Nuclei

The chapter builds in three connected stages. The first stage is pure description: what a nucleus is made of (protons and neutrons, called nucleons), how nuclides are named and classified as isotopes, isobars or isotones, and how big a nucleus is — leading to the striking fact that nuclear density is the same constant for every nucleus. The second stage is energetics: Einstein's mass-energy equivalence, the mass defect, the binding energy and binding energy per nucleon, the famous $E_{bn}$-versus-$A$ curve, and the short-range, saturating nuclear force that holds nucleons together against the Coulomb repulsion of the protons. The third stage is dynamics: radioactivity (alpha, beta and gamma decay), the exponential decay law with half-life and mean life, and finally nuclear energy release through fission and fusion, including the physics of the Sun's proton-proton cycle. Worked examples reuse and extend NCERT's own Examples 13.1 to 13.4.

Composition of the nucleus: protons, neutrons and nucleons

Every atom's positive charge and almost all of its mass sit in a central nucleus built from two kinds of particles collectively called nucleons: the positively charged proton (mass $m_p = 1.00727\ \text{u} = 1.67262\times10^{-27}\ \text{kg}$) and the electrically neutral neutron (mass $m_n = 1.00866\ \text{u} = 1.6749\times10^{-27}\ \text{kg}$), discovered by James Chadwick in 1932 from the anomalous neutral radiation emitted when beryllium was bombarded with alpha particles. A free neutron is unstable, decaying into a proton, an electron and an antineutrino with a mean life of about $1000\ \text{s}$, but inside a nucleus it is stable. The composition of any nucleus is fixed by three numbers:

where: $Z$, the atomic number, is the number of protons (and hence the nuclear charge $+Ze$, and the number of electrons in the neutral atom); $N$, the neutron number, is the number of neutrons; and $A$, the mass number, is the total nucleon count, $A=Z+N$. A nuclear species, or nuclide, is written $^{A}_{Z}\text{X}$, where X is the chemical symbol — for instance gold's principal isotope is $^{197}_{79}\text{Au}$, with $79$ protons and $197-79=118$ neutrons. Because chemical identity is fixed by $Z$ alone, nuclides with the same $Z$ but different $A$ (hence different $N$) are isotopes of the same element and share essentially identical chemistry — hydrogen's isotopes are protium $^1_1\text{H}$, deuterium $^2_1\text{H}$, and tritium $^3_1\text{H}$ (unstable). Nuclides with the same $A$ but different $Z$, such as $^3_1\text{H}$ and $^3_2\text{He}$, are isobars; nuclides with the same $N$ but different $Z$, such as $^{198}_{80}\text{Hg}$ and $^{197}_{79}\text{Au}$, are isotones. The atomic mass unit (u) is defined as one-twelfth the mass of a $^{12}\text{C}$ atom, $1\ \text{u} = 1.660539\times10^{-27}\ \text{kg}$, a far more convenient scale than the kilogram for nuclear-sized masses. Because most elements are naturally occurring mixtures of several isotopes, the atomic mass tabulated for an element is a weighted average over the relative abundances of its isotopes — chlorine's atomic mass of $35.46\ \text{u}$ arises from $75.4\%$ of mass $34.98\ \text{u}$ and $24.6\%$ of mass $36.98\ \text{u}$.

Size and density of the nucleus

Rutherford's alpha-scattering experiments, and later, more precise fast-electron scattering experiments, established that a nucleus behaves as a sphere whose radius depends on its mass number by a strikingly simple rule.

where: $R_0 = 1.2\times10^{-15}\ \text{m} = 1.2\ \text{fm}$ ($1\ \text{fm} = 10^{-15}\ \text{m}$) is an empirical constant common to all nuclei, and $A$ is the mass number. Since the volume of a sphere scales as $R^3$, Equation (2) says the nuclear volume is directly proportional to $A$ — pack in twice as many nucleons and the volume simply doubles, exactly as if nucleons were incompressible droplets of a constant-density liquid packed together. The physical reading is a liquid-drop model of the nucleus, and it makes an immediate, testable prediction: nuclear density should be the same number for every nucleus regardless of size, which is verified experimentally and derived formally below. That constant density, about $2.3\times10^{17}\ \text{kg/m}^3$, is roughly $2\times10^{14}$ times the density of water — a teaspoon of pure nuclear matter would weigh as much as a mountain, which is why neutron stars, whose interiors are compressed to nuclear density, pack more mass than the Sun into a sphere the size of a city.

