass $m$ and $c$ is the speed of light in vacuum. Because the speed of light is very large, close to $3 \times 10^8$ metre per second, even a very small mass is equivalent to an enormous amount of energy. 🔉⇢
The relation means that whenever the energy of a system changes, its mass changes as well, and whenever the mass changes, energy is released or absorbed. In ordinary chemical and mechanical processes the energy changes are so small that the accompanying change in mass is far too small to measure, and mass appears to be separately conserved. In nuclear processes, however, the energy changes are so large that the change in mass becomes measurable, and the equivalence of mass and energy is directly verified. 🔉⇢
To use the relation in nuclear physics it is convenient to express energy in a unit suited to the small masses involved. The natural mass unit is the atomic mass unit $u$, defined as one-twelfth of the mass of a carbon $\,{}^{12}_{6}C$ atom, with the value $1\ u = 1.660539 \times 10^{-27}$ kilogram. Using $E = mc^2$ with this mass gives the energy equivalent of one atomic mass unit, and the result is that $1\ u$ is equivalent to about $931.5$ million electron volt, written $931.5\ \text{MeV}$. 🔉⇢
The result $1\ u = 931.5\ \text{MeV}/c^2$ is one of the most useful numbers in nuclear physics, because it lets any mass measured in atomic mass units be converted at once into an energy in million electron volt. If a nuclear process involves a change of mass of, say, a hundredth of an atomic mass unit, the energy released or absorbed is about $9.3$ million electron volt, a large energy on the nuclear scale. This is why nuclear processes liberate energies millions of times greater than those of chemical reactions between the same numbers of atoms. 🔉⇢
The equivalence of mass and energy also changes the statement of the conservation laws. In relativity it is not mass alone and energy alone that are separately conserved, but the total mass-energy of an isolated system. When we say energy is conserved in a nuclear process, we include the energy equivalent of the masses of the particles taking part. A decrease in the total mass of the particles appears as an equal increase in other forms of energy, such as the kinetic energy of the products, and an increase in mass must be supplied by an equal amount of energy. 🔉⇢
A clear illustration of the interconversion of mass and energy is the behaviour of a particle and its antiparticle. When an electron meets a positron, which is the antiparticle of the electron with the same mass but positive charge, the two particles can annihilate. In this process the whole of their mass disappears and reappears as the energy of the photons that are produced. The reverse process is also possible: a photon of sufficient energy can produce a particle-antiparticle pair, so that energy is converted into mass. These processes show the equivalence directly. 🔉⇢
The energy equivalent of a mass is often written as an energy divided by $c^2$, so that a mass can be quoted in units of $\text{MeV}/c^2$. For example, the mass of an electron corresponds to about $0.51$ million electron volt, and the mass of a proton corresponds to about $938$ million electron volt. Quoting masses in these energy units is convenient in nuclear and particle physics, because the energies released in processes are then obtained simply by taking differences of the masses expressed in the same units. 🔉⇢
It is important to understand that $E = mc^2$ does not mean mass is always turned into energy; rather it means mass is a form of energy. A body at rest already possesses an energy $mc^2$, called its rest energy, simply because it has mass. When the body also moves, it has kinetic energy in addition to its rest energy, and the total energy is greater than $mc^2$. The relation therefore assigns an energy to the mere existence of mass, quite apart from any motion, and it is this rest energy that is released when mass decreases in a nuclear process. 🔉⇢
The enormous factor $c^2$ in the relation explains why the energy equivalent of even a tiny mass is so large. Since $c$ is about $3 \times 10^8$ metre per second, $c^2$ is about $9 \times 10^{16}$ in units of metre squared per second squared. Multiplying a mass of one kilogram by this factor gives an energy of about $9 \times 10^{16}$ joules, which is a truly enormous energy. This is why the complete conversion of a very small amount of mass can supply the energy output of a power station over a long time. 🔉⇢
The mass-energy relation is essential for understanding the binding of nuclei. When protons and neutrons come together to form a nucleus, energy is released, and by the equivalence of mass and energy the mass of the nucleus is less than the total mass of the separate nucleons. The missing mass, called the mass defect, is exactly the mass equivalent of the energy released in forming the nucleus. Without the relation $E = mc^2$ this loss of mass would be a mystery; with it, the loss of mass is understood as energy that has left the system. 🔉⇢
The same relation governs the energy released in the processes that power the stars. In the interior of a star, light nuclei combine to form heavier nuclei, and the mass of the product is slightly less than the mass of the reacting nuclei. The difference in mass, multiplied by $c^2$, is the energy that the star radiates as light and heat. Over the lifetime of a star an astronomical amount of mass is converted into energy in this way, and the relation $E = mc^2$ connects the measured masses of nuclei to the energy output of the star. 🔉⇢
Historically, the equivalence of mass and energy transformed the understanding of the conservation laws. Before relativity, conservation of mass and conservation of energy were regarded as two separate principles, each exactly obeyed. Einstein showed that they are really one principle: the conservation of total mass-energy. The measurement of the energies released in nuclear processes, and their agreement with the mass changes calculated from measured nuclear masses, provided convincing experimental verification of this idea and established $E = mc^2$ as a firmly tested law of nature. 🔉⇢
In practice, the relation is applied by first finding the change in mass in a nuclear process, that is the difference between the total mass before and the total mass after, and then multiplying this mass difference by $c^2$ to obtain the energy released or absorbed. Because the masses of nuclei and particles are measured very accurately in atomic mass units, and because $1\ u$ is equivalent to $931.5$ million electron volt, this calculation can be carried out precisely, and it lies at the heart of every quantitative treatment of nuclear energy. 🔉⇢
In summary, the mass-energy equivalence relation $E = mc^2$ states that mass is a form of energy, so that a mass $m$ carries a rest energy equal to $mc^2$. One atomic mass unit is equivalent to $931.5$ million electron volt. In nuclear processes the energy changes are large enough that the accompanying mass changes are measurable, and the total mass-energy, rather than mass alone or energy alone, is the conserved quantity. This single relation underlies the binding of nuclei, the energy of nuclear processes, and the energy radiated by the stars. 🔉⇢
The mass-energy equivalence relation is best appreciated by contrasting nuclear processes with ordinary chemical ones. In a chemical reaction the energy released is a few electron volt for each atom, and the corresponding change in mass, obtained by dividing this energy by $c^2$, is so small that no measurement can detect it. In a nuclear process the energy released is millions of electron volt for each nucleus, and the corresponding change in mass is a measurable fraction of the mass of a nucleon, so the equivalence of mass and energy shows up plainly in the measured masses. 🔉⇢
Because the speed of light appears squared in the relation, the energy equivalent of a given mass is enormous. One kilogram of matter is equivalent to about $9 \times 10^{16}$ joules, an energy far greater than that stored in any ordinary fuel of the same mass. This is why nuclear sources can supply so much energy from so little material: converting even a small fraction of the mass of the material into energy releases an amount that would require an enormous quantity of ordinary fuel to match. 🔉⇢
The relation also tells us that the mass of a bound system is not simply the sum of the masses of its parts. When particles are bound together and energy is released in the process, the bound system has less energy than the separated parts, and by the equivalence of mass and energy it also has less mass. The nucleus is the outstanding example: its mass is less than the total mass of its free nucleons by an amount equal to the binding energy divided by $c^2$. The relation thus links the measured masses of nuclei to the energy that holds them together. 🔉⇢
It is useful to keep in mind the exact figure for the energy equivalent of the atomic mass unit. Since one atomic mass unit corresponds to $931.5$ million electron volt, a mass change of one atomic mass unit in a nuclear process would release or absorb $931.5$ million electron volt of energy. Typical nuclear processes involve mass changes of a small fraction of an atomic mass unit and therefore energies of the order of a few to a few hundred million electron volt, which is the characteristic scale of nuclear energy. 🔉⇢
The interconversion of mass and energy is seen most directly in the creation and annihilation of particles. When a particle meets its antiparticle, such as an electron meeting a positron, the pair can vanish and their entire mass reappear as the energy of photons, a clear case of mass converted wholly into energy. Conversely, a photon of sufficient energy can convert into a particle and its antiparticle, a case of energy converted wholly into mass. Both processes obey the relation $E = mc^2$ exactly and confirm that mass is a form of energy. 🔉⇢
The equivalence of mass and energy requires the conservation laws to be stated together. In an isolated system it is the total of energy and the energy equivalent of mass that stays constant, not either one separately. When the total mass of the particles in a process decreases, an equal energy appears in other forms, chiefly the kinetic energy of the products; when the total mass increases, an equal energy must be supplied. This unified conservation of mass-energy replaces the two older separate laws and is obeyed exactly in all nuclear processes. 🔉⇢
The energy radiated by the stars provides a grand illustration of the relation. In the interior of a star, light nuclei join to form heavier nuclei whose mass is slightly less than the mass of the reacting nuclei, and the difference in mass, multiplied by $c^2$, is radiated as light and heat. Over the life of a star an enormous amount of mass is steadily converted into energy in this way. The measured masses of the nuclei involved, together with the relation $E = mc^2$, account quantitatively for the energy output of the star. 🔉⇢
In practice the relation is used by computing the change in mass in a process and multiplying by $c^2$. One adds the masses of the products and subtracts them from the sum of the masses of the reactants, all expressed in atomic mass units, and then multiplies the difference by $931.5$ million electron volt to obtain the energy released. Because nuclear masses are known very accurately, this calculation gives precise values for the energy released or absorbed, and it is the basis of every quantitative statement about nuclear energy. 🔉⇢
The relation $E = mc^2$ overturned the older view that mass and energy are conserved separately, replacing it with a single conservation of mass-energy. Before relativity, the total mass of matter and the total energy were each thought to be exactly constant. Einstein showed that the two are aspects of one quantity, so that a change in the energy of a system is always accompanied by a proportional change in its mass. The measured energies released in nuclear processes, and their agreement with the mass changes found from measured masses, give a convincing verification of this idea. 🔉⇢
It is worth stressing that the rest energy $mc^2$ belongs to a body simply because it has mass, quite apart from any motion. A body at rest already possesses this energy, and when it moves it has in addition its kinetic energy, so the total energy is greater than the rest energy. In a nuclear process it is this stored rest energy that is partly released when the total mass of the particles decreases, and the released energy appears as the kinetic energy of the products, which can then be used or observed. 🔉⇢
A bound nucleus weighs LESS than its separate protons and neutrons — the missing mass ΔM (the red sliver, magnified ×30) reappeared as the binding energy E_b = ΔM·c² that glues it together.
Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.
When the mass of a nucleus is measured accurately and compared with the sum of the masses of the protons and neutrons that compose it, a remarkable fact emerges. If one adds the mass of $Z$ separate protons and the mass of $N = A - Z$ separate neutrons, the total is larger than the measured mass of the nucleus they form. In other words, the nuclear mass $M$ is found to be always less than this total mass of the constituents. The nucleus is lighter than the parts from which it is built. 🔉⇢
The difference between the total mass of the free nucleons and the mass of the nucleus is called the mass defect, denoted $\Delta M$. For a nucleus of atomic number $Z$ and mass number $A$, containing $Z$ protons of mass $m_p$ and $A - Z$ neutrons of mass $m_n$, the mass defect is $\Delta M = [Z m_p + (A - Z) m_n] - M$, where $M$ is the measured mass of the nucleus. This quantity is always positive, showing that mass is lost when free nucleons combine to form a nucleus. 🔉⇢
The loss of mass is understood through the mass-energy equivalence relation $E = mc^2$. Since mass is a form of energy, the disappearance of a mass $\Delta M$ means that an energy $\Delta M c^2$ has been released when the nucleons came together to form the nucleus. This released energy is called the binding energy of the nucleus, denoted $E_b$, and it is given by $E_b = \Delta M c^2$. The binding energy is the energy that was liberated in assembling the nucleus from its separate nucleons. 🔉⇢
The same energy has a second, equally important meaning. Because an energy $E_b$ was released when the nucleus formed, exactly that much energy must be supplied from outside to pull the nucleus apart again into its separate protons and neutrons. The binding energy is therefore also the energy required to separate the nucleus completely into free nucleons at rest, far apart from one another. A nucleus with a large binding energy is tightly bound, since a large energy is needed to break it up. 🔉⇢
The existence of a positive binding energy is what makes the nucleus a stable, bound system. If the nucleus had the same mass as its separated constituents, no energy would be needed to take it apart and it would not be bound. The fact that the nucleus is lighter, by the mass defect $\Delta M$, means that the bound state has less energy than the separated state, and the difference is the binding energy. Every stable nucleus has a positive binding energy, and the greater this energy, the more difficult the nucleus is to disassemble. 🔉⇢
To calculate a binding energy in practice, the mass defect is first found in atomic mass units and then converted into an energy using the equivalence $1\ u = 931.5$ million electron volt. Because $\Delta M c^2 = \Delta M \times 931.5$ million electron volt when $\Delta M$ is in atomic mass units, the binding energy in million electron volt follows immediately from the mass defect. Nuclear masses are measured very accurately with a mass spectrometer, so the mass defect and hence the binding energy can be determined precisely. 🔉⇢
Consider the oxygen nucleus $\,{}^{16}_{8}O$ as an example. It contains $8$ protons and $8$ neutrons. Adding the masses of $8$ free protons and $8$ free neutrons gives a total larger than the measured mass of the oxygen nucleus, and the difference, the mass defect, is close to $0.13691\ u$. Multiplying by $931.5$ million electron volt per atomic mass unit gives a binding energy close to $127.5$ million electron volt. This is the energy that would be needed to separate the oxygen nucleus into its sixteen free nucleons. 🔉⇢
The binding energies of nuclei are enormous compared with the energies involved in chemical reactions. Chemical reactions rearrange the outer electrons of atoms and involve energies of only a few electron volt per atom, whereas nuclear binding energies are of the order of millions of electron volt per nucleus. This factor of about a million is the fundamental reason why nuclear processes release so much more energy than chemical processes for the same amount of material, and it follows directly from the size of the mass defect. 🔉⇢
It is important to use the correct masses in calculating the mass defect. The masses that are usually quoted in tables are atomic masses, which include the masses of the electrons in the neutral atom, rather than bare nuclear masses. When atomic masses are used consistently on both sides, the electron masses very nearly cancel, since the number of electrons is the same, and the small binding energy of the electrons is negligible compared with the nuclear binding energy. With this care, the mass defect and binding energy come out correctly. 🔉⇢
The binding energy can be pictured as an energy well. To assemble the nucleus, the nucleons fall together into a bound state of lower energy, releasing the binding energy in the process, just as an object falling into a well releases energy. To take the nucleus apart, that same energy must be supplied to lift the nucleons out of the well. The depth of the well corresponds to the binding energy, and a deeper well, meaning a larger binding energy, corresponds to a more tightly bound and more stable nucleus. 🔉⇢
The concept of binding energy is central to understanding where nuclear energy comes from. If a nucleus can rearrange itself into products that are more tightly bound, meaning products with a larger total binding energy, then the excess energy is released. This is the underlying reason that both the splitting of a very heavy nucleus and the joining of very light nuclei can release energy: in each case the products are more tightly bound than the starting nuclei, so binding energy is set free as the kinetic energy of the products. 🔉⇢
The binding energy of a whole nucleus increases as the nucleus grows larger, simply because there are more nucleons contributing to the binding. A heavy nucleus therefore has a larger total binding energy than a light one. However, the total binding energy by itself does not tell how tightly each individual nucleon is held, because a larger nucleus has both more binding energy and more nucleons. To compare how tightly nucleons are bound in different nuclei, the binding energy is divided by the number of nucleons, a quantity studied separately. 🔉⇢
In every case the calculation follows the same steps: find the total mass of the free protons and neutrons, subtract the measured mass of the nucleus to obtain the mass defect $\Delta M$, and multiply by $c^2$, using $1\ u = 931.5$ million electron volt, to obtain the binding energy $E_b = \Delta M c^2$. This simple procedure, resting on the equivalence of mass and energy, turns a measurement of masses into a statement about the energy that binds the nucleus together and the energy needed to break it apart. 🔉⇢
In summary, the measured mass of a nucleus is always less than the total mass of its free nucleons, and the difference is the mass defect $\Delta M = [Z m_p + (A - Z) m_n] - M$. By the relation $E = mc^2$, this mass defect corresponds to the binding energy $E_b = \Delta M c^2$, the energy released when the nucleus forms and equally the energy needed to separate it into free nucleons. Binding energies are of the order of millions of electron volt, and their existence is what makes nuclei stable, bound systems. 🔉⇢
The observation that a nucleus is lighter than its separate nucleons is completely general: for every nucleus the measured mass is less than the sum of the masses of the free protons and neutrons that compose it. This missing mass, the mass defect, is what makes the nucleus a bound system. If there were no missing mass there would be no binding energy, and the nucleons would not stay together. The positive mass defect of every stable nucleus is therefore the direct evidence that energy was released when the nucleus formed. 🔉⇢
The binding energy obtained from the mass defect is very large on the scale of atomic energies. For a typical nucleus the binding energy amounts to millions of electron volt, whereas the energy binding an electron in an atom is only a few electron volt. This factor of about a million between nuclear and atomic binding energies is the reason that nuclear processes release so much more energy than chemical ones. The large binding energy also means that a large energy must be supplied to break a nucleus into its separate nucleons. 🔉⇢
In using measured masses to find the mass defect, one must be careful about whether the masses quoted are nuclear masses or atomic masses. Tables usually give atomic masses, which include the masses of the electrons of the neutral atom. When atomic masses are used consistently on both sides of the calculation, the electron masses nearly cancel because the number of electrons is unchanged, and the small binding energy of the electrons is negligible compared with the nuclear binding energy. With this care the calculation gives the nuclear binding energy correctly. 🔉⇢
The nitrogen nucleus provides a further numerical example. The nucleus $\,{}^{14}_{7}N$ contains $7$ protons and $7$ neutrons. Adding the masses of these free nucleons and subtracting the measured mass of the nitrogen nucleus gives a mass defect of about $0.11235\ u$. Multiplying by $931.5$ million electron volt per atomic mass unit gives a binding energy close to $104.7$ million electron volt. This is the energy released when seven protons and seven neutrons combine to form the nitrogen nucleus, and equally the energy needed to separate it. 🔉⇢
It is helpful to picture the binding energy as the depth of an energy well into which the nucleons fall when they combine. The separated nucleons, far apart and at rest, are taken as the zero of energy; when they come together to form the nucleus they release the binding energy and settle into a state of lower energy, the bottom of the well. To pull them apart again, the same energy must be supplied to lift them out of the well. A deeper well means a larger binding energy and a more tightly bound, more stable nucleus. 🔉⇢
The concept of binding energy explains where nuclear energy comes from. Energy is released whenever a nucleus rearranges into products that are more tightly bound than the original, that is products whose total binding energy is greater. The gain in binding energy appears as the kinetic energy of the products. This single idea underlies both the splitting of very heavy nuclei and the joining of very light nuclei, since in each case the products turn out to be more tightly bound than the starting nuclei, so energy is released. 🔉⇢
The binding energy of the whole nucleus grows as the number of nucleons grows, since there are more nucleons to contribute to the binding, and a heavy nucleus therefore has a larger total binding energy than a light one. But the total binding energy alone does not tell how firmly each nucleon is held, because the larger nucleus has both more binding energy and more nucleons. To compare how tightly individual nucleons are bound in different nuclei, one divides the binding energy by the number of nucleons, a quantity treated separately. 🔉⇢
Every binding-energy calculation follows the same clear steps. First, add the masses of the $Z$ free protons and the $A - Z$ free neutrons. Second, subtract the measured mass of the nucleus to obtain the mass defect $\Delta M$. Third, multiply the mass defect by $c^2$, using the equivalence $1\ u = 931.5$ million electron volt, to obtain the binding energy $E_b = \Delta M c^2$. This procedure turns accurate measurements of mass into precise values for the energy that binds the nucleus and the energy needed to take it apart. 🔉⇢
The binding energy is a measure of the stability of the nucleus, since a nucleus with a larger binding energy is harder to break apart. A nucleus is stable because its bound state has less energy, and less mass, than the separated nucleons; energy would have to be supplied from outside to raise it to the separated state. This is the same reason that any bound system is stable, and for the nucleus the binding is very strong, so the energy needed to separate the nucleons is very large compared with the energies met in chemistry. 🔉⇢
Because the mass defect is a small difference between two larger masses, it must be found from accurate measured masses. The total mass of the free nucleons and the mass of the nucleus differ only in the second or third decimal place when expressed in atomic mass units, so the masses must be measured to high accuracy for the mass defect, and hence the binding energy, to be reliable. Modern measurements with the mass spectrometer are accurate enough that binding energies obtained in this way are known precisely for a great many nuclei. 🔉⇢
The mass defect and the binding energy are thus two ways of describing the same fact, that the nucleus is a bound system with less energy than its separated parts. The mass defect states the fact in terms of a missing mass, while the binding energy states it in terms of the energy released on formation or required for separation. The relation between the two is simply $E_b = \Delta M c^2$, and this single relation, together with accurate measured masses, gives the binding energy of any nucleus and measures how tightly it is bound. 🔉⇢
The total binding energy of a nucleus grows as the nucleus becomes larger, simply because a larger nucleus contains more nucleons that contribute to the binding. This total by itself, however, does not show how tightly each nucleon is held. To compare different nuclei fairly, the binding energy $E_b$ is divided by the number of nucleons $A$, giving the binding energy per nucleon, $E_{bn} = E_b / A$. This quantity measures, on the average, how strongly a single nucleon is bound in the nucleus, and it is the most useful measure of nuclear stability. 🔉⇢
Full derivation, worked example and interactive 3D on the Binding Energy per Nucleon (BE/A) tab →
The nuclear force is not Coulomb. Push two nucleons closer than ~0.8 fm and it shoves back hard (U>0); pull them a little apart and it grips (deep U<0 well); go beyond a few fm and it vanishes. Short-range, saturating, charge-independent.
Drag the controls and watch the value under each symbol change; the formula bar under the figure substitutes the numbers as you go.
The nucleus is a bound system of protons and neutrons packed closely together in a very small volume. The protons all carry positive charge and therefore repel one another through the electric, or Coulomb, force, which would tend to blow the nucleus apart. Since nuclei nevertheless exist and are stable, there must be another force, stronger than the electric repulsion at short distances, that holds the nucleons together. This is the nuclear force, a strong attractive force acting between nucleons that is responsible for the binding of the nucleus. 🔉⇢
The nuclear force is the strongest of the forces known to act between nucleons at the small distances found inside the nucleus. At these distances it is very much stronger than the electric force between the protons, so it can overcome their repulsion and bind the nucleus together. It is far stronger still than the gravitational force between the nucleons, which is so weak that it plays no part at all in holding the nucleus together. Within its range the nuclear force dominates completely over both the electric and the gravitational forces. 🔉⇢
The most striking property of the nuclear force is its very short range. Unlike the electric and gravitational forces, which reach out to large distances and fall off only slowly, the nuclear force acts only over a distance of a few femtometres, comparable to the size of the nucleus itself. The nuclear force between two nucleons falls rapidly to zero as the distance between them increases beyond a few femtometres, so that two nucleons separated by more than this distance feel almost no nuclear force at all. 🔉⇢
Because the nuclear force has such a short range, a nucleon is attracted only by the nucleons immediately around it, its nearest neighbours, and not by the more distant nucleons in the nucleus. This is quite different from the electric force, through which every proton repels every other proton however far apart they are within the nucleus. The short range of the nuclear force means that as a nucleus grows larger, a nucleon deep inside it still interacts only with the same small number of neighbours, whatever the total number of nucleons. 🔉⇢
At very short distances, closer than about $0.8$ femtometre, the nuclear force becomes repulsive rather than attractive. This repulsive core prevents the nucleons from being pushed too close together and keeps them at a roughly fixed spacing from one another. The force is therefore attractive over most of its short range, drawing nucleons together, but strongly repulsive when they come too close, holding them apart. This combination of attraction at moderate separation and repulsion at very small separation fixes the distance between neighbouring nucleons. 🔉⇢
The behaviour of the nuclear force with distance can be described by its potential energy. When two nucleons are far apart, beyond a few femtometres, the potential energy is nearly zero, showing that there is almost no force. As they approach to within a few femtometres, the potential energy becomes negative, corresponding to attraction, and it reaches a minimum at a separation of about $0.8$ femtometre. If the nucleons are pushed still closer, the potential energy rises steeply and becomes positive, corresponding to the strong repulsion of the repulsive core. 🔉⇢
A further important property of the nuclear force is that it is nearly the same between any two nucleons, whatever their charge. The force between two protons, between two neutrons, and between a proton and a neutron is very nearly equal, provided the nucleons are in the same state of motion. This property is called charge independence, and it means that the nuclear force does not distinguish between protons and neutrons. As far as the nuclear force is concerned, the proton and the neutron behave alike and can be regarded as two forms of the same particle, the nucleon. 🔉⇢
The charge independence of the nuclear force must not be confused with the electric force, which of course does distinguish between protons and neutrons because only the proton is charged. The two protons in a nucleus feel the nuclear attraction like any other pair of nucleons, but they also feel the electric repulsion because they are charged, while a proton and a neutron feel only the nuclear force. The total force between two nucleons is the sum of the charge-independent nuclear force and, for charged nucleons, the electric force. 🔉⇢
The short range of the nuclear force, together with its charge independence, explains the property of saturation. Because each nucleon interacts only with its nearest neighbours, and because a nucleon can be surrounded by only a limited number of neighbours, each nucleon is bound by the same small number of others whatever the size of the nucleus. The binding contributed by each nucleon therefore does not grow with the number of nucleons, and the binding energy per nucleon stays nearly constant. This saturation is a direct consequence of the short range of the nuclear force. 🔉⇢
Saturation also accounts for the constant density of nuclear matter. Since the repulsive core keeps neighbouring nucleons at a fixed spacing and the attractive part binds only nearest neighbours, adding more nucleons simply enlarges the nucleus while keeping the spacing, and hence the density, the same. The same short range of the nuclear force therefore produces both the constant density of nuclei and the nearly constant binding energy per nucleon in the middle range of mass numbers, tying together several nuclear properties in one picture. 🔉⇢
The contrast between the nuclear force and the electric force is important for understanding heavy nuclei. The nuclear force acts only between nearest neighbours, so the total nuclear attraction grows roughly in proportion to the number of nucleons. The electric repulsion, however, acts between every pair of protons however far apart, so it grows faster as the number of protons increases. In a very heavy nucleus with many protons the electric repulsion becomes relatively more important, which is why heavy nuclei are less tightly bound and why there is a limit to how large a stable nucleus can be. 🔉⇢
The nuclear force cannot be described by a simple formula like the electric or gravitational forces, which vary in a simple way with distance. It is a more complicated force, attractive at moderate short range, repulsive at very short range, and vanishing beyond a few femtometres, and it depends on the state of motion of the nucleons as well as their separation. Despite this complexity, its main features, short range, great strength, charge independence, and saturation, are well established from the study of nuclei and of the scattering of nucleons. 🔉⇢
These features of the nuclear force together explain why the nucleus exists as a stable, dense, bound system. The great strength of the force at short range overcomes the electric repulsion of the protons and binds the nucleons; its short range and repulsive core fix the spacing of the nucleons and hence the constant density; its charge independence lets protons and neutrons be bound alike; and its saturation gives the nearly constant binding energy per nucleon. Without a force having just these properties, nuclei as we observe them could not exist. 🔉⇢
In summary, the nuclear force is a strong, short-range, attractive force between nucleons, much stronger than the electric force at distances of a few femtometres but falling rapidly to zero beyond that range. It has a repulsive core at separations closer than about $0.8$ femtometre, it is charge independent, being nearly the same between any pair of nucleons, and it saturates, each nucleon interacting only with its nearest neighbours. These properties account for the binding of the nucleus, its constant density, and the nearly constant binding energy per nucleon. 🔉⇢
The existence of the nuclear force is required by the very existence of nuclei. The protons in a nucleus are packed within a few femtometres of one another and repel one another strongly through the electric force, yet the nucleus holds together. Something must overcome this electric repulsion, and that something is the nuclear force, an attraction between nucleons that is stronger than the electric repulsion at these short distances. Without such a force the protons would fly apart and no nucleus could exist. 🔉⇢
The short range of the nuclear force is what most sharply distinguishes it from the familiar electric and gravitational forces. Those forces reach out to great distances and fall off only slowly, so that a charge or a mass influences others far away. The nuclear force, by contrast, is felt only over a distance of a few femtometres and vanishes beyond, so a nucleon feels the nuclear force only from the nucleons in its immediate neighbourhood. This short range shapes almost every property of the nucleus. 🔉⇢
The repulsive core of the nuclear force, at separations closer than about $0.8$ femtometre, keeps the nucleons from being squeezed together without limit. Although the force is strongly attractive at moderate short range, it turns into a strong repulsion when the nucleons come very close, so there is a preferred separation at which the force changes from attraction to repulsion. This preferred separation fixes the spacing of the nucleons and, together with the short range of the attraction, produces the constant density of nuclear matter. 🔉⇢
The charge independence of the nuclear force is one of its most important properties. Experiments on the binding of nuclei and on the scattering of nucleons show that the force between two protons, between two neutrons, and between a proton and a neutron is very nearly the same, once the electric force between the charged protons is set aside. The nuclear force thus treats protons and neutrons alike, which is why the two are regarded as two states of a single particle, the nucleon, as far as the strong nuclear interaction is concerned. 🔉⇢
The saturation of the nuclear force follows from its short range. Because each nucleon attracts only its nearest neighbours, and because only a limited number of neighbours can be packed around it, each nucleon is bound by the same small number of others no matter how large the nucleus is. The binding contributed by each nucleon therefore does not grow with the size of the nucleus, and the binding energy per nucleon stays nearly constant over the middle range of mass numbers. Saturation is thus a direct consequence of the short range. 🔉⇢
The comparison between the growth of the nuclear attraction and the electric repulsion explains why very heavy nuclei become unstable. The total nuclear attraction grows roughly in proportion to the number of nucleons, because each nucleon binds only its neighbours, whereas the electric repulsion grows faster, because every proton repels every other proton however far apart. In a nucleus with very many protons the electric repulsion eventually becomes large enough to make the nucleus unstable, which sets a limit on how large a stable nucleus can be. 🔉⇢
The nuclear force does not follow a simple law with distance in the way the electric and gravitational forces do. It is attractive at moderate short range, strongly repulsive at very short range, and zero beyond a few femtometres, and it depends on more than just the separation of the nucleons. This complicated behaviour cannot be captured in a simple formula, but its main features, great strength, short range, a repulsive core, charge independence, and saturation, are firmly established from the study of nuclei and of the scattering of nucleons. 🔉⇢
Taken together, these properties of the nuclear force account for the nucleus being a small, dense, strongly bound system of protons and neutrons. The great strength of the force at short range binds the nucleons against the electric repulsion; the repulsive core and short range fix the spacing of the nucleons and give the constant density; the charge independence lets protons and neutrons be bound in the same way; and the saturation gives the nearly constant binding energy per nucleon over the middle range of mass numbers. 🔉⇢
The great strength of the nuclear force can be appreciated by comparing it with the electric force at the same short distance. Two protons a few femtometres apart repel one another electrically with a large force, yet they remain bound inside the nucleus, which shows that the nuclear attraction between them is even larger at that distance. At distances of the size of an atom, however, the nuclear force has fallen to zero while the electric force still acts, so outside the nucleus only the electric force between the charges remains and the nuclear force plays no part. 🔉⇢
The gravitational force between nucleons, though always attractive, is far too weak to bind the nucleus. The masses of the nucleons are so small that the gravitational attraction between two nucleons a few femtometres apart is smaller than the nuclear force by an enormous factor, and it is also far weaker than the electric force between two protons. Gravity therefore plays no part in holding the nucleus together, and the binding is due entirely to the strong, short-range nuclear force acting between neighbouring nucleons. 🔉⇢
The description of the nuclear force in terms of its potential energy brings its features together. The potential energy is nearly zero when the nucleons are more than a few femtometres apart, becomes negative and reaches a minimum at a separation of about $0.8$ femtometre where the attraction is balanced, and rises steeply to positive values when the nucleons come closer, reflecting the repulsive core. This shape of the potential energy with distance captures the short range, the attraction at moderate separation, and the repulsion at very short separation that together define the nuclear force. 🔉⇢
Every decay slides the nucleus across the N–Z chart in a fixed way: α drops both Z and N by 2, β⁻ turns a neutron into a proton (Z+1, N−1), β⁺ does the reverse, and γ only sheds energy — same Z, same A.
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Some nuclei are unstable and do not remain unchanged for ever. Such a nucleus spontaneously emits radiation and transforms into a different nucleus, releasing energy in the process. This is referred to as radioactive decay, and a substance whose nuclei undergo it is said to be radioactive. Radioactivity was discovered when certain substances were found to emit a penetrating radiation without any external cause, and this radiation was traced to changes taking place in the nuclei of the atoms of the substance. 🔉⇢
Three types of radioactive decay occur in nature. In alpha decay the nucleus emits an alpha particle, which is a helium nucleus. In beta decay the nucleus emits a beta particle, which is an electron or its antiparticle the positron, together with a neutral particle. In gamma decay the nucleus emits a gamma ray, which is a high-energy photon of electromagnetic radiation. Each type of decay changes the nucleus in a definite way, and all three obey the conservation of charge and of the number of nucleons. 🔉⇢
In alpha decay the unstable nucleus emits an alpha particle, which is the nucleus of the helium atom and contains two protons and two neutrons. Because two protons and two neutrons are carried away, the atomic number of the nucleus decreases by two and its mass number decreases by four. The original nucleus, called the parent, therefore transforms into a different nucleus, called the daughter, of a different element. Alpha decay is common among the heavy nuclei, which reduce their size by emitting alpha particles. 🔉⇢
The alpha particle emitted in alpha decay carries away a definite amount of kinetic energy, which comes from the energy released in the decay. The energy released is the difference between the rest energy of the parent nucleus and the total rest energy of the daughter nucleus and the alpha particle, in accordance with the equivalence of mass and energy. Because the daughter and the alpha particle share this energy, and the daughter is much heavier, the alpha particle carries away most of the kinetic energy while the daughter recoils with a small energy. 🔉⇢
In beta decay the nucleus emits a beta particle, and there are two kinds. In one kind the nucleus emits an electron, and a neutron in the nucleus changes into a proton, so the atomic number increases by one while the mass number stays the same. In the other kind the nucleus emits a positron, the antiparticle of the electron, and a proton changes into a neutron, so the atomic number decreases by one while the mass number stays the same. In both kinds the number of nucleons is unchanged, only their character changing between proton and neutron. 🔉⇢
A careful study of beta decay showed that the emitted electron or positron does not carry away a single definite energy, but is emitted with a range of energies up to a maximum. This was puzzling, because the energy released in the decay is definite, and it appeared that energy was not conserved. To explain this, it was proposed that a further neutral particle, of very small or zero mass, is emitted together with the beta particle and shares the released energy with it. This particle is the neutrino, and its antiparticle is the antineutrino. 🔉⇢
The neutrino and the antineutrino are neutral particles that carry away part of the energy released in beta decay, so that the beta particle and the neutrino together carry a definite total energy while each may have any share of it. In the beta decay that emits an electron, an antineutrino is emitted with it; in the beta decay that emits a positron, a neutrino is emitted with it. The neutrino interacts very weakly with matter and is difficult to detect, but its existence restores the conservation of energy in beta decay. 🔉⇢
In gamma decay the nucleus emits a gamma ray, a photon of electromagnetic radiation of very high energy and very short wavelength. Gamma decay does not change the number of protons or the number of neutrons, so the atomic number and the mass number both stay the same and the nucleus remains the same element. What changes is the energy of the nucleus: a nucleus in a state of higher energy, called an excited state, drops to a state of lower energy and emits the difference in energy as a gamma ray, much as an atom emits light. 🔉⇢
Gamma decay often follows alpha or beta decay. When a nucleus decays by emitting an alpha or a beta particle, the daughter nucleus is frequently left in an excited state of higher energy rather than in its lowest state. The excited daughter then drops to its lowest state and emits the excess energy as one or more gamma rays. This is why alpha and beta decay are often accompanied by the emission of gamma rays, the gamma rays carrying away the energy by which the daughter nucleus was left excited. 🔉⇢
All three types of radioactive decay obey strict conservation laws. The total charge is conserved, so the atomic numbers balance when the charge of any emitted particle is included, and the total number of nucleons is conserved, so the mass numbers balance. Energy is conserved when the energy equivalent of the masses and the kinetic energies of all the products, including any neutrino, are counted. Momentum is also conserved, which is why the daughter nucleus recoils when a particle is emitted. These conservation laws govern every radioactive decay. 🔉⇢
The energy released in a radioactive decay comes from the difference in mass between the parent nucleus and the products. Because the parent is heavier than the total mass of the products, the difference in mass, multiplied by $c^2$, appears as the kinetic energy of the products and the energy of any emitted photon. A nucleus can decay in a particular way only if the parent is heavier than the products of that decay, so that energy is released; if the products would be heavier, the decay cannot occur spontaneously. 🔉⇢
Radioactive decay is a spontaneous process that takes place without any external cause and cannot be hastened or slowed by ordinary means such as heating or chemical change. Whether a given nucleus decays at a particular moment cannot be predicted, but for a large number of identical nuclei the fraction that decays in a given time is definite. This leads to the law of radioactive decay, which describes how the number of undecayed nuclei falls with time, and which is treated separately. 🔉⇢
The different types of radiation from radioactive substances differ greatly in their ability to penetrate matter. The alpha particles, being heavy and carrying charge, are stopped by a thin sheet of paper. The beta particles, being much lighter, penetrate further and are stopped by a thin sheet of metal. The gamma rays, being uncharged electromagnetic radiation, are the most penetrating and require a thick layer of dense material to stop them. These differences in penetration reflect the different nature of the three kinds of radiation. 🔉⇢
In summary, radioactivity is the spontaneous transformation of an unstable nucleus into another nucleus with the emission of radiation, and there are three types of radioactive decay. Alpha decay emits a helium nucleus, decreasing the atomic number by two and the mass number by four; beta decay emits an electron or a positron together with a neutrino or antineutrino, changing the atomic number by one at constant mass number; and gamma decay emits a high-energy photon from an excited nucleus without changing its atomic number or mass number. All three conserve charge, nucleon number, energy and momentum. 🔉⇢
Radioactivity is a nuclear phenomenon: the radiation comes from changes in the nucleus of the atom, not from the electrons outside. This is shown by the fact that the decay is unaffected by the chemical state of the atom or by ordinary changes of temperature and pressure, which affect only the outer electrons. The emission of alpha, beta and gamma radiation reflects transformations of the nucleus itself, in which the nucleus changes into a different nucleus or drops to a lower energy state. 🔉⇢
In alpha decay the parent nucleus emits a helium nucleus, made of two protons and two neutrons, so the daughter nucleus has two fewer protons and two fewer neutrons than the parent. The atomic number decreases by two and the mass number decreases by four, and the daughter is a nucleus of a different element lower in the periodic table. Alpha decay is common among heavy nuclei, which reduce both their charge and their mass by emitting an alpha particle and so move toward greater stability. 🔉⇢
In beta decay a nucleon changes its character inside the nucleus. In the decay that emits an electron, a neutron changes into a proton, so the atomic number increases by one while the mass number is unchanged, and an electron and an antineutrino are emitted. In the decay that emits a positron, a proton changes into a neutron, so the atomic number decreases by one while the mass number is unchanged, and a positron and a neutrino are emitted. In both, the total number of nucleons stays the same. 🔉⇢
The continuous range of energies of the emitted electrons in beta decay was a puzzle, because the energy released in the decay is definite. If the electron were the only particle emitted along with the daughter nucleus, it would always carry away the same energy. The observed spread of energies showed that another particle must be emitted to share the energy. This particle, the neutrino or antineutrino, is neutral and of very small mass, and it carries away the balance of the energy so that energy is conserved in each decay. 🔉⇢
In gamma decay the nucleus emits a photon of very high energy but keeps the same number of protons and neutrons, so the atomic number and mass number are unchanged. A nucleus in an excited state of higher energy drops to a state of lower energy and emits the difference in energy as a gamma ray. Gamma emission commonly follows alpha or beta decay, because the daughter nucleus is often left in an excited state and then reaches its lowest state by emitting one or more gamma rays. 🔉⇢
Every radioactive decay conserves electric charge and the total number of nucleons, so that the atomic numbers and the mass numbers balance on the two sides of the decay once the emitted particles are included. Energy and momentum are also conserved, the energy released appearing as the kinetic energy of the products and the energy of any emitted photon, and the daughter nucleus recoiling to conserve momentum. These conservation laws hold in every decay and are used to work out the products and the energies involved. 🔉⇢
The energy released in a decay comes from the difference in mass between the parent nucleus and the products, in accordance with the equivalence of mass and energy. Because the parent is heavier than the total mass of the products, the difference in mass, multiplied by $c^2$, is released as energy. A nucleus can decay in a given way only if this difference is positive, that is only if the parent is heavier than the products; if the products would be heavier, the decay cannot happen spontaneously and the nucleus is stable against that mode of decay. 🔉⇢
The three kinds of radiation differ in how deeply they penetrate matter, because they differ in nature. Alpha particles are heavy and charged and lose their energy quickly, so they are stopped by a thin sheet of paper. Beta particles are much lighter and penetrate further, being stopped by a thin sheet of metal. Gamma rays are uncharged electromagnetic radiation and are the most penetrating, requiring a thick layer of dense material to absorb them. These differences are used to identify the type of radiation coming from a source. 🔉⇢
The transformation of one element into another in radioactive decay was one of the most striking discoveries about the nucleus. In alpha and beta decay the parent nucleus becomes a nucleus of a different element, since the number of protons changes, so a radioactive element gradually turns into other elements as its nuclei decay. A heavy radioactive nucleus may decay into a daughter that is itself radioactive, which decays in turn, so that a whole succession of decays takes place until a stable nucleus is finally reached. 🔉⇢
The direction of a beta decay depends on whether the parent nucleus has too many neutrons or too many protons for stability. A nucleus with too many neutrons tends to decay by emitting an electron, turning a neutron into a proton and so lowering the neutron number while raising the proton number. A nucleus with too many protons tends to decay by emitting a positron, turning a proton into a neutron. In each case the beta decay moves the nucleus toward a more balanced and more stable combination of protons and neutrons at the same mass number. 🔉⇢
The energy released in a decay is shared among the products according to the conservation of energy and momentum. In alpha decay the light alpha particle carries away most of the kinetic energy while the heavy daughter nucleus recoils slowly, so the alpha particle is emitted with a definite energy. In beta decay the energy is shared between the beta particle and the neutrino, so the beta particle is emitted with a range of energies. In gamma decay the emitted photon carries away the energy by which the nucleus was excited, leaving the nucleus in its lower state. 🔉⇢
Radioactivity is a nuclear phenomenon in which an unstable nucleus undergoes a decay, transforming spontaneously into another nucleus with the emission of radiation. For a single nucleus it is not possible to say when it will decay, since the decay is a spontaneous process governed by chance. For a large number of identical nuclei, however, a definite law holds: the number of nuclei that decay in a given short time is proportional to the number of undecayed nuclei present at that time. This is the law of radioactive decay. 🔉⇢
Full derivation, worked example and interactive 3D on the The Law of Radioactive Decay tab →
Decay is exponential, not linear. Each half-life T½ the population halves; the mean life τ = T½/0.693 is a bit longer. Shrink T½ and the whole curve plunges faster.