Mass-energy equivalence

Before nuclear physics, mass and energy were believed to be conserved independently. Einstein's special relativity replaced this with a single unified law: mass is itself a form of energy, and the two can convert into one another.

where: $E$ is the energy equivalent of a mass $m$, and $c = 3\times10^8\ \text{m/s}$ is the speed of light in vacuum. The conversion factor $c^2$ is enormous, so even a tiny mass carries a huge amount of energy — one gram of matter, fully converted, releases $E = 10^{-3}\times(3\times10^8)^2 = 9\times10^{13}\ \text{J}$, comparable to a large power station's output over several days (NCERT Example 13.2). In nuclear physics the practical unit conversion is $1\ \text{u} = 931.5\ \text{MeV}/c^2$, obtained by expressing $1\ \text{u} = 1.6605\times10^{-27}\ \text{kg}$ in joules via Equation (3) and converting to electronvolts (NCERT Example 13.3) — a conversion so central to this chapter that every binding-energy and Q-value calculation ultimately runs through it.

Mass defect and binding energy

If a nucleus were simply its protons and neutrons sitting side by side, its mass would equal the sum of their individual masses. It never does — the nucleus is always lighter. This missing mass is the single most important quantity in nuclear physics.

where: $M$ is the actual (measured) mass of the nucleus, and $\Delta M$ is the mass defect — the difference between the mass of the separated, free nucleons and the mass of the assembled nucleus. For $^{16}_8\text{O}$, eight protons and eight neutrons in isolation weigh $16.12744\ \text{u}$, but the actual nuclear mass is only $15.99053\ \text{u}$, a defect of $\Delta M = 0.13691\ \text{u}$ — small as a fraction of the total, but, through Equation (3), it corresponds to a large amount of energy:

where: $E_b$ is the binding energy of the nucleus — physically, the energy that would have to be supplied to pull the nucleus apart into its separate, free nucleons, or equivalently the energy released when free nucleons assemble into that nucleus. For $^{16}_8\text{O}$, $E_b = 0.13691\times931.5 = 127.5\ \text{MeV}$ (NCERT Example 13.3). Because bigger nuclei naturally have bigger total binding energies simply by having more nucleons, a fairer measure of how tightly a nucleus is bound is the binding energy divided by the number of nucleons:

where: $E_{bn}$, the binding energy per nucleon, is the average energy needed to knock one nucleon out of the nucleus — it is the quantity that actually decides nuclear stability, not the total $E_b$.

The binding energy curve and nuclear stability

Plotting $E_{bn}$ against $A$ for every known nuclide produces one of the most information-dense graphs in physics. It rises steeply for the lightest nuclei, climbs to a broad maximum of about $8.75\ \text{MeV}$ near $A=56$ (iron), stays nearly flat at about $8\ \text{MeV}$ across the middle range $30 < A < 170$, and then declines slowly to about $7.6\ \text{MeV}$ at $A=238$ (uranium).

Two features of the curve carry all the physics. First, the near-constancy of $E_{bn}$ across the middle range is a direct signature of the short range of the nuclear force: a nucleon deep inside a large nucleus interacts only with its immediate neighbours (a saturation property), so adding more nucleons at the surface barely changes the binding felt by nucleons already inside, and $E_{bn}$ stops growing with $A$. Second, because $E_{bn}$ falls away on both sides of the iron peak, any process that moves nuclei toward the peak — splitting a heavy nucleus into two medium ones (fission), or fusing two light nuclei into one heavier one (fusion) — leaves the nucleons more tightly bound than before, and the resulting decrease in total mass is released as energy. This one curve is the entire reason nuclear power plants and stars both work, and it is worked out quantitatively for fission in the derivation below.