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The rate at which a radioactive substance decays is described by its decay constant, but two other measures, the half-life and the mean life, are often more convenient because they are times that can be pictured directly. The half-life is the time in which half of the undecayed nuclei present decay, so that the number remaining falls to one-half of its value. The mean life is the average time for which a nucleus survives before it decays. Both are fixed for a particular kind of nucleus and are simply related to the decay constant. 🔉⇢
The half-life follows directly from the exponential decay law $N = N_0 e^{-\lambda t}$. After a time equal to one half-life the number of undecayed nuclei has fallen to one-half of the number present at the start. This condition fixes the half-life in terms of the decay constant, giving $T_{1/2} = 0.693/\lambda$, where the number $0.693$ is the natural logarithm of two. A nucleus with a large decay constant has a short half-life and decays quickly, while a nucleus with a small decay constant has a long half-life and decays slowly. 🔉⇢
The meaning of the half-life is seen clearly by following the number of undecayed nuclei through successive half-lives. After one half-life, one-half of the original nuclei remain. After two half-lives, one-half of these remain, that is one-quarter of the original. After three half-lives, one-eighth remain, and so on, the number being halved with each further half-life. This repeated halving is the same exponential fall described by the decay law, now expressed in equal steps each lasting one half-life. 🔉⇢
Because the number is halved in each half-life, the number of undecayed nuclei falls quickly at first and then more and more slowly, but it never quite reaches zero. After ten half-lives only about one part in a thousand of the original nuclei remain, and after twenty half-lives only about one part in a million. The half-life therefore gives a convenient measure of how long a radioactive substance lasts: after a number of half-lives the substance is almost entirely decayed, though a very small fraction always remains. 🔉⇢
The mean life is the average time for which a nucleus exists before it decays, taken over all the nuclei in the sample. Some nuclei decay very soon and others survive much longer, and the mean life is the average of all these survival times. Working out this average from the exponential decay law gives the simple result that the mean life is the reciprocal of the decay constant, $\tau = 1/\lambda$. The mean life, like the half-life, is fixed for a particular kind of nucleus and is determined entirely by the decay constant. 🔉⇢
The half-life and the mean life are closely related, since both are determined by the same decay constant. Combining $T_{1/2} = 0.693/\lambda$ with $\tau = 1/\lambda$ gives the relation $T_{1/2} = 0.693\,\tau$, so the half-life is about seven-tenths of the mean life. The mean life is therefore always somewhat longer than the half-life. Knowing any one of the three quantities, the decay constant, the half-life, or the mean life, at once determines the other two, so they are three ways of expressing the same information about the rate of decay. 🔉⇢
The mean life can be understood as the time over which the number of undecayed nuclei would fall to a certain fraction of its value if the decay continued at its initial rate. In one mean life the number of undecayed nuclei falls to about one over the base of natural logarithms, that is to about thirty-seven per cent of the initial number. This is a larger fraction than one-half, which is why the mean life is longer than the half-life, and it gives another way of picturing the mean life directly in terms of the decay. 🔉⇢
The half-lives of radioactive nuclei vary over an enormous range. Some nuclei have half-lives of a tiny fraction of a second, decaying almost as soon as they are formed, while others have half-lives of thousands, millions, or even billions of years, decaying so slowly that they persist over the age of the earth. Whatever the half-life, the same exponential decay law describes the fall in the number of undecayed nuclei, and the half-life simply sets the time scale over which that fall takes place for a particular kind of nucleus. 🔉⇢
A striking example of a decay time is provided by the free neutron. A neutron outside a nucleus is itself unstable and decays into a proton, an electron and an antineutrino. The free neutron has a mean life of about 1000s. It is, however, stable inside the nucleus, where the presence of the other nucleons and the binding of the nucleus make the neutron stable. Thus a neutron that would decay in about a thousand seconds when free can remain unchanged indefinitely when it is bound in a stable nucleus. 🔉⇢
The contrast between the free neutron and the bound neutron shows that stability depends on the whole nucleus, not on the individual nucleon alone. A neutron bound in a stable nucleus does not decay, because the decay would lead to a nucleus of higher energy, which is not allowed. Only when the decay leads to a nucleus of lower total energy can it occur. This is why some nuclei are stable while others are radioactive, and why the same neutron can be unstable when free but stable when bound. 🔉⇢
The half-life is used in practice to describe how long a radioactive source remains useful or hazardous. A source with a short half-life loses its activity quickly, becoming much weaker after a few half-lives, while a source with a long half-life remains active over a very long time. In estimating how long a radioactive material must be stored before its activity falls to a safe level, or how long a source will remain strong enough to be useful, the half-life is the natural measure to use. 🔉⇢
The half-life also underlies the use of radioactive decay to measure the ages of old objects. Because the number of undecayed nuclei falls by one-half in each half-life, the fraction of the original nuclei remaining tells how many half-lives have passed, and hence the age of the object. A radioactive nucleus with a suitable half-life therefore serves as a clock, and by measuring how much of it remains, compared with the amount originally present, the time that has passed can be estimated over periods ranging up to billions of years. 🔉⇢
It should be remembered that the half-life and the mean life are statistical quantities, describing the behaviour of a large number of nuclei. For a single nucleus it is impossible to say when it will decay; only the chance of decay per unit time is known. But for a large number of nuclei, half will have decayed after one half-life and the average survival time will be the mean life, and these statements become more and more exact the larger the number of nuclei present. The half-life and mean life thus describe the sample as a whole. 🔉⇢
In summary, the half-life $T_{1/2}$ is the time for half of the undecayed nuclei to decay and equals $0.693/\lambda$, while the mean life $\tau$ is the average survival time of a nucleus and equals $1/\lambda$. The two are related by $T_{1/2} = 0.693\,\tau$, so the mean life is somewhat longer than the half-life. The number of undecayed nuclei is halved in each half-life, and both quantities, determined by the decay constant, describe the statistical behaviour of a large number of nuclei and vary enormously from one kind of nucleus to another. 🔉⇢
The half-life and the mean life are two ways of stating the same information about the rate of decay, and both follow from the exponential decay law. The half-life answers the question of how long it takes for half of the undecayed nuclei present to decay, while the mean life answers the question of how long a nucleus survives on the average. Because both are fixed by the single decay constant, they always stand in the same ratio to one another, the mean life being longer than the half-life. 🔉⇢
The result $T_{1/2} = 0.693/\lambda$ shows that the half-life is inversely related to the decay constant, so a nucleus that decays rapidly, with a large decay constant, has a short half-life, and a nucleus that decays slowly, with a small decay constant, has a long half-life. This inverse relation is the reason that measuring the half-life is equivalent to measuring the decay constant, since one determines the other at once, and either can be used to describe how fast the substance decays. 🔉⇢
The mean life $\tau = 1/\lambda$ has a direct meaning in terms of the exponential fall of the number of undecayed nuclei. In one mean life the number of undecayed nuclei falls to a definite fraction, about thirty-seven per cent, of its value, which is a larger fraction than the one-half remaining after one half-life. This is why the mean life is longer than the half-life, and it gives a second natural time associated with the decay, alongside the half-life, both fixed by the decay constant. 🔉⇢
The behaviour of the number of undecayed nuclei through successive half-lives shows how quickly a radioactive substance is used up. Half remains after one half-life, one-quarter after two, one-eighth after three, and so on, so that after ten half-lives only about one part in a thousand remains. A substance therefore becomes almost entirely decayed after a number of half-lives, though a very small fraction always survives, since the number is only halved, never brought to zero, in each further half-life. 🔉⇢
The enormous range of half-lives found among radioactive nuclei is remarkable. Some nuclei decay in a tiny fraction of a second, having a very large decay constant and a very short half-life, while others decay only over millions or billions of years, having a very small decay constant and a very long half-life. The same exponential decay law governs all of them, so the only difference is the value of the decay constant and hence the time scale of the decay, which varies over an enormous range from one kind of nucleus to another. 🔉⇢
The example of the free neutron shows clearly that stability is a property of the whole system, not of an isolated nucleon. A neutron by itself is unstable and decays with a mean life of about a thousand seconds, yet the same neutron is stable when bound in a stable nucleus, because there the decay would lead to a nucleus of higher energy and so cannot occur. Whether a nucleus is stable or radioactive depends on whether a decay would lead to products of lower total energy, which in turn depends on the masses involved. 🔉⇢
The half-life is the natural measure to use when deciding how long a radioactive material remains active. A source with a short half-life is strong at first but loses its activity quickly, becoming much weaker after only a few half-lives, while a source with a long half-life remains active over a long period. In deciding how long a radioactive material must be kept before its activity has fallen to a low level, the half-life gives the time scale, since the activity is halved with each half-life that passes. 🔉⇢
The use of radioactive decay as a clock rests entirely on the fixed half-life of the nucleus concerned. Because the number of undecayed nuclei falls by one-half in each half-life, the fraction remaining tells how many half-lives have passed, and hence the time. A nucleus with a half-life suited to the period to be measured serves as a reliable clock, since the decay is unaffected by the external conditions the sample has experienced, and by measuring the fraction remaining the age of the sample can be estimated. 🔉⇢
The half-life and the mean life are statistical quantities that describe a large number of nuclei rather than a single one. For one nucleus only the chance of decay per unit time is known, and it is impossible to say when that nucleus will decay. But for a large number of nuclei half will have decayed after one half-life, and the average survival time will be the mean life, and these statements become more exact the larger the number of nuclei present. The half-life and the mean life thus describe the behaviour of the sample as a whole. 🔉⇢
The relation between the half-life and the mean life can be seen numerically. Since $T_{1/2} = 0.693/\lambda$ and $\tau = 1/\lambda$, the half-life is about seven-tenths of the mean life, so for a nucleus with a mean life of a thousand seconds the half-life is about six hundred and ninety-three seconds. Whichever of the two is measured, the other follows at once, and both give the same information about the rate of decay expressed in terms of a time rather than the decay constant. 🔉⇢
The fact that a bound neutron can be stable while a free neutron decays shows that the stability of a nucleus depends on the total energy of the whole system. A decay takes place only if it leads to products of lower total energy, and inside a stable nucleus the decay of a neutron would lead to a nucleus of higher energy, which is not allowed. This is why the same neutron is unstable when free, with a mean life of about a thousand seconds, but stable when bound in a stable nucleus. 🔉⇢
The half-life and the mean life are measures of the phenomenon of radioactivity, the nuclear phenomenon in which an unstable nucleus undergoes disintegration into another nucleus with the emission of radiation. The instability of the nucleus is expressed by its decay constant, and the half-life and mean life follow from it. Each radioactive nuclide has its own decay constant, and hence its own half-life and mean life, so these quantities are properties of the particular nuclide and are determined by the nature of the nucleus that undergoes the disintegration. 🔉⇢
The measurement of a half-life is in effect a measurement of the decay constant, since the two are related by $T_{1/2} = 0.693/\lambda$. By observing the radioactive substance and determining how the number of undecayed nuclei, or the radiation emitted, falls with time, the decay constant is found, and the half-life follows. For a nuclide whose nuclei transform slowly the half-life is long and the substance persists, while for a nuclide whose nuclei transform quickly the half-life is short and the substance is soon depleted of its undecayed nuclei. 🔉⇢
Activity is decays per second, R = λN — so it falls off with exactly the same exponential as the population. A short half-life (large λ) means a fiercely active but short-lived source.
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When a radioactive sample is studied, what is usually measured is not the number of undecayed nuclei directly but the rate at which the nuclei decay, since each decay produces a particle or a photon that can be detected. The number of nuclei that decay per unit time is called the activity of the sample. The activity is a measure of how strongly the sample is decaying, and it is the quantity that a detector responds to when it counts the particles emitted by the sample. 🔉⇢
The activity follows directly from the law of radioactive decay. Because the number of nuclei decaying per unit time is proportional to the number of undecayed nuclei present, with the decay constant as the constant of proportionality, the activity equals the decay constant multiplied by the number of undecayed nuclei. Writing the activity as $R$, the number of undecayed nuclei as $N$, and the decay constant as $\lambda$, this relation is $R = \lambda N$. The activity is thus large when many undecayed nuclei are present and small when few remain. 🔉⇢
Since the number of undecayed nuclei falls exponentially with time according to $N = N_0 e^{-\lambda t}$, the activity falls in exactly the same exponential way. Multiplying the decay law by the decay constant gives the activity at time $t$ as $R = \lambda N_0 e^{-\lambda t}$, which can be written $R = R_0 e^{-\lambda t}$, where $R_0 = \lambda N_0$ is the activity at the starting time. The activity of a sample therefore decreases exponentially, halving in each half-life just as the number of undecayed nuclei does. 🔉⇢
The activity is measured in a unit called the becquerel, defined so that one becquerel is one decay per second. A sample has an activity of a certain number of becquerel if that many of its nuclei decay each second. Because a real sample contains an enormous number of nuclei, its activity in becquerel is usually a very large number, so activities are often quoted in multiples of the becquerel. The becquerel is the unit of activity in the same way that other units measure other physical quantities. 🔉⇢
An older unit of activity, still often used, is the curie. One curie is defined as an activity of $3.7 \times 10^{10}$ decays per second, that is $3.7 \times 10^{10}$ becquerel. The curie was originally chosen to be close to the activity of a certain amount of a natural radioactive substance, and because it corresponds to a large number of decays per second it is a convenient unit for strong sources. The activity of a sample can be expressed equally in becquerel or in curie, the two being related by this fixed factor. 🔉⇢
The relation $R = \lambda N$ shows that the activity depends both on the number of undecayed nuclei present and on the decay constant of the particular kind of nucleus. Two samples containing the same number of undecayed nuclei will have different activities if their nuclei have different decay constants, the sample with the larger decay constant being more active. A nucleus with a short half-life has a large decay constant and therefore gives a high activity for a given number of nuclei, while a nucleus with a long half-life gives a low activity. 🔉⇢
Because the activity is proportional to the number of undecayed nuclei, a measurement of the activity gives a way of finding the number of undecayed nuclei present, provided the decay constant is known. Conversely, if the number of nuclei is known, a measurement of the activity gives the decay constant. In this way the easily measured activity is linked, through the relation $R = \lambda N$, to the number of undecayed nuclei and to the decay constant, which are harder to measure directly. 🔉⇢
The exponential fall of the activity with time is used to determine the decay constant and hence the half-life of a radioactive substance. By measuring the activity at different times and observing how it decreases, the decay constant can be found from the rate of the fall, since the activity halves in each half-life. This is one of the most common ways of measuring the half-life of a radioactive nucleus, especially for nuclei whose half-lives are neither too short nor too long to follow in this way. 🔉⇢
The activity of a sample decreases as time goes on because the undecayed nuclei are steadily used up. As nuclei decay, fewer undecayed nuclei remain, so fewer decays occur per unit time and the activity falls. After each half-life the number of undecayed nuclei, and therefore the activity, is halved. A sample that is strongly active when fresh becomes weaker and weaker as time passes, and after many half-lives its activity has fallen to a very small fraction of its original value. 🔉⇢
It is important to distinguish the activity from the energy of the radiation. The activity measures only how many nuclei decay per unit time, not how much energy each decay releases or how penetrating the radiation is. Two sources may have the same activity yet emit radiation of very different energies, or of different types such as alpha, beta or gamma. The activity is therefore a measure of the rate of decay alone, while the effect of a source depends also on the nature and energy of the radiation it emits. 🔉⇢
The activity of a sample, being the number of decays per unit time, is also the rate at which the number of undecayed nuclei decreases, so it can be written as the decrease in the number of undecayed nuclei per unit time. This is why the activity is equal to the decay constant multiplied by the number present: the decay constant gives the fraction of the undecayed nuclei that decay per unit time, and multiplying by the number present gives the number that decay per unit time, which is the activity. 🔉⇢
Since a sample loses activity as it decays, a radioactive source has a limited useful life. A source with a short half-life must be used soon after it is prepared, because its activity falls quickly, while a source with a long half-life remains active over a long period. In choosing a source for a particular use, the half-life, and hence how fast the activity falls, is an important consideration, as is the initial activity, which depends on the number of undecayed nuclei present at the start. 🔉⇢
In summary, the activity of a radioactive sample is the number of nuclei that decay in it per unit time and equals the decay constant multiplied by the number of undecayed nuclei present, $R = \lambda N$. It falls exponentially with time as $R = R_0 e^{-\lambda t}$, halving in each half-life. The activity is measured in becquerel, one becquerel being one decay per second, and the older unit the curie equals $3.7 \times 10^{10}$ becquerel. The activity measures only the rate of decay, not the energy or type of the radiation emitted. 🔉⇢
The activity is the quantity that is measured when a radioactive sample is studied, because each decay produces a particle or a photon that a detector can count. What the detector records is the number of decays per unit time, which is the activity, rather than the number of undecayed nuclei present. The relation $R = \lambda N$ then links this measured activity to the number of undecayed nuclei, so that from the activity the number of nuclei can be found if the decay constant is known. 🔉⇢
Because the activity is the decay constant multiplied by the number of undecayed nuclei, and the number of undecayed nuclei falls exponentially, the activity falls exponentially in the same way, halving in each half-life. A graph of the activity against time therefore has the same exponential shape as a graph of the number of undecayed nuclei against time, and the rate at which the activity falls gives the decay constant, and hence the half-life, of the substance being studied. 🔉⇢
The becquerel, defined as one decay per second, is a very small unit compared with the activity of a typical sample, because even a small sample contains an enormous number of nuclei. For this reason the activity of a real sample, measured in becquerel, is usually a very large number. The older unit the curie, equal to $3.7 \times 10^{10}$ becquerel, was chosen to be close to the activity of a certain amount of a natural radioactive substance and is a convenient unit for describing strong sources. 🔉⇢
The activity of a sample depends on two things: the number of undecayed nuclei present and the decay constant of the particular kind of nucleus. Two samples with the same number of undecayed nuclei but different decay constants have different activities, the one with the larger decay constant, and hence the shorter half-life, being the more active. A small amount of a nucleus with a short half-life can therefore be as active as a much larger amount of a nucleus with a long half-life. 🔉⇢
Since the activity falls as the undecayed nuclei are used up, a radioactive source becomes weaker as time goes on. After each half-life the number of undecayed nuclei, and therefore the activity, is halved, so a source that is strong when fresh becomes much weaker after a few half-lives. This is why a source with a short half-life must be used soon after it is prepared, while a source with a long half-life keeps its activity over a much longer period of time. 🔉⇢
It is important to keep the activity distinct from the energy and the type of the radiation. The activity measures only how many nuclei decay per unit time, not how much energy each decay releases nor whether the radiation is alpha, beta or gamma. Two sources of the same activity may emit radiation of quite different energies or of different types, so the effect of a source on its surroundings depends not only on its activity but also on the nature and energy of the radiation it emits. 🔉⇢
The measurement of the activity at different times is one of the most common ways of finding the half-life of a radioactive substance. By observing how the activity decreases, and using the fact that it halves in each half-life, the half-life can be determined directly. This works best for nuclei whose half-lives are neither so short that the activity falls before it can be measured nor so long that the fall is too slow to observe over a reasonable time. 🔉⇢
The relation $R = \lambda N$ can also be read as a statement about the fraction of nuclei that decay per unit time. The decay constant is the fraction of the undecayed nuclei that decay per unit time, so multiplying it by the number of undecayed nuclei present gives the number that decay per unit time, which is the activity. This makes clear why the activity is proportional to the number of undecayed nuclei and why it falls exponentially as those nuclei are used up. 🔉⇢
The activity of a sample and the number of undecayed nuclei present are linked so closely that a measurement of one gives the other, provided the decay constant is known. Since the activity is the decay constant multiplied by the number of undecayed nuclei, dividing the measured activity by the decay constant gives the number of undecayed nuclei present in the sample. In this way the activity, which is easy to measure by counting the particles emitted, gives access to the number of nuclei, which cannot be counted directly. 🔉⇢
Because the activity falls exponentially with time, halving in each half-life, the activity after a number of half-lives can be found by repeated halving. After one half-life the activity is one-half of its original value, after two half-lives one-quarter, after three one-eighth, and so on. After ten half-lives the activity has fallen to about one part in a thousand of its original value, so a source becomes very weak after a number of half-lives have passed, though its activity never falls exactly to zero. 🔉⇢
The activity of a sample determines how strong a source it is, but not how harmful or useful the radiation is, which depends also on the energy and type of the radiation. A source of high activity emitting weak, easily absorbed radiation may have less effect than a source of lower activity emitting penetrating radiation. For this reason the activity, the energy of the radiation, and its type must all be considered together when the effect of a radioactive source is being judged, the activity giving only the rate of decay. 🔉⇢
The activity is a direct measure of the phenomenon of radioactivity, since radioactivity is the disintegration of an unstable nucleus and the activity counts the disintegrations that occur per unit time. The instability of the nucleus of a radioactive nuclide is expressed by its decay constant, and the number of disintegrations per unit time in a sample is the activity. Each radioactive nuclide has its own decay constant, so a sample of one nuclide has a definite activity for a given number of undecayed nuclei, determined by how unstable that nucleus is. 🔉⇢
When a nucleus in the sample undergoes disintegration it is transformed into a nucleus of another nuclide, and it is this transformation that the detector registers as one count of the activity. The activity therefore measures the rate at which the nuclei of the sample are transformed by radioactivity into other nuclei. Because the instability of the nucleus is fixed by its decay constant, and the number of undecayed nuclei falls as the disintegrations proceed, the activity is largest when the sample is fresh and falls as more of its nuclei are transformed. 🔉⇢
The number of disintegrations per unit time observed for a radioactive substance is thus a measurement of both the number of undecayed nuclei present and the instability of the particular nuclide. A substance whose nuclei are very unstable has a large decay constant and shows a high activity even for a small number of nuclei, while a substance whose nuclei are only slightly unstable has a small decay constant and shows a low activity for the same number of nuclei, since fewer disintegrations occur per unit time. 🔉⇢
In the disintegration of the nuclei of a radioactive element the sample emits particles and photons: an alpha particle, or a beta particle together with an antineutrino, or gamma photons, and the nucleus is left as a nucleus of another element with a different number of protons and neutrons. The detector counts these particles and photons, so the activity of the sample is the number of such disintegrations per unit time, whatever nuclides the sample contains. The activity does not depend on which element the sample is, only on how many undecayed nuclei it contains and on the decay constant of that nuclide. 🔉⇢
A most important neutron-induced nuclear reaction is fission, in which a heavy nucleus breaks into two lighter nuclei of middle size. Fission is usually brought about by a neutron: when a heavy nucleus such as a uranium nucleus absorbs a neutron, it becomes unstable and splits into two fragments of comparable size, together with a few free neutrons. A large amount of energy is released in the process, far more than in any chemical reaction, which is why fission is so important as a source of energy. 🔉⇢
Full derivation, worked example and interactive 3D on the Nuclear Fission and Chain Reaction tab →
To fuse, two nuclei must climb their Coulomb barrier — hundreds of keV. Yet the Sun's core thermal energy is only a couple of keV. Classically impossible; it happens because quantum tunnelling lets nuclei slip through the barrier.
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Nuclear fusion is the process in which two light nuclei join together to form a heavier nucleus. When two light nuclei fuse to form a larger nucleus, energy is released, because the heavier nucleus that is formed is more tightly bound than the two light nuclei from which it is made. Fusion is therefore the opposite of fission: in fission a heavy nucleus splits into middle-sized fragments, while in fusion light nuclei join into a heavier one, and both release energy by moving toward more tightly bound nuclei. 🔉⇢
The reason fusion releases energy is again found in the binding energy per nucleon curve. The lightest nuclei have a small binding energy per nucleon, and the nucleus formed when they join lies higher on the steeply rising part of the curve, so it has a larger binding energy per nucleon. The nucleons are therefore more tightly bound after fusion than before, and the gain in binding energy per nucleon, multiplied by the number of nucleons, is released as energy. Because the curve rises steeply for light nuclei, the energy released per nucleon in fusion can be large. 🔉⇢
Although fusion releases energy, it does not happen easily, because the two nuclei that are to join both carry positive charge and therefore repel one another through the electric force. To fuse, the two nuclei must come close enough for the short-range nuclear force to act, which means they must approach to within a few femtometres. To do this they must overcome the electric repulsion that grows stronger as they approach, and this repulsion forms a barrier that the nuclei must surmount before they can fuse. 🔉⇢
The electric repulsion between the nuclei is called the Coulomb barrier, and its height sets the energy the nuclei must have to come close enough to fuse. For two light nuclei the barrier is of the order of a few hundred thousand electron volt. The nuclei must be moving fast enough, that is they must have enough kinetic energy, to overcome this barrier and approach within the range of the nuclear force. Only then can the attractive nuclear force take over and bind them into a single heavier nucleus. 🔉⇢
The way to give the nuclei enough energy to overcome the Coulomb barrier is to raise them to a very high temperature. At a high enough temperature the nuclei move so fast, on the average, that many of them have enough kinetic energy to overcome the barrier when they collide. The temperature needed is enormous, of the order of many million degrees, because the barrier is high. Fusion brought about by such high temperatures is called thermonuclear fusion, since it is the heat that supplies the energy to overcome the repulsion. 🔉⇢
The very high temperatures required for fusion are found in the interior of the stars, and fusion is the process that supplies the energy the stars radiate. In the interior of a star the temperature and the density are high enough that light nuclei fuse into heavier ones, releasing energy that flows outward and is radiated as light and heat. Over the long life of a star, an enormous amount of energy is released in this way, and it is fusion that keeps the stars shining for such long times. 🔉⇢
In the Sun and similar stars the main energy-releasing process is the fusion of hydrogen nuclei into helium. Through a sequence of steps, four hydrogen nuclei, that is four protons, are in effect converted into one helium nucleus, together with other light particles, and energy is released. The overall result of this proton-proton cycle is that four hydrogen nuclei fuse into one helium nucleus with the release of about $26.7$ million electron volt of energy. This is the process that powers the Sun. 🔉⇢
The energy released when hydrogen fuses into helium comes from the mass difference between the four hydrogen nuclei and the helium nucleus. The helium nucleus is lighter than the four hydrogen nuclei from which it is formed, and this difference in mass, multiplied by $c^2$, is the energy released, in accordance with the equivalence of mass and energy. Because a very large number of such fusions take place every second in the interior of a star, the total energy released is enormous and sustains the star over billions of years. 🔉⇢
Fusion has an important advantage as a possible source of energy on the earth, because the light nuclei that fuse, such as the heavy isotopes of hydrogen, are abundant, and the energy released per unit mass is large. If fusion could be controlled and sustained, it would provide an almost unlimited supply of energy. The great difficulty is that the very high temperature needed to overcome the Coulomb barrier must be produced and the hot material held together long enough for fusion to release more energy than is supplied. 🔉⇢
At the very high temperatures needed for fusion, matter exists in a state called a plasma, in which the atoms are broken up into nuclei and free electrons. To achieve controlled fusion, this hot plasma must be confined and kept hot long enough for many fusions to occur, which is very difficult because no ordinary container can hold matter at such temperatures. Overcoming this difficulty is the central problem in the effort to harness fusion as a controlled source of energy on the earth. 🔉⇢
The comparison between fusion and fission is instructive. Both release energy by moving toward more tightly bound nuclei, fission from the heavy side of the binding energy per nucleon curve and fusion from the light side, and both can release energy that appears as the kinetic energy of the products. Fusion requires the nuclei to overcome their electric repulsion and therefore needs very high temperatures, while fission is brought about by a neutron, which carries no charge and so is not repelled by the nucleus and can be absorbed even at ordinary temperatures. 🔉⇢
The energy released per nucleon in fusion can be larger than in fission, because the binding energy per nucleon curve rises steeply for light nuclei. For a given mass of fuel, therefore, fusion can release even more energy than fission. This, together with the abundance of the light nuclei that serve as fuel, is what makes controlled fusion such an attractive goal, despite the great difficulty of reaching and holding the very high temperatures needed to overcome the Coulomb barrier between the nuclei. 🔉⇢
Thus fusion is the joining of light nuclei into a heavier one, releasing energy because the product is more tightly bound. It requires the nuclei to overcome their electric repulsion, which needs very high temperatures, and it is the process that powers the stars through the fusion of hydrogen into helium. On the earth, controlled fusion promises an almost unlimited source of energy but remains difficult to achieve because of the high temperatures required and the difficulty of confining the hot plasma. 🔉⇢
In summary, nuclear fusion joins two light nuclei into a heavier one, releasing energy because the heavier nucleus is more tightly bound, as shown by the steep rise of the binding energy per nucleon curve for light nuclei. The nuclei must overcome the Coulomb barrier of their electric repulsion, which requires very high temperatures, so fusion is a thermonuclear process. It powers the stars through the fusion of four hydrogen nuclei into helium with the release of about $26.7$ million electron volt, and it is a promising but difficult source of energy on the earth. 🔉⇢
Fusion is the opposite of fission in the way it moves along the binding energy per nucleon curve, but it releases energy for the same underlying reason. In fission a heavy nucleus moves toward the peak of the curve by splitting into middle-sized fragments, while in fusion light nuclei move toward the peak by joining into a heavier nucleus. In both the products are more tightly bound than the reactants, so the total binding energy increases and the difference is released as energy. 🔉⇢
The difficulty of fusion lies in bringing the two light nuclei close enough for the short-range nuclear force to act. Both nuclei carry positive charge, so as they approach they repel one another more and more strongly through the electric force, and this repulsion forms a barrier. To fuse, the nuclei must have enough kinetic energy to overcome this Coulomb barrier and approach within a few femtometres, where the attractive nuclear force can bind them into a single nucleus. 🔉⇢
The height of the Coulomb barrier for two light nuclei is of the order of a few hundred thousand electron volt, and to give the nuclei this much energy by their thermal motion requires a temperature of the order of many million degrees. At such temperatures the nuclei move fast enough that many collisions bring them close enough to fuse. Because the energy is supplied as heat, fusion at these high temperatures is called thermonuclear fusion, and it is the process at work in the interior of the stars. 🔉⇢
In the Sun the main process is the fusion of hydrogen into helium through the proton-proton cycle, in which four hydrogen nuclei are in effect converted into one helium nucleus with the release of about $26.7$ million electron volt. The helium nucleus is lighter than the four hydrogen nuclei, and this difference in mass, multiplied by $c^2$, is the energy released. Because a very large number of such fusions take place every second, the Sun radiates an enormous amount of energy and does so over billions of years. 🔉⇢
Fusion is attractive as a possible source of energy on the earth because the light nuclei that serve as fuel are abundant and the energy released per unit mass is large. If the very high temperature needed to overcome the Coulomb barrier could be produced and the hot material held together long enough, fusion would provide an almost unlimited supply of energy. The great difficulty is reaching and holding these very high temperatures, at which matter exists as a plasma of nuclei and free electrons. 🔉⇢
At the temperatures needed for fusion, no ordinary container can hold the hot material, since any container would be destroyed and would cool the plasma. The central problem of achieving controlled fusion on the earth is therefore to confine the hot plasma and keep it hot long enough for fusion to release more energy than is supplied to heat and confine it. This is very difficult, which is why controlled fusion, despite its promise, has been so hard to achieve. 🔉⇢
The comparison of fusion with fission brings out the role of charge. Fission is brought about by a neutron, which carries no charge and so is not repelled by the nucleus and can be absorbed even at ordinary temperatures. Fusion, by contrast, requires two charged nuclei to be forced together against their electric repulsion, which is why it needs the very high temperatures of a thermonuclear process. Both release energy by forming more tightly bound nuclei, but they differ greatly in the conditions they require. 🔉⇢
Because the binding energy per nucleon curve rises steeply for light nuclei, the energy released per nucleon in fusion can be larger than in fission, so for a given mass of fuel fusion can release even more energy. This, together with the abundance of the light nuclei that serve as fuel, is what makes controlled fusion such an attractive goal, and it is the same increase in binding energy that powers the stars through the fusion of hydrogen into helium. 🔉⇢
The very high temperature needed for fusion is required because the two nuclei must be given enough kinetic energy to overcome the Coulomb barrier of their electric repulsion. Only when the nuclei approach within a few femtometres can the short-range nuclear force act and bind them, and to reach this separation against the repulsion they must be moving very fast. At a temperature of many million degrees a large number of the nuclei have enough kinetic energy for this, so fusion proceeds; at lower temperatures too few nuclei can overcome the barrier. 🔉⇢
The energy released in fusion, like that in fission, comes from the difference in mass between the reactants and the product, in accordance with the equivalence of mass and energy. The heavier nucleus formed in fusion is lighter than the light nuclei from which it is made, and this difference in mass, multiplied by $c^2$, is the energy released. Because the product nucleus is more tightly bound, its mass defect is greater, and the increase in binding energy is the source of the energy set free in the fusion. 🔉⇢
In the stars the balance between the energy released by fusion and the inward pull of gravity keeps the star stable over long times. The fusion of light nuclei into heavier ones in the hot, dense interior releases energy that flows outward and is radiated, while gravity holds the star together and keeps the interior hot and dense enough for fusion to continue. Over the long life of a star an enormous amount of light nuclear fuel is converted into heavier nuclei, releasing the energy that the star radiates as light and heat. 🔉⇢
The fusion of hydrogen into helium in the Sun proceeds through a series of steps rather than in a single event, but the net result of the series is that four hydrogen nuclei are converted into one helium nucleus with the release of about $26.7$ million electron volt. The energy released at each step appears as the kinetic energy of the particles produced, and this kinetic energy keeps the interior of the Sun hot enough for the fusion to continue. In this way the Sun has been releasing energy by the fusion of hydrogen into helium for billions of years and will continue to do so for billions more, so long as its store of hydrogen fuel lasts. 🔉⇢
The Q-value is just the rest-mass energy that the reaction gives up (or must be fed). If the products weigh less than the reactants, Q>0 and that missing mass flies off as energy; if they weigh more, Q<0 and the reaction absorbs energy.