The nuclear force

Something must hold protons together against their mutual Coulomb repulsion, and bind neutral neutrons too — an attractive force of a completely different character from both gravity and electromagnetism. Experiments carried out between 1930 and 1950 established its main features.

The nuclear force is, first, far stronger than the Coulomb force at nuclear separations — strong enough to overcome the electrical repulsion between protons packed within a few femtometres of one another, which is why it must dominate for the nucleus to hold together at all. Second, it is short-ranged: it falls rapidly to essentially zero once nucleons are separated by more than a few femtometres, and the potential energy between a pair of nucleons is a minimum (most attractive) near $r_0 \approx 0.8\ \text{fm}$, turning strongly repulsive at smaller separations (nucleons resist being crushed together) and attractive beyond it. Third, it is charge-independent: the proton-proton, neutron-neutron and proton-neutron nuclear interactions are all approximately equal, unlike the Coulomb force, which only protons feel. Unlike Coulomb's law or Newton's law of gravitation, the nuclear force has no simple algebraic form — it is described instead through the potential-energy curve above and, at a deeper level, through the exchange of particles called mesons.

Radioactivity: alpha, beta and gamma decay

Discovered by accident in 1896 when Henri Becquerel found that uranium salts fogged a photographic plate through black paper, radioactivity is the spontaneous transformation of an unstable nucleus into a different, more stable configuration, with the emission of radiation. Three types occur in nature. In alpha decay, the nucleus emits a helium nucleus $^4_2\text{He}$ (an alpha particle), so the parent's mass number falls by 4 and atomic number by 2: $^{A}_{Z}\text{X} \rightarrow {}^{A-4}_{Z-2}\text{Y} + {}^4_2\text{He}$. In beta decay, the nucleus emits an electron ($\beta^-$, with a neutron converting to a proton) or a positron ($\beta^+$, with a proton converting to a neutron), always accompanied by a neutrino or antineutrino; mass number $A$ is unchanged, while $Z$ shifts by $\pm 1$. In gamma decay, a nucleus left in an excited state after alpha or beta emission drops to a lower energy state by emitting a high-energy photon (hundreds of keV or more), with neither $A$ nor $Z$ changing. Stability itself follows a pattern: light stable nuclei need a neutron-to-proton ratio near $1{:}1$, rising to about $3{:}2$ for heavy nuclei (extra neutrons dilute the mutually repelling protons without adding more Coulomb repulsion), and nuclei that stray from this band are the ones that decay.

The law of radioactive decay: half-life, mean life and activity

However a nucleus decays, the rate at which a large sample of identical unstable nuclei decays follows one universal statistical law, since each nucleus decays independently of every other and independently of its own past — decay has no memory.

where: $N_0$ is the number of undecayed nuclei at $t=0$, $N(t)$ is the number still undecayed at time $t$, and $\lambda$ (units $\text{s}^{-1}$) is the decay constant, the probability per unit time that any one nucleus decays — a fixed property of the nuclide, unaffected by temperature, pressure or chemical state. Two practical timescales follow directly from Equation (7):

where: the half-life $T_{1/2}$ is the time for exactly half of any sample to decay, and the mean life $\tau$ is the time for the sample to fall to $1/e$ ($\approx 37\%$) of its initial size — also the true statistical average lifetime of the nuclei. The rate of decay itself, called the activity, is:

where: activity is measured in becquerel ($1\ \text{Bq} = 1$ decay per second) in SI, or in the older non-SI unit curie ($1\ \text{Ci} = 3.7\times10^{10}\ \text{Bq}$). Because $R$ is directly proportional to $N$, activity decays with exactly the same exponential law and the same half-life as the population itself — this is the practical quantity a Geiger counter actually measures, since individual undecayed nuclei cannot be counted directly.