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When nuclei take part in a reaction, energy may be released or absorbed, and the amount is measured by a quantity called the Q value of the reaction. The Q value of a nuclear reaction $A + b \to C + d$, in which a nucleus $A$ is struck by a particle $b$ and produces a nucleus $C$ and a particle $d$, is the energy released in the reaction. It is found from the masses of the nuclei and particles involved, using the equivalence of mass and energy. 🔉⇢
The Q value is defined as the difference between the total rest energy of the reactants and the total rest energy of the products. In terms of masses, it is the sum of the masses of the reactants minus the sum of the masses of the products, multiplied by $c^2$. Writing the reactants as $A$ and $b$ and the products as $C$ and $d$, the Q value is $Q = [(m_A + m_b) - (m_C + m_d)]c^2$. It measures how much rest energy is converted into other forms, or the reverse, in the reaction. 🔉⇢
The sign of the Q value tells whether energy is released or must be supplied. If the total mass of the reactants is greater than the total mass of the products, the Q value is positive, and energy is released in the reaction, appearing as the kinetic energy of the products. Such a reaction is called exothermic, and it can proceed on its own once started, since it gives out energy. The released energy comes from the difference in mass, in accordance with the equivalence of mass and energy. 🔉⇢
If, on the other hand, the total mass of the products is greater than the total mass of the reactants, the Q value is negative, and energy must be supplied for the reaction to take place. Such a reaction is called endothermic, and it cannot proceed unless the reactants bring in enough energy, usually as the kinetic energy of the incoming particle, to make up the difference in mass. The energy that must be supplied is at least the amount by which the rest energy of the products exceeds that of the reactants. 🔉⇢
The Q value is calculated from the measured masses of the nuclei and particles taking part in the reaction. The masses are expressed in atomic mass units, the total mass of the products is subtracted from the total mass of the reactants, and the difference is multiplied by $c^2$, using the equivalence $1\ u = 931.5$ million electron volt, to obtain the Q value in million electron volt. Because the masses are known accurately, the Q value of a reaction can be found precisely, and it predicts the energy released or required. 🔉⇢
The Q value is closely related to the binding energies of the nuclei involved. Since the mass of a nucleus is less than the total mass of its nucleons by an amount equal to its binding energy divided by $c^2$, the difference in mass between the reactants and the products can be expressed in terms of the difference in their binding energies. A reaction releases energy, having a positive Q value, when the products are more tightly bound than the reactants, so that the total binding energy increases in the reaction. 🔉⇢
This connection between the Q value and the binding energies shows why fission and fusion release energy. In fission a heavy nucleus splits into more tightly bound fragments, so the products have a greater total binding energy than the reactants and the Q value is positive. In fusion light nuclei join into a more tightly bound nucleus, so again the products are more tightly bound and the Q value is positive. In both cases the positive Q value, coming from the increase in binding energy, is the energy released. 🔉⇢
For an endothermic reaction, with a negative Q value, there is a smallest energy that the incoming particle must have for the reaction to take place, since energy must be supplied to make up the difference in rest energy between the products and the reactants. The incoming particle must bring at least this much energy as kinetic energy, and in fact somewhat more, because some kinetic energy must remain with the products to conserve momentum. Below this least energy the reaction cannot occur, whatever the incoming particle does. 🔉⇢
The Q value is used together with the conservation laws to work out the outcome of a nuclear reaction. The charge and the number of nucleons are conserved, so the atomic numbers and the mass numbers balance on the two sides of the reaction, and the total energy, including the rest energy of the masses and the kinetic energies, is conserved, which is expressed by the Q value. Using these together, the products of a reaction and the energies released or required can be found from the masses of the particles involved. 🔉⇢
An example of an exothermic reaction is one in which the products are more tightly bound than the reactants, so that the Q value is positive and energy is released. The energy released appears as the extra kinetic energy of the products compared with the reactants, so the products fly apart with more kinetic energy than the reactants brought in. Such reactions are the basis of the release of nuclear energy, since they convert some of the rest energy of the nuclei into kinetic energy that can be used. 🔉⇢
An example of an endothermic reaction is one in which the products are less tightly bound than the reactants, so that the Q value is negative and energy must be supplied. Such a reaction takes place only if the incoming particle brings enough kinetic energy, and it removes energy from the motion of the particles, leaving the products with less kinetic energy than the reactants brought in. Many reactions used to produce particular nuclei in the laboratory are of this endothermic kind and require fast incoming particles. 🔉⇢
The Q value applies not only to reactions in which one nucleus is struck by a particle but also to radioactive decays, which can be regarded as reactions in which a single nucleus spontaneously produces the decay products. For a decay the Q value is the energy released, equal to the difference in rest energy between the parent nucleus and the products, and it appears as the kinetic energy of the products. A decay can occur spontaneously only if its Q value is positive, that is only if the parent is heavier than the products. 🔉⇢
Thus the Q value is a single quantity that measures the energy released or required in any nuclear process, whether a reaction brought about by an incoming particle or a spontaneous decay. It is found from the masses of the nuclei and particles involved, it is positive when energy is released and negative when energy must be supplied, and it is closely related to the change in the binding energies of the nuclei. The Q value therefore ties together the masses, the binding energies, and the energies of nuclear processes. 🔉⇢
In summary, the Q value of a nuclear reaction $A + b \to C + d$ is the energy released, equal to the difference between the total rest energy of the reactants and that of the products, $Q = [(m_A + m_b) - (m_C + m_d)]c^2$. A positive Q value means energy is released, the reaction being exothermic, and a negative Q value means energy must be supplied, the reaction being endothermic. The Q value is found from measured masses, is related to the change in binding energy, and applies to nuclear reactions and to radioactive decays alike. 🔉⇢
The Q value gives a single measure of the energy change in any nuclear process, found entirely from the masses of the nuclei and particles taking part. For the reaction $A + b \to C + d$ it is the difference between the total rest energy of the reactants and that of the products, so it states how much rest energy is turned into other forms, chiefly the kinetic energy of the products, or how much must be supplied. The masses are measured accurately, so the Q value can be calculated precisely. 🔉⇢
The sign of the Q value decides the character of the reaction. A positive Q value means the reactants are heavier than the products, so energy is released and the reaction is exothermic; the released energy appears as the extra kinetic energy of the products. A negative Q value means the products are heavier than the reactants, so energy must be supplied and the reaction is endothermic; the incoming particle must bring in enough kinetic energy to make up the difference in rest energy. 🔉⇢
The Q value is closely tied to the binding energies of the nuclei involved, because the mass of a nucleus is less than the mass of its free nucleons by an amount equal to its binding energy divided by $c^2$. A reaction has a positive Q value, releasing energy, when the products are more tightly bound than the reactants, so that the total binding energy increases. This is why the Q value can be worked out either from the masses of the nuclei or from the change in their binding energies. 🔉⇢
This link between the Q value and the binding energies explains at once why both fission and fusion release energy. In fission a heavy nucleus splits into more tightly bound fragments, so the total binding energy increases and the Q value is positive. In fusion light nuclei join into a more tightly bound nucleus, so again the total binding energy increases and the Q value is positive. In each case the positive Q value is the energy released, coming from the increase in binding energy of the nuclei. 🔉⇢
For an endothermic reaction, with a negative Q value, there is a least energy that the incoming particle must bring for the reaction to be possible. Energy at least equal to the amount by which the rest energy of the products exceeds that of the reactants must be supplied, and in fact somewhat more, because some kinetic energy must remain with the products to conserve momentum. Below this least energy the reaction cannot take place, however the incoming particle is directed at the target. 🔉⇢
The Q value is used together with the conservation laws to work out the outcome of a reaction. The charge and the number of nucleons are conserved, so the atomic numbers and mass numbers balance on the two sides, and the total energy, including the rest energy of the masses and the kinetic energies, is conserved, which the Q value expresses. Using these together, the products and the energies released or required in a reaction can be found from the masses of the particles taking part. 🔉⇢
The Q value applies equally to radioactive decays, which can be regarded as reactions in which a single nucleus spontaneously produces the decay products. For a decay the Q value is the energy released, equal to the difference in rest energy between the parent nucleus and the products, and it appears as the kinetic energy of the products. A decay can occur spontaneously only if its Q value is positive, that is only if the parent nucleus is heavier than the products it would produce. 🔉⇢
Thus the Q value ties together the masses, the binding energies, and the energies of nuclear processes into a single quantity. It is found from the measured masses, it is positive when energy is released and negative when energy must be supplied, and it is related to the change in the binding energies of the nuclei. Whether the process is a reaction brought about by an incoming particle or a spontaneous decay, the Q value measures the energy released or required in the same way. 🔉⇢
The calculation of the Q value from measured masses is straightforward but requires accurate values, because the Q value is a difference between large rest energies. The total mass of the products is subtracted from the total mass of the reactants, and the small difference, which may be only a fraction of an atomic mass unit, is multiplied by $931.5$ million electron volt per atomic mass unit to give the Q value. Since the masses are measured to high accuracy, this small difference, and hence the Q value, is found precisely. 🔉⇢
The Q value determines not only whether a reaction releases or requires energy but also how much kinetic energy the products carry. In an exothermic reaction the products share, as extra kinetic energy, the energy released, so they move faster than the reactants; in an endothermic reaction the products carry less kinetic energy than the reactants brought in, since some has been used to make up the difference in rest energy. The Q value thus governs the energies of the particles that come out of a nuclear reaction. 🔉⇢
Because the Q value depends only on the masses of the reactants and products, it can be predicted before a reaction is carried out, using measured masses, and then compared with the energy actually released. The close agreement between the Q value calculated from the masses and the energy observed in the reaction is a direct confirmation of the equivalence of mass and energy, and it shows that the Q value correctly measures the energy change in nuclear reactions and decays alike. 🔉⇢
🔬 Interactive 3D · Interactive binding-energy-per-nucleon curve: drag the mass number A to place a nucleus on the curve and read the energy released by fusion (toward the peak from the left) or fission (toward the peak from the right) relative to the iron maximum. Draggable slider for mass number A (2 to 240); the 3D nucleus rebuilds with that many nucleons; arrows show energy released moving each way from the A=56 iron peak; live readout of E_b/A in MeV.
The total binding energy of a nucleus grows as the nucleus becomes larger, simply because a larger nucleus contains more nucleons that contribute to the binding. This total by itself, however, does not show how tightly each nucleon is held. To compare different nuclei fairly, the binding energy $E_b$ is divided by the number of nucleons $A$, giving the binding energy per nucleon, $E_{bn} = E_b / A$. This quantity measures, on the average, how strongly a single nucleon is bound in the nucleus, and it is the most useful measure of nuclear stability. 🔉⇢
We can think of binding energy per nucleon as the average energy needed to remove one nucleon from the nucleus. A nucleus with a large binding energy per nucleon is one in which each nucleon is held tightly, so a large energy is required to remove a nucleon, and such a nucleus is very stable. A nucleus with a smaller binding energy per nucleon is more loosely bound and less stable. Comparing the binding energy per nucleon of different nuclei therefore tells us which nuclei are more stable and which are less. 🔉⇢
When the binding energy per nucleon is calculated for nuclei across the whole range of mass numbers and plotted against the mass number $A$, a smooth curve is obtained. This curve is one of the most important results in nuclear physics, because its shape explains why energy is released both when very heavy nuclei split and when very light nuclei join. The curve rises steeply at small mass numbers, reaches a broad maximum in the middle range, and then falls slowly for the heaviest nuclei. 🔉⇢
For the lightest nuclei the binding energy per nucleon is small, and it increases rapidly as nucleons are added. The very light nuclei are loosely bound, but as the mass number increases the nucleons become more tightly bound, so the curve rises steeply over the first part of its range. There are a few exceptions where certain light nuclei are unusually tightly bound for their size, appearing as peaks above the general trend, but the overall behaviour at small mass numbers is a steep rise. 🔉⇢
In the middle range of mass numbers, roughly from mass number thirty to mass number one hundred and twenty, the binding energy per nucleon is nearly constant, staying close to a value of about $8.5$ million electron volt. The curve is flat and broad in this region, so all these nuclei have nearly the same binding energy per nucleon and are all strongly bound. The maximum of the curve lies in this middle region, near the iron nucleus, where the binding energy per nucleon is greatest. 🔉⇢
The binding energy per nucleon reaches its maximum value, close to $8.75$ million electron volt, near a mass number of about fifty-six, which is the region of the iron nucleus. Nuclei near this peak are the most tightly bound of all nuclei, and therefore the most stable. Because these nuclei sit at the top of the curve, they cannot release energy either by splitting or by joining, since any such change would move them to nuclei with a smaller binding energy per nucleon and would require energy rather than release it. 🔉⇢
For nuclei heavier than the iron region the binding energy per nucleon slowly decreases as the mass number increases, falling to about $7.6$ million electron volt for the heaviest nuclei such as uranium. This gentle decline at large mass numbers is caused by the growing effect of the repulsion between the many protons in a heavy nucleus, which works against the binding and makes each nucleon slightly less tightly bound. The heaviest nuclei are therefore somewhat less stable than those in the middle of the curve. 🔉⇢
The near constancy of the binding energy per nucleon over the middle range carries a deep meaning about the nuclear force. If a nucleon interacted with every other nucleon in the nucleus, the binding energy per nucleon would grow with the number of nucleons, since each added nucleon would attract all the others. The fact that the binding energy per nucleon stays nearly constant instead shows that a nucleon interacts only with its nearest neighbours, a property of the nuclear force called saturation. Each nucleon is bound by the same few neighbours whatever the size of the nucleus. 🔉⇢
This saturation of the nuclear force is closely connected with the constant density of nuclear matter. Because each nucleon interacts only with those nucleons immediately around it, and because the nucleons keep a fixed spacing, the binding contributed by each nucleon is the same in a small nucleus and in a large one, giving a constant binding energy per nucleon. The same short range of the force that produces the constant density also produces the flat middle portion of the binding energy per nucleon curve, so the two facts are two views of the same underlying property. 🔉⇢
The shape of the curve immediately explains why energy is released when a very heavy nucleus splits into two lighter ones. A heavy nucleus near uranium has a binding energy per nucleon of about $7.6$ million electron volt, while the two middle-sized fragments into which it splits have a binding energy per nucleon of about $8.5$ million electron volt. The fragments are therefore more tightly bound than the original nucleus, and the increase in binding energy per nucleon, multiplied by the number of nucleons, appears as the energy released in the splitting. 🔉⇢
In the same way, the curve explains why energy is released when very light nuclei join to form a heavier one. The lightest nuclei have a small binding energy per nucleon, but the nucleus they form by joining has a larger binding energy per nucleon, further up the steep rising part of the curve. The nucleons are more tightly bound after joining than before, so binding energy is released. Because the rise is steep at small mass numbers, the energy released per nucleon in joining light nuclei can be even larger than in splitting heavy ones. 🔉⇢
Both of these energy-releasing processes move nuclei toward the peak of the curve, near the iron region, where the binding energy per nucleon is greatest. Heavy nuclei move toward the peak by splitting into middle-sized fragments, and light nuclei move toward the peak by joining into larger nuclei. In each case the products lie higher on the curve, meaning they are more tightly bound, and the gain in binding energy per nucleon is the source of the energy released. The peak near iron is thus the point of greatest stability that both processes approach from opposite sides. 🔉⇢
To find the energy released in such a process from the curve, one multiplies the change in binding energy per nucleon by the number of nucleons involved. For example, if a nucleus of mass number close to $240$ splits into fragments whose binding energy per nucleon is greater by about $0.8$ million electron volt, the energy released is about $240 \times 0.8$, which is close to $200$ million electron volt. This simple estimate, read directly from the binding energy per nucleon curve, gives the characteristic energy released when a heavy nucleus splits. 🔉⇢
The binding energy per nucleon curve summarises in a single picture the stability of all nuclei. Its steep rise at small mass numbers, its flat maximum near iron, and its gentle fall for heavy nuclei together determine which nuclei are stable and which can release energy by rearranging. The constancy of the curve over the middle range reveals the saturation of the nuclear force, and the slope of the curve on either side of the peak governs the energy released in the joining of light nuclei and the splitting of heavy ones. 🔉⇢
In summary, the binding energy per nucleon $E_{bn} = E_b / A$ is the average energy needed to remove a nucleon and is the best measure of nuclear stability. Its curve against mass number rises steeply for light nuclei, is nearly constant at about $8.5$ million electron volt in the middle range, reaches a maximum of about $8.75$ million electron volt near mass number fifty-six, and falls slowly to about $7.6$ million electron volt for the heaviest nuclei. The flat middle shows the saturation of the nuclear force, and the slopes explain why splitting heavy nuclei and joining light nuclei both release energy. 🔉⇢
The binding energy per nucleon is obtained for each nucleus by first finding its total binding energy from the mass defect and then dividing by the mass number. The total binding energy is $E_b = \Delta M c^2$, where the mass defect is the difference between the total mass of the free nucleons and the mass of the nucleus, and dividing by $A$ gives $E_{bn} = E_b / A$. Carrying out this calculation for many nuclei and plotting the results against the mass number produces the binding energy per nucleon curve. 🔉⇢
The steep rise of the curve at small mass numbers shows that the very light nuclei gain a great deal of stability as nucleons are added. A nucleus with only a few nucleons has few neighbours contributing to the binding of each nucleon, so the binding energy per nucleon is small. As more nucleons are added, each nucleon acquires more neighbours and is bound more tightly, so the binding energy per nucleon rises quickly until the nucleons have as many neighbours as the short range of the nuclear force allows. 🔉⇢
Once each nucleon has its full complement of nearest neighbours, adding still more nucleons does not increase the binding of each one, because the nuclear force is short-ranged and a nucleon interacts only with those immediately around it. This is why the curve becomes flat in the middle range of mass numbers, where the binding energy per nucleon stays close to $8.5$ million electron volt. The flatness is a direct sign of the saturation of the nuclear force and mirrors the constancy of the density of nuclear matter. 🔉⇢
The slow fall of the curve for the heaviest nuclei is caused by the electric repulsion between the protons. Although the nuclear force binds only nearest neighbours, the electric repulsion acts between every pair of protons, however far apart, and in a heavy nucleus with many protons this repulsion grows large. The repulsion works against the binding and lowers the binding energy per nucleon, so the heaviest nuclei are slightly less tightly bound than those in the middle range, and the curve slopes gently downward beyond the peak near iron. 🔉⇢
The position of the maximum near mass number fifty-six, in the region of the iron nucleus, marks the most stable nuclei. Because these nuclei sit at the top of the curve, no rearrangement can increase their binding energy per nucleon, so they can release energy neither by splitting nor by joining. All other nuclei lie below the peak and can in principle move toward it, the heavy nuclei by splitting into middle-sized fragments and the light nuclei by joining into larger nuclei, releasing energy as they become more tightly bound. 🔉⇢
The energy released when a heavy nucleus splits can be read directly from the curve. A heavy nucleus near uranium has a binding energy per nucleon of about $7.6$ million electron volt, while the middle-sized fragments have about $8.5$ million electron volt. The gain of about $0.9$ million electron volt for each of the roughly two hundred and forty nucleons gives an energy release of the order of $200$ million electron volt, which is the characteristic large energy released in the splitting of a heavy nucleus. 🔉⇢
The energy released when light nuclei join can be read from the curve in the same way, and it can be even larger per nucleon. Because the curve rises steeply at small mass numbers, the increase in binding energy per nucleon when very light nuclei join into a heavier one is large, so a great deal of energy is released for each nucleon involved. This is why the joining of light nuclei is such a powerful source of energy, and it is the process that supplies the energy radiated by the stars. 🔉⇢
The binding energy per nucleon curve therefore contains, in a single picture, the reason for both of the great sources of nuclear energy. Its downward slope on the heavy side shows that splitting heavy nuclei releases energy, and its steep upward slope on the light side shows that joining light nuclei releases even more energy per nucleon. Between them lies the peak near iron, the region of greatest stability, toward which both processes move, and the height of the curve measures how tightly the nucleons are bound in each nucleus. 🔉⇢
It is worth stressing that the binding energy per nucleon, rather than the total binding energy, is the true measure of stability, because it removes the effect of the size of the nucleus. A heavy nucleus has a larger total binding energy than a light one simply because it has more nucleons, yet it may be less stable, since each of its nucleons is bound less tightly. By dividing the total binding energy by the number of nucleons, the binding energy per nucleon allows nuclei of very different sizes to be compared on the same footing, and it is this quantity whose variation with mass number governs which nuclei can release energy. 🔉⇢
Source: NCERT Section 13.7.1 (adapted)
🔬 Interactive 3D · Interactive radioactive decay: a live population of nuclei decays in real time while the number-versus-time graph traces the exponential N = N0 e^(-lambda t); adjust the half-life to see fast versus slow decay. Slider for half-life T-half; the 3D population of glowing nuclei decays stochastically at rate lambda = 0.693/T-half; live N vs t curve and a running count of decayed vs remaining nuclei update each frame.
Radioactivity is a nuclear phenomenon in which an unstable nucleus undergoes a decay, transforming spontaneously into another nucleus with the emission of radiation. For a single nucleus it is not possible to say when it will decay, since the decay is a spontaneous process governed by chance. For a large number of identical nuclei, however, a definite law holds: the number of nuclei that decay in a given short time is proportional to the number of undecayed nuclei present at that time. This is the law of radioactive decay. 🔉⇢
The basis of the law is that every undecayed nucleus has the same chance of decaying in the next interval of time, independent of how long it has already existed and independent of the other nuclei. Because each of the undecayed nuclei has the same chance of decaying, the total number of decays in a short time is proportional to the number of undecayed nuclei present. If there are twice as many undecayed nuclei, twice as many will decay in the same short time, and this simple proportionality is the content of the decay law. 🔉⇢
Let $N$ be the number of undecayed nuclei present at time $t$. In a short time interval the number that decay is proportional to $N$ and to the length of the interval. The constant of proportionality is called the decay constant, denoted $\lambda$. The decay constant measures the chance that a given nucleus decays per unit time, and it is a fixed number for a particular kind of nucleus. A large decay constant means the nuclei decay quickly, and a small decay constant means they decay slowly. 🔉⇢
Because the number of undecayed nuclei decreases as nuclei decay, the change in $N$ in a short time is negative and proportional to $N$. Writing the rate of change of the number of undecayed nuclei as proportional to the number present, with the decay constant $\lambda$ as the constant of proportionality, gives a simple relation whose solution shows how the number of undecayed nuclei falls with time. The solution is an exponential decrease, in which the number remaining falls by the same fraction in each equal interval of time. 🔉⇢
The solution of the decay law is $N = N_0 e^{-\lambda t}$, where $N_0$ is the number of undecayed nuclei present at the starting time $t = 0$ and $N$ is the number remaining at a later time $t$. The number of undecayed nuclei therefore decreases exponentially with time. At the start there are $N_0$ nuclei; as time goes on the number falls, rapidly at first when many nuclei are present and more slowly later when few remain, approaching zero but never quite reaching it. 🔉⇢
The exponential form of the decay means that the number of undecayed nuclei falls by the same factor in each equal interval of time. In one fixed interval the number is multiplied by a certain fraction less than one; in the next equal interval it is multiplied by the same fraction again, and so on. This is the characteristic feature of exponential decrease and distinguishes radioactive decay from a process in which a fixed number, rather than a fixed fraction, decays in each interval. 🔉⇢
The decay constant $\lambda$ completely determines how fast a particular kind of nucleus decays. It appears in the exponent of the decay law, so a nucleus with a large decay constant has its number of undecayed nuclei fall quickly, while a nucleus with a small decay constant has its number fall slowly. The decay constant is a property of the particular nucleus and does not depend on the amount of substance present, on temperature, or on chemical state, since radioactive decay is a nuclear process unaffected by these external conditions. 🔉⇢
The number of decays per unit time is called the activity of the sample. Since the number of decays per unit time is proportional to the number of undecayed nuclei, with the decay constant as the constant of proportionality, the activity is equal to the decay constant multiplied by the number of undecayed nuclei present. As the number of undecayed nuclei falls exponentially with time, the activity falls in the same exponential way, so a sample becomes less active as time goes on. The activity is treated in detail separately. 🔉⇢
Two useful measures of the rate of decay are defined from the decay constant. The half-life is the time in which half of the undecayed nuclei present decay, so that the number remaining falls to one-half of its value. The mean life is the average time for which a nucleus survives before it decays. Both the half-life and the mean life are fixed for a particular kind of nucleus and are simply related to the decay constant, so that any one of the three determines the other two. These measures are treated separately. 🔉⇢
The law of radioactive decay applies to a very wide range of nuclei, from those that decay in a tiny fraction of a second to those that decay only over billions of years. Whatever the decay constant, the same exponential law $N = N_0 e^{-\lambda t}$ describes the fall in the number of undecayed nuclei, only the value of the decay constant differing from one kind of nucleus to another. This universality of the law reflects the fact that it rests only on the constant chance of decay of each nucleus. 🔉⇢
The exponential decay law can be used to find the age of old samples that contain radioactive nuclei. Because the number of undecayed nuclei falls in a known way with time, a measurement of how many undecayed nuclei remain, compared with the number originally present, gives the time that has passed. This method is used to estimate the ages of very old objects, and it works because the decay constant is fixed and the decay is unaffected by the external conditions the sample has experienced over its history. 🔉⇢
It is important to understand that the decay law is a statistical law, describing the behaviour of a large number of nuclei, not the fate of any single nucleus. For one nucleus only the chance of decay per unit time is known; it is impossible to say when that particular nucleus will decay. But when a very large number of nuclei are present, the fraction that decays in a given time is definite, and the number remaining follows the smooth exponential curve very closely. The larger the number of nuclei, the more exactly the law is obeyed. 🔉⇢
The decay law also shows that the number of undecayed nuclei never quite reaches zero, but continues to fall by the same fraction in each interval. After one half-life half the nuclei remain, after two half-lives one-quarter remain, after three half-lives one-eighth remain, and so on, the number being halved with each further half-life. This repeated halving is another way of expressing the exponential fall described by $N = N_0 e^{-\lambda t}$, and it shows how the sample becomes steadily depleted of undecayed nuclei as time goes on. 🔉⇢
In summary, the law of radioactive decay states that the number of nuclei decaying per unit time is proportional to the number of undecayed nuclei present, the constant of proportionality being the decay constant $\lambda$. The number of undecayed nuclei therefore falls exponentially as $N = N_0 e^{-\lambda t}$, halving in each half-life. The decay constant is fixed for a particular kind of nucleus and is unaffected by external conditions, and the law is a statistical law describing the behaviour of a large number of nuclei. 🔉⇢
The starting point of the decay law is the fact that each undecayed nucleus has the same fixed chance of decaying in the next unit of time, whatever its past history. Because the nuclei decay independently and each has the same chance, the total number decaying in a short time is proportional to the number of undecayed nuclei present. This proportionality, with the decay constant as the constant of proportionality, is the whole content of the law, and everything else follows from it by working out how the number changes with time. 🔉⇢
The solution $N = N_0 e^{-\lambda t}$ describes an exponential fall in the number of undecayed nuclei. At the start, when $t = 0$, the exponential is one and $N$ equals $N_0$, the initial number. As time goes on the exponent becomes more negative and $N$ falls, quickly at first while many nuclei remain and more slowly later when few remain. The number approaches zero as time becomes very large but never quite reaches it, since in each further interval it falls only by the same fraction. 🔉⇢
The decay constant $\lambda$ fixes the rate of the fall. A large decay constant makes the exponent large, so the number of undecayed nuclei falls quickly, while a small decay constant makes the number fall slowly. The decay constant is a property of the particular kind of nucleus and is the same for every nucleus of that kind, so a sample of a given radioactive substance always decays with the same decay constant, whatever the amount present or the external conditions of temperature and chemical state. 🔉⇢
The number of decays per unit time, the activity, is proportional to the number of undecayed nuclei present, with the decay constant as the constant of proportionality, so the activity equals the decay constant multiplied by the number of undecayed nuclei. Because the number of undecayed nuclei falls exponentially, the activity also falls exponentially with the same decay constant, and a sample therefore becomes steadily less active as time goes on. The activity is measured as the number of decays per unit time and is treated in detail separately. 🔉⇢
The half-life and the mean life are two convenient measures of the rate of decay, both fixed by the decay constant. The half-life is the time in which half of the undecayed nuclei present decay, so that the number remaining falls to one-half. The mean life is the average time for which a nucleus survives before decaying. Each of these is simply related to the decay constant, so knowing any one of the decay constant, the half-life, or the mean life determines the other two. They are treated separately. 🔉⇢
The exponential nature of the decay can be stated in terms of repeated halving. After one half-life, half of the original nuclei remain; after two half-lives, one-quarter remain; after three half-lives, one-eighth remain, and so on, the number being halved with each further half-life. This repeated halving is the same behaviour as the exponential fall $N = N_0 e^{-\lambda t}$ described in another way, and it shows clearly how the sample is steadily depleted of undecayed nuclei as one half-life follows another. 🔉⇢
The decay law is a statistical law, describing the behaviour of a large number of nuclei rather than the fate of a single nucleus. For one nucleus only the chance of decay per unit time is known, and it is impossible to say when that nucleus will decay. But for a very large number of nuclei the fraction decaying in a given time is definite, so the number remaining follows the smooth exponential curve closely, and the larger the number of nuclei present, the more exactly the law is obeyed. 🔉⇢
The decay law is used to estimate the ages of very old objects that contain radioactive nuclei. Because the number of undecayed nuclei falls in a known exponential way with time, a measurement of the number of undecayed nuclei remaining, compared with the number originally present, gives the time that has passed. The method works because the decay constant is fixed and radioactive decay is unaffected by the external conditions the sample has experienced, so the decay serves as a reliable clock over very long periods of time. 🔉⇢
The value of the decay constant varies enormously from one kind of nucleus to another, and with it the time over which the sample decays. Some nuclei have a very large decay constant and decay almost completely in a tiny fraction of a second, while others have a very small decay constant and decay only over millions or billions of years. Whatever the value of the decay constant, the same exponential law governs the fall in the number of undecayed nuclei, so the behaviour is always the same in form and differs only in the time scale set by the decay constant. 🔉⇢
The decay law can be written equally in terms of the number of undecayed nuclei or in terms of the activity, since the activity is simply the decay constant multiplied by the number of undecayed nuclei. Both fall exponentially with time and with the same decay constant, so a graph of the number of undecayed nuclei against time and a graph of the activity against time have the same exponential shape. Measuring how the activity falls with time is therefore a direct way of finding the decay constant of a radioactive substance and hence its half-life. 🔉⇢
The independence of the decay constant from external conditions is a key feature of the law. Because radioactive decay is a nuclear process, taking place in the nucleus and not among the outer electrons, it is not affected by heating, by pressure, or by the chemical combination in which the atom is found. The decay constant is therefore a fixed property of the nucleus, and the exponential decay proceeds at the same rate whatever is done to the sample by ordinary means. This is why the decay law provides such a dependable measure of the passage of time. 🔉⇢
Source: JEE Physics — Nuclei
🔬 Interactive 3D · Interactive fission chain reaction: a neutron strikes a U-235 nucleus, which splits into fragments and releases free neutrons that strike further nuclei; a multiplication-factor slider shows sub-critical (k<1, dies out), critical (k=1, steady), and super-critical (k>1, runaway) behaviour. Slider for multiplication factor k from 0.5 to 2.0; incoming neutron triggers fission of a 3D U-235 nucleus into two fragments plus 2-3 neutrons; those neutrons propagate and induce further fissions at the chosen k; live generation counter shows the chain growing, holding steady, or dying.
A most important neutron-induced nuclear reaction is fission, in which a heavy nucleus breaks into two lighter nuclei of middle size. Fission is usually brought about by a neutron: when a heavy nucleus such as a uranium nucleus absorbs a neutron, it becomes unstable and splits into two fragments of comparable size, together with a few free neutrons. A large amount of energy is released in the process, far more than in any chemical reaction, which is why fission is so important as a source of energy. 🔉⇢
The reason fission releases energy is found in the binding energy per nucleon curve. A heavy nucleus near uranium has a binding energy per nucleon of about $7.6$ million electron volt, while the middle-sized fragments into which it splits have a binding energy per nucleon of about $8.5$ million electron volt. The fragments are therefore more tightly bound than the original heavy nucleus, and the gain in binding energy per nucleon, multiplied by the number of nucleons, is released as energy. This is the source of the large energy of fission. 🔉⇢
The energy released in a single fission is about $200$ million electron volt. This can be estimated from the binding energy per nucleon curve: the gain in binding energy per nucleon of nearly one million electron volt, multiplied by the roughly two hundred and forty nucleons of the heavy nucleus, gives an energy of the order of $200$ million electron volt for each fission. This is an enormous energy compared with the few electron volt released in a chemical reaction involving a single atom, which is why fission of even a small mass of material releases so much energy. 🔉⇢
Most of the energy released in fission appears as the kinetic energy of the two fragments, which fly apart at high speed because they strongly repel one another electrically once the nucleus has split. The rest of the energy is carried by the neutrons that are released, by the beta particles and gamma rays emitted as the fragments, which are themselves unstable, decay toward stability, and by neutrinos. When the fragments are stopped in the surrounding material, their kinetic energy appears as heat, which is the energy that is finally harnessed. 🔉⇢
A key feature of fission is that, besides the two fragments, a few free neutrons are released in each fission. These neutrons are important because they can go on to cause the fission of other heavy nuclei. If each fission releases neutrons that produce further fissions, the process can continue on its own, each fission leading to more fissions in a self-sustaining sequence. This possibility of a self-sustaining sequence of fissions is what makes fission useful as a large-scale source of energy. 🔉⇢
A self-sustaining sequence of fissions, in which the neutrons from one fission cause further fissions, is called a chain reaction. Whether a chain reaction can be sustained depends on how many of the neutrons released in each fission go on to cause further fissions, rather than escaping from the material or being absorbed without causing fission. If on the average at least one neutron from each fission causes a further fission, the chain reaction continues; if fewer do so, the chain reaction dies out. 🔉⇢
The number that measures whether a chain reaction is sustained is called the multiplication factor. It is the average number of neutrons from each fission that go on to cause a further fission. When the multiplication factor is equal to one, each fission leads on the average to exactly one further fission, and the chain reaction proceeds at a steady rate. When it is greater than one, the number of fissions grows from one generation to the next, and when it is less than one, the number of fissions falls and the chain reaction stops. 🔉⇢
For a chain reaction to be sustained, enough of the neutrons released must remain in the material and cause further fissions. In a small piece of material many neutrons escape from the surface before causing fission, so the multiplication factor is less than one and no chain reaction occurs. In a large enough piece, fewer neutrons escape in proportion, so the multiplication factor can reach one and the chain reaction is sustained. There is thus a minimum size of material needed for a self-sustaining chain reaction. 🔉⇢
In a nuclear reactor the chain reaction is controlled so that the multiplication factor is kept equal to one and the fissions proceed at a steady rate, releasing energy steadily as heat. The heat is carried away and used to generate power. The control is achieved by adjusting how many neutrons are allowed to cause further fissions, removing neutrons when the rate is too high and allowing more when it is too low, so that the multiplication factor is held at one and the reaction neither grows nor dies out. 🔉⇢
The neutrons released in fission are fast, but the fission of the heavy nucleus is caused more readily by slow neutrons. For this reason a reactor contains a material, called a moderator, whose light nuclei slow the fast neutrons down through repeated collisions without absorbing many of them. The slowed neutrons are then more effective in causing further fissions. The moderator is therefore an essential part of a reactor, since it keeps the chain reaction going by turning the fast neutrons of fission into slow neutrons that readily cause new fissions. 🔉⇢
The steady rate of the controlled chain reaction is maintained by devices that can absorb neutrons. These are made of a material that readily absorbs neutrons, and by moving them into or out of the reactor the number of neutrons available to cause further fissions is adjusted. Moving them in absorbs more neutrons and lowers the multiplication factor, slowing the reaction, while moving them out allows more neutrons to cause fissions and raises the multiplication factor, speeding it up. In this way the reaction is held at a steady rate. 🔉⇢
The energy released in the fission of a heavy nucleus is so large that the fission of the nuclei in even a small mass of material can supply an enormous amount of energy. Because each fission releases about $200$ million electron volt, the complete fission of the nuclei in one kilogram of a heavy element would release an energy of the order of that produced by burning many thousands of tonnes of ordinary fuel. This is why nuclear fission is such a concentrated source of energy compared with the burning of chemical fuels. 🔉⇢
Fission thus combines several features into one important process: a heavy nucleus is split by absorbing a neutron, the fragments are more tightly bound so that a large energy is released, and a few neutrons are set free that can sustain a chain reaction. When the chain reaction is controlled, as in a reactor, the energy is released steadily and can be used to generate power, and the whole process rests on the difference in binding energy per nucleon between the heavy nucleus and its middle-sized fragments. 🔉⇢
In summary, nuclear fission is a neutron-induced reaction in which a heavy nucleus such as uranium absorbs a neutron and splits into two middle-sized fragments, releasing a few neutrons and about $200$ million electron volt of energy. The energy comes from the greater binding energy per nucleon of the fragments compared with the heavy nucleus. The released neutrons can cause further fissions, giving a chain reaction whose growth is measured by the multiplication factor, and in a reactor this chain reaction is controlled, with a moderator to slow the neutrons, so that energy is released steadily. 🔉⇢
Fission is a neutron-induced reaction because it is most readily brought about when a heavy nucleus absorbs a neutron. The neutron carries no charge, so it is not repelled by the positive charge of the nucleus and can approach and enter the nucleus even when moving slowly. Once the heavy nucleus has absorbed the neutron it becomes unstable and splits into two fragments of middle size, and this is why the neutron is so effective in causing fission of heavy nuclei. 🔉⇢
The large energy released in fission is a direct consequence of the shape of the binding energy per nucleon curve. The heavy nucleus lies on the part of the curve where the binding energy per nucleon is about $7.6$ million electron volt, while the middle-sized fragments lie near the top of the curve where it is about $8.5$ million electron volt. The fragments are more tightly bound, and the gain in binding energy per nucleon, multiplied by the number of nucleons, is released as the energy of the fission. 🔉⇢
Most of the energy released in a fission appears as the kinetic energy of the two fragments. Once the nucleus has split, the two fragments both carry positive charge and repel one another strongly through the electric force, so they fly apart at high speed. When they are stopped in the surrounding material, their kinetic energy is turned into heat, and this heat is the form in which the energy of fission is finally collected and used. The neutrons, beta particles and gamma rays released carry away the rest of the energy. 🔉⇢
The release of a few free neutrons in each fission is the feature that makes a chain reaction possible. These neutrons can be absorbed by other heavy nuclei and cause them to undergo fission in turn, releasing still more neutrons, so that the process can continue on its own. Without the release of neutrons in each fission there could be no chain reaction, and fission could not be used as a large-scale source of energy; it is the extra neutrons that carry the reaction forward from one fission to the next. 🔉⇢
Whether a chain reaction grows, stays steady, or dies out is decided by the multiplication factor, the average number of neutrons from each fission that cause a further fission. When the multiplication factor is exactly one, the reaction proceeds at a steady rate; when it is greater than one, the number of fissions grows from one generation to the next; and when it is less than one, the number falls and the reaction stops. Controlling a chain reaction means keeping the multiplication factor equal to one. 🔉⇢
Not all the neutrons released in a fission go on to cause further fissions. Some escape from the surface of the material without being absorbed, and some are absorbed by nuclei without causing fission. Only the remaining neutrons cause further fissions, and whether these are enough to keep the reaction going depends on the size and composition of the material. In a piece that is too small, too many neutrons escape and the reaction dies out, so there is a least size needed for a sustained chain reaction. 🔉⇢