Nuclear energy: fission and fusion

The binding energy curve makes the direction of energy release predictable, and NCERT's own quantitative estimate for fission illustrates the scale involved. In nuclear fission, a heavy nucleus such as $^{235}_{92}\text{U}$, on absorbing a slow neutron, splits into two intermediate-mass fragments plus two or three free neutrons, for example $^1_0\text{n} + {}^{235}_{92}\text{U} \rightarrow {}^{236}_{92}\text{U} \rightarrow {}^{144}_{56}\text{Ba} + {}^{89}_{36}\text{Kr} + 3\,^1_0\text{n}$, releasing roughly $200\ \text{MeV}$ per fission event (derived quantitatively below), enough that one kilogram of uranium releases about $10^{14}\ \text{J}$ — a million times more energy, per kilogram, than burning coal ($\sim 10^7\ \text{J/kg}$). The freed neutrons can trigger further fissions in neighbouring nuclei, producing the chain reaction that powers both nuclear reactors (controlled) and atomic bombs (uncontrolled). In nuclear fusion, light nuclei combine into a heavier, more tightly bound nucleus, again releasing energy, but only once the nuclei have enough kinetic energy to overcome their mutual Coulomb repulsion at very close range — the Coulomb barrier, roughly $400\ \text{keV}$ for two protons — which is why fusion needs extremely high temperatures ($T \sim 10^7$ to $10^9\ \text{K}$) and is called thermonuclear fusion. This is the Sun's power source: the proton-proton cycle converts four protons into one helium-4 nucleus in a multi-step sequence, $4\,{}^1_1\text{H} + 2e^- \rightarrow {}^4_2\text{He} + 2\nu + 6\gamma + 26.7\ \text{MeV}$, releasing $26.7\ \text{MeV}$ per helium nucleus formed, and it is estimated the Sun has enough hydrogen fuel to continue for roughly another five billion years. Controlled thermonuclear fusion on Earth — confining a plasma at $\sim 10^8\ \text{K}$ long enough to sustain net fusion — remains an active engineering frontier, pursued by international collaborations including India, because it promises power generation with abundant fuel and no long-lived radioactive waste.

Board level - binding energy and binding energy per nucleon of nitrogen-14 (NCERT Exercise 13.1 pattern)

Find the binding energy and the binding energy per nucleon of a $^{14}_7\text{N}$ nucleus, given its measured atomic mass $m(^{14}_7\text{N}) = 14.00307\ \text{u}$. Use $m_H = 1.007825\ \text{u}$ (hydrogen atom, proton plus one electron) and $m_n = 1.008665\ \text{u}$.

Step 1. Identify the composition. Nitrogen-14 has $Z=7$ protons and $N=A-Z=14-7=7$ neutrons. Because the given masses are atomic masses (nucleus plus electrons), it is consistent to compute the mass defect using atomic hydrogen mass on one side and the atomic nitrogen mass on the other — the $7$ electrons that would appear on each side cancel automatically, since a neutral nitrogen atom also has $7$ electrons.

Step 2. Compute the mass of the separated constituents. $7\,m_H + 7\,m_n = 7(1.007825) + 7(1.008665) = 7.054775 + 7.060655 = 14.115430\ \text{u}$.

Step 3. Compute the mass defect and binding energy. $\Delta M = 14.115430 - 14.00307 = 0.112360\ \text{u}$. Converting with $1\ \text{u}=931.5\ \text{MeV}/c^2$: $E_b = 0.112360 \times 931.5 \approx 104.66\ \text{MeV}$.