In a nuclear reactor the chain reaction is controlled so that the multiplication factor is held at one and the energy is released steadily as heat. A moderator, made of light nuclei, slows the fast neutrons of fission down through repeated collisions, because slow neutrons cause the fission of the heavy nuclei more readily than fast ones. Devices that absorb neutrons are moved into or out of the reactor to adjust how many neutrons cause further fissions, and so keep the reaction steady, and the heat produced is carried away to generate power. 🔉⇢
The concentration of energy in fission is enormous compared with chemical burning. Because each fission releases about $200$ million electron volt, while the burning of a single atom of ordinary fuel releases only a few electron volt, the fission of the nuclei in a small mass of a heavy element releases as much energy as the burning of thousands of tonnes of ordinary fuel. This is why fission is such a concentrated source of energy, and why a reactor can produce a great deal of power from a small amount of material. 🔉⇢
The fragments produced in fission are themselves unstable, because they have too many neutrons for their size compared with stable nuclei of the same mass number. They therefore decay, chiefly by emitting beta particles and gamma rays, moving step by step toward stable nuclei. This is why the material left after fission remains radioactive for a time, and why the energy released in fission includes, besides the kinetic energy of the fragments, the energy carried by the beta particles, gamma rays and neutrinos emitted as the fragments decay. 🔉⇢
The slowing of the neutrons by the moderator is essential because the fast neutrons released in fission cause the fission of the heavy nuclei much less readily than slow neutrons do. In the moderator the fast neutrons collide repeatedly with the light nuclei and lose energy at each collision, without being absorbed in large numbers, so that they emerge as slow neutrons. These slow neutrons are then much more effective in causing further fissions, and it is by this slowing that the moderator keeps the chain reaction going in the reactor. 🔉⇢
The control of the chain reaction depends on keeping the multiplication factor equal to one at all times. If the factor were allowed to rise above one, the number of fissions, and with it the rate of release of energy, would grow rapidly from one generation to the next; if it fell below one, the reaction would die out. By moving the neutron-absorbing devices in or out, the number of neutrons available to cause further fissions is adjusted so that on the average exactly one neutron from each fission causes a further fission, and the reaction proceeds steadily. 🔉⇢
The steady release of energy in a controlled chain reaction makes fission useful for producing power. The heat released as the fragments are stopped in the surrounding material is carried away and used, while the multiplication factor is held at one so that the energy is released at a constant rate rather than growing without limit. In this way the large energy of fission, arising from the greater binding energy per nucleon of the fragments, is turned into a steady and controllable supply of energy. 🔉⇢
The neutrons released in fission, besides carrying the chain reaction forward, also make possible the fission of a heavy nucleus that does not itself split readily. Such a nucleus can absorb a slow neutron and be transformed into a nucleus that then undergoes fission, so that the moderator, by slowing the neutrons, again plays a part in keeping the fission going. The energy released in each fission of the heavy nucleus is again about $200$ million electron volt, most of it appearing as the kinetic energy of the two fragments and the rest carried by the neutrons, beta particles and gamma rays that the fission and the decay of the fragments release. 🔉⇢
Source: NCERT Exercise 13.7 (adapted)
Source: NCERT Section 13.2 (adapted)
Source: NCERT Example 13.1 (adapted)
Source: NCERT Example 13.3 (adapted)
Source: NCERT Exercise 13.1 (adapted)
Source: JEE Physics — Nuclei
Source: JEE Physics — Nuclei
Source: JEE Physics — Nuclei
Source: JEE Physics — Nuclei
Source: NCERT Exercise 13.9 (adapted)
Source: NCERT Exercise 13.5 (adapted)
Source: JEE Physics - Nuclei
Source: JEE Physics - Nuclei
Source: NCERT Exercise 13.4 (adapted)
Source: NCERT Example 13.1 (adapted)
Source: NCERT Exercise 13.10 (adapted)
Source: NCERT Example 13.2 (adapted)
Source: NCERT Example 13.3 (adapted)
Source: NCERT Example 13.3 (adapted)
Source: NCERT Exercise 13.1 (adapted)
Source: NCERT Exercise 13.2 (adapted)
Source: NCERT Exercise 13.2 (adapted)
Source: NCERT Exercise 13.3 (adapted)
Source: JEE Physics - Nuclei
Source: JEE Physics - Nuclei
Source: JEE Physics - Nuclei
Source: JEE Physics - Nuclei
Source: JEE Physics - Nuclei
Source: JEE Physics - Nuclei
Source: JEE Physics - Nuclei
Source: NCERT Exercise 13.7 (adapted)
Source: NCERT Exercise 13.8 (adapted)
Source: NCERT Exercise 13.9 (adapted)
Source: NCERT Exercise 13.5 (adapted)
Source: NCERT Exercise 13.5 (adapted)
Source: JEE Physics - Nuclei
These worked examples are taught in full alongside their interactive scene:
| Quantity | Formula | What it means / when to use | Source |
|---|---|---|---|
| Mass number 🔉⇢ | A = Z + N | Mass number: understand what each symbol means and when this applies — see the concept tab for the derivation. | NCERT Class XII Physics — Nuclei |
| Atomic mass unit 🔉⇢ | 1\,u = 1.660539\times10^{-27}\,\text{kg} = 931.5\ \text{MeV}/c^{2} | Atomic mass unit: understand what each symbol means and when this applies — see the concept tab for the derivation. | NCERT Class XII Physics — Nuclei |
| Nuclear radius 🔉⇢ | R = R_{0}A^{1/3},\quad R_{0}=1.2\ \text{fm} | Nuclear radius: understand what each symbol means and when this applies — see the concept tab for the derivation. | NCERT Class XII Physics — Nuclei |
| Nuclear density (independent of A) 🔉⇢ | \rho = \dfrac{3m_{\text{nucleon}}}{4\pi R_{0}^{3}} \approx 2.3\times10^{17}\ \text{kg\,m^{-3}} | Nuclear density (independent of A): understand what each symbol means and when this applies — see the concept tab for the derivation. | NCERT Class XII Physics — Nuclei |
| Ratio of nuclear radii 🔉⇢ | \dfrac{R_{1}}{R_{2}} = \left(\dfrac{A_{1}}{A_{2}}\right)^{1/3} | Ratio of nuclear radii: understand what each symbol means and when this applies — see the concept tab for the derivation. | NCERT Class XII Physics — Nuclei |
| Quantity | Formula | What it means / when to use | Source |
|---|---|---|---|
| Mass–energy equivalence 🔉⇢ | E = mc^{2} | Mass–energy equivalence: understand what each symbol means and when this applies — see the concept tab for the derivation. | NCERT Class XII Physics — Nuclei |
| Energy of 1 u 🔉⇢ | 1\,u\,c^{2} = 931.5\ \text{MeV} | Energy of 1 u: understand what each symbol means and when this applies — see the concept tab for the derivation. | NCERT Class XII Physics — Nuclei |
| Mass defect 🔉⇢ | \Delta M = [\,Z m_{p} + (A-Z)m_{n}\,] - M | Mass defect: understand what each symbol means and when this applies — see the concept tab for the derivation. | NCERT Class XII Physics — Nuclei |
| Binding energy 🔉⇢ | E_{b} = \Delta M\,c^{2} | Binding energy: understand what each symbol means and when this applies — see the concept tab for the derivation. | NCERT Class XII Physics — Nuclei |
| Binding energy per nucleon 🔉⇢ | E_{bn} = \dfrac{E_{b}}{A} | Binding energy per nucleon: understand what each symbol means and when this applies — see the concept tab for the derivation. | NCERT Class XII Physics — Nuclei |
| Quantity | Formula | What it means / when to use | Source |
|---|---|---|---|
| Radioactive decay law 🔉⇢ | N = N_{0}\,e^{-\lambda t} | Radioactive decay law: understand what each symbol means and when this applies — see the concept tab for the derivation. | NCERT Class XII Physics — Nuclei |
| Halving form 🔉⇢ | N = N_{0}\left(\dfrac{1}{2}\right)^{t/T_{1/2}} | Halving form: understand what each symbol means and when this applies — see the concept tab for the derivation. | NCERT Class XII Physics — Nuclei |
| Half-life 🔉⇢ | T_{1/2} = \dfrac{\ln 2}{\lambda} = \dfrac{0.693}{\lambda} | Half-life: understand what each symbol means and when this applies — see the concept tab for the derivation. | NCERT Class XII Physics — Nuclei |
| Mean life 🔉⇢ | \tau = \dfrac{1}{\lambda},\qquad T_{1/2} = 0.693\,\tau | Mean life: understand what each symbol means and when this applies — see the concept tab for the derivation. | NCERT Class XII Physics — Nuclei |
| Activity 🔉⇢ | R = -\dfrac{dN}{dt} = \lambda N = \lambda N_{0}e^{-\lambda t} | Activity: understand what each symbol means and when this applies — see the concept tab for the derivation. | NCERT Class XII Physics — Nuclei |
| Units of activity 🔉⇢ | 1\ \text{Bq} = 1\ \text{decay/s},\quad 1\ \text{Ci} = 3.7\times10^{10}\ \text{Bq} | Units of activity: understand what each symbol means and when this applies — see the concept tab for the derivation. | NCERT Class XII Physics — Nuclei |
| Quantity | Formula | What it means / when to use | Source |
|---|---|---|---|
| Alpha decay 🔉⇢ | ^{A}_{Z}X \to\ ^{A-4}_{Z-2}Y + \,^{4}_{2}\text{He} | Alpha decay: understand what each symbol means and when this applies — see the concept tab for the derivation. | NCERT Class XII Physics — Nuclei |
| Beta-minus decay 🔉⇢ | ^{A}_{Z}X \to\ ^{A}_{Z+1}Y + e^{-} + \bar{\nu} | Beta-minus decay: understand what each symbol means and when this applies — see the concept tab for the derivation. | NCERT Class XII Physics — Nuclei |
| Beta-plus decay 🔉⇢ | ^{A}_{Z}X \to\ ^{A}_{Z-1}Y + e^{+} + \nu | Beta-plus decay: understand what each symbol means and when this applies — see the concept tab for the derivation. | NCERT Class XII Physics — Nuclei |
| Gamma decay 🔉⇢ | ^{A}_{Z}X^{*} \to\ ^{A}_{Z}X + \gamma | Gamma decay: understand what each symbol means and when this applies — see the concept tab for the derivation. | NCERT Class XII Physics — Nuclei |
| Quantity | Formula | What it means / when to use | Source |
|---|---|---|---|
| Q-value (from masses) 🔉⇢ | Q = \left(\textstyle\sum m_{\text{initial}} - \sum m_{\text{final}}\right)c^{2} | Q-value (from masses): understand what each symbol means and when this applies — see the concept tab for the derivation. | NCERT Class XII Physics — Nuclei |
| Q-value (from kinetic energy) 🔉⇢ | Q = K_{\text{final}} - K_{\text{initial}} | Q-value (from kinetic energy): understand what each symbol means and when this applies — see the concept tab for the derivation. | NCERT Class XII Physics — Nuclei |
| Fission of uranium-235 🔉⇢ | ^{1}_{0}n + \,^{235}_{92}\text{U} \to\ ^{144}_{56}\text{Ba} + \,^{89}_{36}\text{Kr} + 3\,^{1}_{0}n,\quad Q\approx 200\ \text{MeV} | Fission of uranium-235: understand what each symbol means and when this applies — see the concept tab for the derivation. | NCERT Class XII Physics — Nuclei |
| Proton–proton cycle (net) 🔉⇢ | 4\,^{1}_{1}\text{H} + 2e^{-} \to\ ^{4}_{2}\text{He} + 2\nu + 6\gamma + 26.7\ \text{MeV} | Proton–proton cycle (net): understand what each symbol means and when this applies — see the concept tab for the derivation. | NCERT Class XII Physics — Nuclei |
| Coulomb barrier temperature (order) 🔉⇢ | \dfrac{3}{2}kT \approx K \approx 400\ \text{keV}\ \Rightarrow\ T \approx 3\times10^{9}\ \text{K} | Coulomb barrier temperature (order): understand what each symbol means and when this applies — see the concept tab for the derivation. | NCERT Class XII Physics — Nuclei |
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.
A radioactive sample S1 having an activity of $5\ \mu$Ci has twice the number of nuclei as another sample S2 which has an activity of $10\ \mu$Ci. The half lives of S1 and S2 can be
Scientists are working hard to develop nuclear fusion reactor. Nuclei of heavy hydrogen, $^{2}_{1}\text{H}$, known as deuteron and denoted by D, can be thought of as a candidate for fusion reactor. The D-D reaction is $^{2}_{1}\text{H} + {}^{2}_{1}\text{H} \rightarrow {}^{3}_{2}\text{He} + n + $ energy. In the core of fusion reactor, a gas of heavy hydrogen is fully ionized into deuteron nuclei and electrons. This collection of $^{2}_{1}\text{H}$ nuclei and electrons is known as plasma. The nuclei move randomly in the reactor core and occasionally come close enough for nuclear fusion to take place. Usually, the temperatures in the reactor core are too high and no material wall can be used to confine the plasma. Special techniques are used which confine the plasma for a time $t_0$ before the particles fly away from the core. If $n$ is the density (number/volume) of deuterons, the product $nt_0$ is called Lawson number. In one of the criteria, a reactor is termed successful if Lawson number is greater than $5\times10^{14}$ s/cm$^3$. It may be helpful to use the following: Boltzmann constant $k = 8.6\times10^{-5}$ eV/K; $\dfrac{e^2}{4\pi\varepsilon_0} = 1.44\times10^{-9}$ eV m. In the core of nuclear fusion reactor, the gas becomes plasma because of
Scientists are working hard to develop nuclear fusion reactor. Nuclei of heavy hydrogen, $^{2}_{1}\text{H}$, known as deuteron and denoted by D, can be thought of as a candidate for fusion reactor. The D-D reaction is $^{2}_{1}\text{H} + {}^{2}_{1}\text{H} \rightarrow {}^{3}_{2}\text{He} + n + $ energy. It may be helpful to use the following: Boltzmann constant $k = 8.6\times10^{-5}$ eV/K; $\dfrac{e^2}{4\pi\varepsilon_0} = 1.44\times10^{-9}$ eV m. Assume that two deuteron nuclei in the core of fusion reactor at temperature $T$ are moving towards each other, each with kinetic energy $1.5\,kT$, when the separation between them is large enough to neglect Coulomb potential energy. Also neglect any interaction from other particles in the core. The minimum temperature $T$ required for them to reach a separation of $4\times10^{-15}$ m is in the range
In a D-D fusion reactor, if $n$ is the density (number/volume) of deuterons and $t_0$ is the confinement time, the product $nt_0$ is called the Lawson number. In one of the criteria, a reactor is termed successful if the Lawson number is greater than $5\times10^{14}$ s/cm$^3$. Results of calculations for four different designs of a fusion reactor using D-D reaction are given below. Which of these is most promising based on Lawson criterion?
The activity of a freshly prepared radioactive sample is $10^{10}$ disintegrations per second, whose mean life is $10^{9}$ s. The mass of an atom of this radioisotope is $10^{-25}$ kg. The mass (in mg) of the radioactive sample is
Paragraph for Questions 11 and 12: The $\beta$-decay process, discovered around 1900, is basically the decay of a neutron ($n$). In the laboratory, a proton ($p$) and an electron ($e^-$) are observed as the decay products of the neutron. Therefore, considering the decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant. But experimentally, it was observed that the electron kinetic energy has a continuous spectrum. Considering a three-body decay process, i.e. $n\rightarrow p+e^-+\bar{\nu}_e$, around 1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino ($\bar{\nu}_e$) to be massless and possessing negligible energy, and the neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is $0.8\times10^{6}$ eV. The kinetic energy carried by the proton is only the recoil energy. What is the maximum energy of the anti-neutrino?
Paragraph for Questions 11 and 12: The $\beta$-decay process, discovered around 1900, is basically the decay of a neutron ($n$). In the laboratory, a proton ($p$) and an electron ($e^-$) are observed as the decay products of the neutron. Therefore, considering the decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant. But experimentally, it was observed that the electron kinetic energy has a continuous spectrum. Considering a three-body decay process, i.e. $n\rightarrow p+e^-+\bar{\nu}_e$, around 1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino ($\bar{\nu}_e$) to be massless and possessing negligible energy, and the neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is $0.8\times10^{6}$ eV. The kinetic energy carried by the proton is only the recoil energy. If the anti-neutrino had a mass of $3$ eV/c$^{2}$ (where c is the speed of light) instead of zero mass, what should be the range of the kinetic energy, $K$, of the electron?
A freshly prepared sample of a radioisotope of half-life $1386$ s has activity $10^{3}$ disintegrations per second. Given that $\ln 2=0.693$, the fraction of the initial number of nuclei (expressed in nearest integer percentage) that will decay in the first $80$ s after preparation of the sample is
Paragraph for Questions 11 and 12: The mass of a nucleus $^{A}_{Z}X$ is less than the sum of the masses of $(A-Z)$ number of neutrons and $Z$ number of protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. A heavy nucleus of mass $M$ can break into two light nuclei of masses $m_1$ and $m_2$ only if $(m_1+m_2)<M$. Also two light nuclei of masses $m_3$ and $m_4$ can undergo complete fusion and form a heavy nucleus of mass $M'$ only if $(m_3+m_4)>M'$. The masses of some neutral atoms are given in the table below: $^{1}_{1}H$: $1.007825$ u; $^{2}_{1}H$: $2.014102$ u; $^{3}_{1}H$: $3.016050$ u; $^{4}_{2}He$: $4.002603$ u; $^{6}_{3}Li$: $6.015123$ u; $^{7}_{3}Li$: $7.016004$ u; $^{70}_{30}Zn$: $69.925325$ u; $^{82}_{34}Se$: $81.916709$ u; $^{152}_{64}Gd$: $151.919803$ u; $^{206}_{82}Pb$: $205.974455$ u; $^{209}_{83}Bi$: $208.980388$ u; $^{210}_{84}Po$: $209.982876$ u. ($1$ u $=932\ \text{MeV}/c^{2}$) The correct statement is
Paragraph for Questions 11 and 12: The mass of a nucleus $^{A}_{Z}X$ is less than the sum of the masses of $(A-Z)$ number of neutrons and $Z$ number of protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. A heavy nucleus of mass $M$ can break into two light nuclei of masses $m_1$ and $m_2$ only if $(m_1+m_2)<M$. Also two light nuclei of masses $m_3$ and $m_4$ can undergo complete fusion and form a heavy nucleus of mass $M'$ only if $(m_3+m_4)>M'$. The masses of some neutral atoms are given in the table below: $^{1}_{1}H$: $1.007825$ u; $^{2}_{1}H$: $2.014102$ u; $^{3}_{1}H$: $3.016050$ u; $^{4}_{2}He$: $4.002603$ u; $^{6}_{3}Li$: $6.015123$ u; $^{7}_{3}Li$: $7.016004$ u; $^{70}_{30}Zn$: $69.925325$ u; $^{82}_{34}Se$: $81.916709$ u; $^{152}_{64}Gd$: $151.919803$ u; $^{206}_{82}Pb$: $205.974455$ u; $^{209}_{83}Bi$: $208.980388$ u; $^{210}_{84}Po$: $209.982876$ u. ($1$ u $=932\ \text{MeV}/c^{2}$) The kinetic energy (in keV) of the alpha particle, when the nucleus $^{210}_{84}Po$ at rest undergoes alpha decay, is
Match List I of the nuclear processes with List II containing parent nucleus and one of the end products of each process and then select the correct answer using the codes given below the lists: List I -- P. Alpha decay; Q. $\beta^{+}$ decay; R. Fission; S. Proton emission. List II -- 1. $^{15}_{8}O\rightarrow{}^{15}_{7}N+\ldots$; 2. $^{238}_{92}U\rightarrow{}^{234}_{90}Th+\ldots$; 3. $^{185}_{83}Bi\rightarrow{}^{184}_{82}Pb+\ldots$; 4. $^{239}_{94}Pu\rightarrow{}^{140}_{57}La+\ldots$
A nuclear power plant supplying electrical power to a village uses a radioactive material of half life $T$ years as the fuel. The amount of fuel at the beginning is such that the total power requirement of the village is $12.5\%$ of the electrical power available from the plant at that time. If the plant is able to meet the total power needs of the village for a maximum period of $nT$ years, then the value of $n$ is
For a radioactive material, its activity $A$ and rate of change of its activity $R$ are defined as $A = -\dfrac{dN}{dt}$ and $R = -\dfrac{dA}{dt}$, where $N(t)$ is the number of nuclei at time $t$. Two radioactive sources $P$ (mean life $\tau$) and $Q$ (mean life $2\tau$) have the same activity at $t = 0$. Their rates of change of activities at $t = 2\tau$ are $R_P$ and $R_Q$, respectively. If $\dfrac{R_P}{R_Q} = \dfrac{n}{e}$, then the value of $n$ is
A fission reaction is given by $^{236}_{92}\text{U} \to\ ^{140}_{54}\text{Xe} + ^{94}_{38}\text{Sr} + x + y$, where $x$ and $y$ are two particles. Considering $^{236}_{92}\text{U}$ to be at rest, the kinetic energies of the products are denoted by $K_{\text{Xe}}$, $K_{\text{Sr}}$, $K_x$ ($2$ MeV) and $K_y$ ($2$ MeV), respectively. Let the binding energies per nucleon of $^{236}_{92}\text{U}$, $^{140}_{54}\text{Xe}$ and $^{94}_{38}\text{Sr}$ be $7.5$ MeV, $8.5$ MeV and $8.5$ MeV, respectively. Considering different conservation laws, the correct option(s) is(are)
The electrostatic energy of $Z$ protons uniformly distributed throughout a spherical nucleus of radius $R$ is given by $$E = \frac{3}{5}\frac{Z(Z-1)e^2}{4\pi\varepsilon_0 R}.$$ The measured masses of the neutron, $^{1}_{1}$H, $^{15}_{7}$N and $^{15}_{8}$O are $1.008665$ u, $1.007825$ u, $15.000109$ u and $15.003065$ u, respectively. Given that the radii of both the $^{15}_{7}$N and $^{15}_{8}$O nuclei are same, $1$ u $= 931.5$ MeV/$c^2$ ($c$ is the speed of light) and $e^2/(4\pi\varepsilon_0) = 1.44$ MeV fm. Assuming that the difference between the binding energies of $^{15}_{7}$N and $^{15}_{8}$O is purely due to the electrostatic energy, the radius of either of the nuclei is ($1$ fm $= 10^{-15}$ m)
An accident in a nuclear laboratory resulted in deposition of a certain amount of radioactive material of half-life $18$ days inside the laboratory. Tests revealed that the radiation was $64$ times more than the permissible level required for safe operation of the laboratory. What is the minimum number of days after which the laboratory can be considered safe for use?
$^{131}\text{I}$ is an isotope of Iodine that $\beta$ decays to an isotope of Xenon with a half-life of 8 days. A small amount of a serum labelled with $^{131}\text{I}$ is injected into the blood of a person. The activity of the amount of $^{131}\text{I}$ injected was $2.4 \times 10^{5}$ Becquerel (Bq). It is known that the injected serum will get distributed uniformly in the blood stream in less than half an hour. After 11.5 hours, 2.5 ml of blood is drawn from the person's body, and gives an activity of 115 Bq. The total volume of blood in the person's body, in liters is approximately (you may use $e^{x} \approx 1 + x$ for $x \ll 1$ and $\ln 2 \approx 0.7$).
In a radioactive decay chain, $^{232}_{90}\text{Th}$ nucleus decays to $^{212}_{82}\text{Pb}$ nucleus. Let $N_\alpha$ and $N_\beta$ be the number of $\alpha$ and $\beta^-$ particles, respectively, emitted in this decay process. Which of the following statements is (are) true?
In a radioactive sample, $^{40}_{19}\text{K}$ nuclei either decay into stable $^{40}_{20}\text{Ca}$ nuclei with decay constant $4.5 \times 10^{-10}$ per year or into stable $^{40}_{18}\text{Ar}$ nuclei with decay constant $0.5 \times 10^{-10}$ per year. Given that in this sample all the stable $^{40}_{20}\text{Ca}$ and $^{40}_{18}\text{Ar}$ nuclei are produced by the $^{40}_{19}\text{K}$ nuclei only. In time $t \times 10^{9}$ years, if the ratio of the sum of stable $^{40}_{20}\text{Ca}$ and $^{40}_{18}\text{Ar}$ nuclei to the radioactive $^{40}_{19}\text{K}$ nuclei is 99, the value of $t$ will be, [Given: $\ln 10 = 2.3$]
Suppose a $^{226}_{88}\text{Ra}$ nucleus at rest and in ground state undergoes $\alpha$-decay to a $^{222}_{86}\text{Rn}$ nucleus in its excited state. The kinetic energy of the emitted $\alpha$ particle is found to be 4.44 MeV. $^{222}_{86}\text{Rn}$ nucleus then goes to its ground state by $\gamma$-decay. The energy of the emitted $\gamma$ photon is ____ keV. [Given: atomic mass of $^{226}_{88}\text{Ra} = 226.005$ u, atomic mass of $^{222}_{86}\text{Rn} = 222.000$ u, atomic mass of $\alpha$ particle $= 4.000$ u, 1 u $= 931$ MeV/c$^{2}$, c is speed of the light]
There are $10^{10}$ radioactive nuclei in a given radioactive element, its half-life time is 1 minute. How many nuclei will remain after 30 seconds? $\left( {\sqrt 2 = 1.414} \right)$
A heavy nucleus $Q$ of half-life 20 minutes undergoes alpha-decay with probability of 60% and beta-decay with probability of 40%. Initially, the number of $Q$ nuclei is 1000. The number of alpha-decays of $Q$ in the first one hour is
The half life period of radioactive element x is same as the mean life time of another radioactive element y. Initially they have the same number of atoms. Then :
A heavy nucleus $N$, at rest, undergoes fission $N \to P + Q$, where $P$ and $Q$ are two lighter nuclei. Let $\delta = M_N - M_P - M_Q$, where $M_P$, $M_Q$ and $M_N$ are the masses of $P$, $Q$ and $N$, respectively. $E_P$ and $E_Q$ are the kinetic energies of $P$ and $Q$, respectively. The speeds of $P$ and $Q$ are $v_P$ and $v_Q$, respectively. If $c$ is the speed of light, which of the following statement(s) is(are) correct?
The minimum kinetic energy needed by an alpha particle to cause the nuclear reaction $^{16}_{7}\text{N} + {}^{4}_{2}\text{He} \rightarrow {}^{1}_{1}\text{H} + {}^{19}_{8}\text{O}$ in a laboratory frame is $n$ (in $MeV$). Assume that $^{16}_{7}\text{N}$ is at rest in the laboratory frame. The masses of $^{16}_{7}\text{N}$, $^{4}_{2}\text{He}$, $^{1}_{1}\text{H}$ and $^{19}_{8}\text{O}$ can be taken to be $16.006\ u$, $4.003\ u$, $1.008\ u$ and $19.003\ u$, respectively, where $1\ u = 930\ MeV\,c^{-2}$. The value of $n$ is _____.
The binding energy of nucleons in a nucleus can be affected by the pairwise Coulomb repulsion. Assume that all nucleons are uniformly distributed inside the nucleus. Let the binding energy of a proton be $E_b^{p}$ and the binding energy of a neutron be $E_b^{n}$ in the nucleus. Which of the following statement(s) is(are) correct?
In a radioactive decay chain reaction, $^{230}_{90}\text{Th}$ nucleus decays into $^{214}_{84}\text{Po}$ nucleus. The ratio of the number of $\alpha$ to number of $\beta^{-}$ particles emitted in this process is _____.
Substance A has atomic mass number 16 and half life of 1 day. Another substance B has atomic mass number 32 and half life of $\frac{1}{2}$ day. If both A and B simultaneously start undergo radio activity at the same time with initial mass 320 g each, how many total atoms of A and B combined would be left after 2 days.
Given below are two statements: one is labelled as Assertion $\mathbf{A}$ and the other is labelled as Reason $\mathbf{R}$ Assertion A: The nuclear density of nuclides ${ }_{5}^{10} \text{~B},{ }_{3}^{6} \text{Li},{ }_{26}^{56} \text{Fe},{ }_{10}^{20} \text{Ne}$ and ${ }_{83}^{209} \text{Bi}$ can be arranged as $\rho_{\text{Bi}}^{\text{N}}>\rho_{\text{Fe}}^{\text{N}}>\rho_{\text{Ne}}^{\text{N}}>\rho_{\text{B}}^{\text{N}}>\rho_{\text{Li}}^{\text{N}}$ Reason R: The radius $R$ of nucleus is related to its mass number $A$ as $R=R_{0} A^{1 / 3}$, where $R_{0}$ is a constant. In the light of the above statements, choose the correct answer from the options given below
A free neutron decays into a proton but a free proton does not decay into neutron. This is because
The ratio of the density of oxygen nucleus ($_8^{16}O$) and helium nucleus ($_2^{4}\text{He}$) is
List-I shows different radioactive decay processes and List-II provides possible emitted particles. Match each entry in List-I with an appropriate entry from List-II, and choose the correct option. List-I: (P) $^{238}_{92}\text{U} \to {}^{234}_{91}\text{Pa}$ (Q) $^{214}_{82}\text{Pb} \to {}^{210}_{82}\text{Pb}$ (R) $^{210}_{81}\text{Tl} \to {}^{206}_{82}\text{Pb}$ (S) $^{228}_{91}\text{Pa} \to {}^{224}_{88}\text{Ra}$ List-II: (1) one $\alpha$ particle and one $\beta^{+}$ particle (2) three $\beta^{-}$ particles and one $\alpha$ particle (3) two $\beta^{-}$ particles and one $\alpha$ particle (4) one $\alpha$ particle and one $\beta^{-}$ particle (5) one $\alpha$ particle and two $\beta^{+}$ particles
The mass of proton, neutron and helium nucleus are respectively $1.0073~u,1.0087~u$ and $4.0015~u$. The binding energy of helium nucleus is :
If a radioactive element having half-life of $30 \text{~min}$ is undergoing beta decay, the fraction of radioactive element remains undecayed after $90 \text{~min}$. will be
The energy released per fission of nucleus of $^{240}$X is 200 MeV. The energy released if all the atoms in 120g of pure $^{240}$X undergo fission is ____________ $\times$ 10$^{25}$ MeV. (Given $\text{N_A=6\times10^{23}}$)
A nucleus disintegrates into two smaller parts, which have their velocities in the ratio 3 : 2. The ratio of their nuclear sizes will be ${\left( {{x \over 3}} \right)^{{1 \over 3}}}$. The value of '$x$' is :-
In a radioactive decay process, the activity is defined as $A = -\dfrac{dN}{dt}$, where $N(t)$ is the number of radioactive nuclei at time $t$. Two radioactive sources, $S_1$ and $S_2$ have same activity at time $t = 0$. At a later time, the activities of $S_1$ and $S_2$ are $A_1$ and $A_2$, respectively. When $S_1$ and $S_2$ have just completed their 3rd and 7th half-lives, respectively, the ratio $A_1/A_2$ is __________.
Two radioactive substances A and B of mass numbers 200 and 212 respectively, shows spontaneous $\alpha$-decay with same $Q$ value of 1 MeV . The ratio of energies of $\alpha$-rays produced by A and B is $\_\_\_\_$ .
A nuclear reactor starts producing a radioactive nuclide $X$ from $t = 0$, at a constant rate of $\alpha$ per second. Each decay of $X$ produces energy $E_0$, which is utilized to heat a liquid of mass $m$ and specific heat $s$. Assuming no heat loss from the liquid and taking $\lambda$ as the decay constant of $X$, the rate of increase in the temperature of the liquid is:
A radioactive element $_{92}^{242}$X emits two $\alpha$-particles, one electron and two positrons. The product nucleus is represented by $_{\text{P}}^{234}$Y. The value of P is __________.
A radioactive nucleus decays by two different process. The half life of the first process is 5 minutes and that of the second process is $30 \text{~s}$. The effective half-life of the nucleus is calculated to be $\frac{\alpha}{11} \text{~s}$. The value of $\alpha$ is __________.
Assume that protons and neutrons have equal masses. Mass of a nucleon is $1.6\times10^{-27}$ kg and radius of nucleus is $1.5\times10^{-15}~\text{A^{1/3}}$ m. The approximate ratio of the nuclear density and water density is $n\times10^{13}$. The value of $n$ is __________.
Two radioactive elements A and B initially have same number of atoms. The half life of A is same as the average life of B. If $\lambda_{A}$ and $\lambda_{B}$ are decay constants of A and B respectively, then choose the correct relation from the given options.
The half-life of a radioactive nucleus is 5 years. The fraction of the original sample that would decay in 15 years is:
Given below are two statements: one is labelled as Assertion $\mathbf{A}$ and the other is labelled as Reason $\mathbf{R}$ Assertion A : The binding energy per nucleon is practically independent of the atomic number for nuclei of mass number in the range 30 to 170 . Reason R : Nuclear force is short ranged. In the light of the above statements, choose the correct answer from the options given below
For a nucleus ${ }_{\text{A}}^{\text{A}} \text{X}$ having mass number $\text{A}$ and atomic number $\text{Z}$ A. The surface energy per nucleon $\left(b_{\text{s}}\right)=-a_{1} A^{2 / 3}$. B. The Coulomb contribution to the binding energy $\text{b}_{\text{c}}=-a_{2} \frac{Z(Z-1)}{A^{4 / 3}}$ C. The volume energy $\text{b}_{\text{v}}=a_{3} A$ D. Decrease in the binding energy is proportional to surface area. E. While estimating the surface energy, it is assumed that each nucleon interacts with 12 nucleons. ( $a_{1}, a_{2}$ and $a_{3}$ are constants) Choose the most appropriate answer from the options given below:
The half life of a radioactive substance is T. The time taken, for disintegrating $\frac{7}{8}$th part of its original mass will be:
A radio active material is reduced to $1 / 8$ of its original amount in 3 days. If $8 \times 10^{-3} \text{~kg}$ of the material is left after 5 days the initial amount of the material is
A nucleus with mass number 242 and binding energy per nucleon as $7.6~ \text{MeV}$ breaks into two fragment each with mass number 121. If each fragment nucleus has binding energy per nucleon as $8.1 ~\text{MeV}$, the total gain in binding energy is _________ $\text{MeV}$.
A nucleus disintegrates into two nuclear parts, in such a way that ratio of their nuclear sizes is $1: 2^{1 / 3}$. Their respective speed have a ratio of $n: 1$. The value of $n$ is __________.
The decay constant for a radioactive nuclide is 1.5 $\times$ 10$^{-5}$ s$^{-1}$. Atomic weight of the substance is 60 g mole$^{-1}$, ($N_A=6\times10^{23}$). The activity of 1.0 $\mu$g of the substance is ___________ $\times$ 10$^{10}$ Bq.
Two nuclei of mass number 3 combine with another nucleus of mass number 4 to yield a nucleus of mass number 10. If the binding energy per nucleon for the mass numbers 3,4 and 10 are $5.6 \text{MeV}, 7.4 \text{MeV}$ and 6.1 MeV , respectively, then in the process, $\Delta \text{Mc}^2= \_\_\_\_$ MeV .
Assuming the experimental mass of ${ }_6^{12} C$ as $12 u$, the mass defect of ${ }_6^{12} C$ atom is $\_\_\_\_ \text{MeV} / \text{c}^2$. (Mass of proton $=1.00727 \text{u}$. mass of neutron $=1.00866 \text{u}, 1 \text{u}=931.5 \text{MeV} / \text{c}^2$ and c is the speed of the light in vacuum).
Distribution — advanced: 10 · easy: 40 · hard: 20 · medium: 30. Every question carries a source trace; each ends in an SME-verify solution.
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MIT OCW / Walter Lewin, 3Blue1Brown, Khan, Veritasium. Each video is broken into clips — click a clip to play just that segment.
👁 Observe: How the nucleus is built from protons and neutrons; notation A, Z, N and isotopes.
📚 Teaches: Composition of the nucleus and nuclear notation.
👁 Observe: How nuclear radius scales as R=R0*A^(1/3) and why density is nearly constant.
📚 Teaches: Nuclear size and the near-constant nuclear density.
👁 Observe: How mass defect between nucleus and its constituents gives binding energy via E=mc^2.
📚 Teaches: Mass defect, mass-energy equivalence and binding energy.
👁 Observe: How an unstable heavy nucleus splits (fission) and the link to radioactivity.
📚 Teaches: Nuclear fission and radioactivity of heavy nuclei.
👁 Observe: NPTEL lecture on radioactivity and the mechanism of alpha decay.
📚 Teaches: Radioactivity and alpha decay (NPTEL).
👁 Observe: Composition of the nucleus and basic nuclear quantities in Hindi (NCERT PM eVidya).
📚 Teaches: Nabhik part 1: nuclear composition.
👁 Observe: Mass defect and binding energy per nucleon in Hindi (NCERT PM eVidya).
📚 Teaches: Nabhik part 2: mass defect and binding energy.
👁 Observe: Radioactivity, fission and fusion in Hindi (NCERT PM eVidya).
📚 Teaches: Nabhik part 3: decay, fission and fusion.
👁 Observe: Deriving binding energy with the semi-empirical (liquid-drop) mass formula.
📚 Teaches: Binding energy and the liquid-drop model.
👁 Observe: Characterising the nuclear force: short-range, attractive, saturating, charge-independent.
📚 Teaches: The nuclear force.
👁 Observe: How fission liberates energy and the physics of the fragment distribution.
📚 Teaches: Nuclear fission physics.
👁 Observe: How fusion powers stars and the energy released per reaction.
📚 Teaches: Nuclear fusion physics.
👁 Observe: How fusion in the Sun core converts hydrogen to helium, releasing binding energy.
📚 Teaches: Stellar nuclear fusion as the Sun energy source.
👁 Observe: Deriving nuclear size from R=R0 A^(1/3) and computing the huge constant nuclear density.
📚 Teaches: Nuclear size and density, worked numerically.
👁 Observe: Reading the binding-energy-per-nucleon curve and why iron (A~56) is most stable.
📚 Teaches: Binding energy per nucleon curve and stability peak.
👁 Observe: Computing mass defect and converting it to binding energy with E=mc^2 (931.5 MeV/u).
📚 Teaches: Mass defect to binding energy conversion.
👁 Observe: What radioactive decay is and how unstable nuclei emit radiation.
📚 Teaches: Introduction to radioactivity.
👁 Observe: How alpha decay changes Z by -2 and A by -4, emitting a helium nucleus.
📚 Teaches: Alpha decay and the daughter nucleus.
👁 Observe: How beta-minus decay converts a neutron to a proton, raising Z by 1.
📚 Teaches: Beta decay and neutrino emission.
👁 Observe: How gamma emission releases energy without changing Z or A.
📚 Teaches: Gamma decay from excited nuclei.
👁 Observe: How to decide which decay mode (alpha/beta/gamma) an unstable nucleus undergoes.
📚 Teaches: Choosing the decay mode from the N-Z plot.
👁 Observe: Meaning of half-life and how a sample halves every T-half.
📚 Teaches: Half-life of a radioactive sample.
👁 Observe: Worked half-life problem: fraction remaining after several half-lives.
📚 Teaches: Half-life numerical practice.
👁 Observe: Deriving the exponential decay law N=N0 e^(-lambda t) from first-order kinetics.
📚 Teaches: The radioactive decay law.
👁 Observe: Defining activity R=lambda N and mean life tau=1/lambda of a sample.
📚 Teaches: Activity and mean life.
👁 Observe: How neutron-induced fission of U-235 releases energy and sustains a chain reaction.
📚 Teaches: Nuclear fission and chain reaction.
👁 Observe: How light nuclei fuse, releasing energy, and why huge temperatures are needed.
📚 Teaches: Nuclear fusion and the Coulomb barrier.
👁 Observe: Balancing nuclear equations and using mass differences to find reaction energy.
📚 Teaches: Nuclear stability and Q-value of reactions.
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.
Compute the binding energy and the binding energy per nucleon of $^{16}_{8}\text{O}$ given the atomic masses $m(^{1}\text{H})=1.007825\,\text{u}$, $m_n=1.008665\,\text{u}$ and $m(^{16}\text{O})=15.994915\,\text{u}$. Take $1\,\text{u}=931.5\ \text{MeV}/c^2$.
JEE Physics — Nuclei
Assuming each fission of $^{235}\text{U}$ releases about $200\ \text{MeV}$, estimate the energy released by the complete fission of $1\ \text{kg}$ of $^{235}\text{U}$, and compare it with the energy from burning coal ($\approx3\times10^{7}\ \text{J/kg}$). Take $N_A=6.022\times10^{23}\,\text{mol}^{-1}$.
JEE Physics — Nuclei
Find the Q-value of the alpha decay $^{238}_{92}\text{U}\to{}^{234}_{90}\text{Th}+{}^{4}_{2}\text{He}$ given $m(^{238}\text{U})=238.05079\,\text{u}$, $m(^{234}\text{Th})=234.04363\,\text{u}$ and $m(^{4}\text{He})=4.00260\,\text{u}$. Is the decay energetically allowed?
JEE Physics — Nuclei
The activity of a radioactive sample falls from $8000\ \text{Bq}$ to $1000\ \text{Bq}$ in $15$ days. Find the half-life, the decay constant $\lambda$, and the mean life $\tau$ of the nuclide.
JEE Physics — Nuclei
Calculate the activity of $1\ \text{g}$ of pure $^{226}\text{Ra}$, whose half-life is $1600\ \text{years}$. Take $1\ \text{year}=3.156\times10^{7}\ \text{s}$ and $N_A=6.022\times10^{23}\,\text{mol}^{-1}$. Express the result in becquerel and in curie.
JEE Physics — Nuclei
Using $R=R_0A^{1/3}$ with $R_0=1.2\ \text{fm}$ and taking the nucleon mass as $1.66\times10^{-27}\ \text{kg}$, show that nuclear matter density is about $2.3\times10^{17}\ \text{kg/m}^3$ and independent of the mass number $A$.
JEE Physics — Nuclei
The Sun radiates energy at $L=3.8\times10^{26}\ \text{W}$ mainly through the proton–proton cycle, $4\,{}^{1}\text{H}\to{}^{4}\text{He}+2e^{+}+2\nu+6\gamma+26.7\ \text{MeV}$. Estimate (a) the number of protons consumed per second and (b) the mass converted into energy per second.
JEE Physics — Nuclei
Find the energy released in the fusion reaction $^{2}_{1}\text{H}+{}^{3}_{1}\text{H}\to{}^{4}_{2}\text{He}+n$ given $m(^{2}\text{H})=2.014102\,\text{u}$, $m(^{3}\text{H})=3.016049\,\text{u}$, $m(^{4}\text{He})=4.002603\,\text{u}$ and $m_n=1.008665\,\text{u}$. Why does this reaction need millions of kelvin to start?
JEE Physics — Nuclei
A radioactive sample has a half-life of $20\ \text{minutes}$. Starting with $N_0$ undecayed nuclei, find the fraction that remains after one hour, the fraction that has decayed, the decay constant $\lambda$, and the mean life $\tau$ of the nuclide.
JEE Physics — Nuclei
Estimate the age of an ancient wooden artefact whose carbon-14 activity is measured to be only one-eighth of the activity found in a living sample of the same mass. Take the half-life of carbon-14 to be $5730\ \text{years}$, and explain briefly why the method works.
JEE Physics — Nuclei
Calculate the Q-value, in kilo-electron-volts, of the beta-minus decay of tritium, $^{3}_{1}\text{H}\to{}^{3}_{2}\text{He}+e^-+\bar{\nu}$, given the atomic masses $m(^{3}\text{H})=3.016049\ \text{u}$ and $m(^{3}\text{He})=3.016029\ \text{u}$. Explain why the electron mass need not be subtracted separately.
JEE Physics — Nuclei
Determine the binding energy of the deuteron $^{2}_{1}\text{H}$, and its binding energy per nucleon, using the atomic masses $m(^{1}\text{H})=1.007825\ \text{u}$ and $m(^{2}\text{H})=2.014102\ \text{u}$ together with the neutron mass $m_n=1.008665\ \text{u}$.
JEE Physics — Nuclei
In the alpha decay $^{238}_{92}\text{U}\to{}^{234}_{90}\text{Th}+{}^{4}_{2}\text{He}$ the released energy is $Q=4.27\ \text{MeV}$. Find how this energy is shared between the emitted alpha particle and the recoiling thorium nucleus, and use the result to explain why alpha particles emerge with a sharp, definite energy.
JEE Physics — Nuclei
Estimate the temperature at which the average thermal kinetic energy of nuclei becomes comparable to a Coulomb barrier of about $400\ \text{keV}$, using $E=\dfrac{3}{2}kT$ with Boltzmann's constant $k=8.62\times10^{-5}\ \text{eV K}^{-1}$. Compare your answer with the Sun's core temperature of about $1.5\times10^{7}\ \text{K}$ and comment on how fusion nonetheless proceeds there.
JEE Physics — Nuclei
A nuclear reactor delivers a thermal power of $1000\ \text{MW}$ from the fission of $^{235}\text{U}$, with each fission releasing about $200\ \text{MeV}$. Find the number of fissions required per second and the mass of $^{235}\text{U}$ consumed per day. Take Avogadro's number as $6.02\times10^{23}\ \text{mol}^{-1}$.
JEE Physics — Nuclei
Using the binding-energy-per-nucleon curve, estimate the energy released when a heavy nucleus of mass number $A=240$, with a binding energy per nucleon of about $7.6\ \text{MeV}$, splits into two equal fragments each of mass number $120$, whose binding energy per nucleon is about $8.5\ \text{MeV}$. Explain why this is where the roughly $200\ \text{MeV}$ of a fission comes from.
JEE Physics — Nuclei
| Chapter-mock score | Percentile band | Projected AIR band |
|---|---|---|
| 281–300 | 99.9+ | 1 – 200 |
| 251–280 | 99.5 – 99.9 | 200 – 3,000 |
| 221–250 | 99.0 – 99.5 | 3,000 – 8,000 |
| 181–220 | 98.0 – 99.0 | 8,000 – 20,000 |
| 141–180 | 95.0 – 98.0 | 20,000 – 55,000 |
| 101–140 | 90.0 – 95.0 | 55,000 – 1,10,000 |
| 61–100 | 78 – 90 | 1,10,000 – 2,60,000 |
| 31–60 | 55 – 78 | 2,60,000 – 5,30,000 |
Historical NTA JEE Main percentile trends (indicative).
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Authoritative & comprehensive JEE Main + Advanced resource · sources traced Tier 1–3 · SME-review state (append ?review=1)