Step 4. Divide by the nucleon count. $E_{bn} = E_b/A = 104.66/14 \approx 7.48\ \text{MeV}$ per nucleon. Answer: $E_b \approx 104.7\ \text{MeV}$, $E_{bn} \approx 7.48\ \text{MeV}$ per nucleon, consistent with nitrogen sitting on the rising, light-mass side of the binding energy curve, below the iron peak of $8.75\ \text{MeV}$.

Competitive level (JEE / NEET) - Q-value of a fusion reaction and exothermic versus endothermic (NCERT Exercise 13.5(i) pattern)

Find the Q-value of the reaction $^1_1\text{H} + {}^3_1\text{H} \rightarrow {}^2_1\text{H} + {}^2_1\text{H}$ and state whether it is exothermic or endothermic. Given $m(^1_1\text{H})=1.007825\ \text{u}$, $m(^3_1\text{H})=3.016049\ \text{u}$, $m(^2_1\text{H})=2.014102\ \text{u}$.

Step 1. Recall the Q-value definition for a reaction $A+b\rightarrow C+d$: $Q = [m_A+m_b-m_C-m_d]\,c^2$, the mass "lost" in the reaction converted to energy — a positive $Q$ releases energy (exothermic), a negative $Q$ requires energy input (endothermic).

Step 2. Sum the initial masses. $m(^1_1\text{H}) + m(^3_1\text{H}) = 1.007825 + 3.016049 = 4.023874\ \text{u}$.

Step 3. Sum the final masses. Two deuterium nuclei: $2\times m(^2_1\text{H}) = 2\times 2.014102 = 4.028204\ \text{u}$.

Step 4. Compute the mass difference and convert to energy. $\Delta m = 4.023874 - 4.028204 = -0.004330\ \text{u}$, so $Q = -0.004330\times931.5 \approx -4.03\ \text{MeV}$. Answer: $Q \approx -4.0\ \text{MeV}$, so the reaction is endothermic — despite both products being "light nuclei fusing," this particular combination actually requires an energy input, because a single triton is already more tightly bound, nucleon for nucleon, than a deuteron; not every rearrangement of light nuclei is downhill on the binding-energy curve. Timing note: about ninety seconds. The entire calculation is bookkeeping of six-decimal masses — write initial minus final without rounding early, since the answer is the difference of two numbers each near 4 u, and premature rounding can flip the sign of Q and hence the exothermic/endothermic conclusion. Always sanity-check the sign against the binding-energy curve: fusing light nuclei is not automatically exothermic unless the products truly sit higher on the curve than the reactants.

Competitive level (JEE / NEET) - half-life and decay constant from a falling activity reading

The activity of a radioactive sample falls from $5000$ disintegrations per minute to $1250$ disintegrations per minute in exactly $30$ minutes. Find (a) the half-life, (b) the decay constant in $\text{s}^{-1}$, and (c) the activity after a further $45$ minutes from the start.

Step 1. Recognise the ratio as a power of one-half. Since activity is proportional to $N$, and both obey the same exponential law, $\dfrac{R(t)}{R_0} = \dfrac{1250}{5000} = \dfrac{1}{4} = \left(\dfrac{1}{2}\right)^2$. A drop to exactly $(1/2)^n$ means precisely $n$ half-lives have elapsed — here $n=2$.

Step 2. Solve for the half-life. Two half-lives take $30$ minutes, so $T_{1/2} = 30/2 = 15\ \text{minutes} = 900\ \text{s}$.

Step 3. Convert to the decay constant. $\lambda = \dfrac{0.693}{T_{1/2}} = \dfrac{0.693}{900} \approx 7.70\times10^{-4}\ \text{s}^{-1}$.

Step 4. Find the activity at $t=45$ minutes using the half-life shortcut rather than re-deriving the exponential. Forty-five minutes is exactly $45/15=3$ half-lives, so $R = R_0\,(1/2)^3 = 5000/8 = 625$ disintegrations per minute. Answer: $T_{1/2}=15\ \text{min}$, $\lambda \approx 7.70\times10^{-4}\ \text{s}^{-1}$, and $R(45\ \text{min}) = 625$ disintegrations per minute. Timing note: under two minutes if you spot the power-of-half shortcut immediately. The trap is reaching for $N=N_0e^{-\lambda t}$ and a calculator's natural log when the given ratio is already a clean power of $1/2$ — always check whether $R/R_0$ is $1/2$, $1/4$, $1/8$, and so on, before doing any logarithm at all, since whole numbers of half-lives solve instantly by inspection.

• Confusing atomic mass with nuclear mass. NCERT problems almost always give atomic masses (nucleus plus electrons). When computing $\Delta M$ for an atom-to-atom reaction, use atomic hydrogen mass $m_H$ (not bare proton mass $m_p$) on the constituent side, so the electron counts cancel automatically; mixing bare nuclear masses with atomic masses silently introduces an error of several electron masses.
• Reading total binding energy where per-nucleon binding energy is needed. A uranium nucleus has a larger total $E_b$ than an iron nucleus simply because it has more nucleons, but a smaller $E_{bn}$ — it is $E_{bn}$, not $E_b$, that decides which way energy flows in fission or fusion.
• Forgetting the cube root in the radius law. $R = R_0A^{1/3}$, not $R_0A$ or $R_0A^3$; a common slip is to treat volume (which is proportional to $A$) as if it were radius.
• Believing beta decay changes the mass number. $\beta^-$ and $\beta^+$ decay change $Z$ by $\pm 1$ while $A$ stays exactly fixed (a neutron converts to a proton, or vice versa, inside the same nucleon count); only alpha decay changes $A$ (by $4$).
• Treating half-life as "when the sample is completely gone." After one half-life, half remains — not zero. After $n$ half-lives, a fraction $(1/2)^n$ remains, and this fraction never reaches exactly zero in finite time.
• Mixing up half-life and mean life. $\tau = T_{1/2}/0.693 \approx 1.44\,T_{1/2}$; the mean life is always the larger number, and substituting one for the other in an exam answer is a common lost mark.
• Assuming a negative Q-value means "fusion never releases energy." Some specific light-nucleus reactions genuinely require energy input (endothermic), as in the worked example above; the general fusion-releases-energy statement is about the broad slope of the binding-energy curve, not a guarantee for every individual reaction.
• Ignoring units when converting $H$ or activity. Curie and becquerel differ by a factor of $3.7\times10^{10}$; forgetting to convert before comparing two activities is a frequent numerical slip.

• Write the standard nuclide notation $^{A}_{Z}\text{X}$ for any nucleus, and correctly classify pairs of nuclides as isotopes, isobars or isotones.
• State and apply the nuclear radius law $R=R_0A^{1/3}$, and explain why it forces nuclear density to be constant across all nuclei.
• Compute mass defect and binding energy from given atomic masses, being careful to keep atomic (not bare nuclear) masses consistent on both sides of the calculation.
• Sketch the binding energy per nucleon versus mass number curve from memory, locate its peak near iron, and use its shape to justify why both fission and fusion release energy.
• List the defining properties of the nuclear force (strong, short-range, charge-independent, saturating) and connect saturation directly to the flatness of the binding energy curve.
• Distinguish alpha, beta and gamma decay by what is emitted and by how each changes $A$ and $Z$.
• Derive and apply the exponential decay law, and convert fluently between decay constant, half-life, mean life and activity.
• Solve half-life word problems by recognising whole numbers of half-lives before reaching for logarithms.
• Compute the Q-value of a nuclear reaction from given masses and correctly classify it as exothermic or endothermic.
• Explain, in physical terms, why fusion requires extremely high temperatures (the Coulomb barrier) while fission does not.

Ask the tutor: Want to work through a mixed problem that combines a mass-defect calculation with a half-life question on the same radioactive sample, step by step, using your own numbers?