Nuclei

Chapter 12 ended with Rutherford's picture: almost the whole mass of an atom, and all of its positive charge, sits in a nucleus about 10⁻¹⁴ m across, inside an atom about 10⁻¹⁰ m across. The nucleus is therefore roughly 10⁴ times smaller in radius than the atom, so its volume is about 10¹² times smaller. Nearly all the atom's mass is packed into that tiny volume, which already tells you the nucleus must be extraordinarily dense.

This chapter asks the next set of questions. What is the nucleus made of? How big is it, and does its size follow a rule? Why is the measured mass of a nucleus always less than the sum of the masses of its constituents, and what does that missing mass have to do with energy? What force holds positively charged protons together against their Coulomb repulsion? And finally, how do fission and fusion release energy on a scale that powers reactors and stars?

In the current rationalised syllabus the chapter has six parts: composition of the nucleus, its size, mass-energy and binding energy, the nuclear force, and nuclear energy through fission and fusion. Radioactivity (decay law, half-life, alpha, beta and gamma decay) and the nuclear reactor were removed from the CBSE syllabus in 2023-24, so board questions from before that year on half-life are no longer in scope. Almost every mark in this chapter is earned by understanding one graph, the binding energy per nucleon against mass number, and by being fluent with the conversion 1 u = 931.5 MeV/c².

Atomic mass unit. Atomic masses are tiny, so they are measured in the atomic mass unit u, defined as 1/12 of the mass of a carbon-12 atom: 1 u = 1.660539 × 10⁻²⁷ kg. Mass spectrometers measure atomic masses very precisely, and they show that most elements are mixtures of isotopes, atoms with the same atomic number but different masses. Chlorine, for example, has isotopes of masses 34.98 u (75.4%) and 36.98 u (24.6%); the weighted average, (75.4 × 34.98 + 24.6 × 36.98)/100 = 35.47 u, is the atomic mass you see in the periodic table. Hydrogen has three isotopes: protium (99.985%), deuterium (0.015%) and tritium (unstable, made artificially).

Constituents. The lightest hydrogen nucleus is the proton, with charge +e and mass m_p = 1.00727 u = 1.67262 × 10⁻²⁷ kg, about 1836 times the electron mass. The number of protons is the atomic number Z, and it fixes the element. Since deuterium and tritium have one proton each but roughly two and three times the mass, the nucleus must also contain neutral particles. Chadwick discovered the neutron in 1932 by bombarding beryllium with alpha particles: ⁴₂He + ⁹₄Be → ¹²₆C + ¹₀n. The neutron is neutral, with m_n = 1.00866 u = 1.6749 × 10⁻²⁷ kg, very slightly heavier than the proton. A free neutron is unstable (it decays into a proton, an electron and an antineutrino with a mean life of about 1000 s) but it is stable inside a nucleus.

Notation. A nuclide is written ^A_ZX where Z is the number of protons, N is the number of neutrons and A = Z + N is the mass number, the total number of nucleons. Gold-197 has Z = 79, so N = 118. The three families you must be able to name: isotopes have the same Z (¹H, ²H, ³H); isobars have the same A (³H and ³He); isotones have the same N (¹⁹⁸Hg and ¹⁹⁷Au, both with 118 neutrons).

Rutherford's alpha-scattering gave the first estimate of nuclear size. Later experiments that scattered fast electrons off many different nuclei showed a clean empirical rule: a nucleus of mass number A has radius

R = R₀A^1/3, with R₀ = 1.2 × 10⁻¹⁵ m = 1.2 fm.

Radius grows only as the cube root of A, so volume, (4/3)πR³ = (4/3)πR₀³A, grows in direct proportion to A. Each nucleon takes up about the same volume no matter how large the nucleus. This is the fact behind the most common 3-mark derivation in the chapter.

Nuclear density is independent of A. Take the mass of the nucleus as A × m, where m is the average nucleon mass (about 1.67 × 10⁻²⁷ kg, ignoring the small binding-energy correction). Then

ρ = mass/volume = A m / [(4/3)πR₀³A] = 3m / (4πR₀³).

A cancels, so every nucleus has the same density. Putting in the numbers: ρ = 3 × 1.67 × 10⁻²⁷ / [4 × 3.14 × (1.2 × 10⁻¹⁵)³] = 5.01 × 10⁻²⁷ / 2.17 × 10⁻⁴⁴ ≈ 2.3 × 10¹⁷ kg m⁻³. Compare water at 10³ kg m⁻³: nuclear matter is about 10¹⁴ times denser. This is why the atom is mostly empty space.

Radius ratios. Since R ∝ A^1/3, R₁/R₂ = (A₁/A₂)^1/3. The 2020 board question gave the radius of ²⁷Al as 3.6 fm and asked for ⁶⁴Cu: R = 3.6 × (64/27)^1/3 = 3.6 × 4/3 = 4.8 fm. Examiners choose mass numbers that are perfect cubes or ratios of cubes (1, 8, 27, 64, 125, 216), so look for that before reaching for a calculator.

Before Einstein, mass and energy were conserved separately. In 1905 Einstein showed that mass is a form of energy, and that a mass m is equivalent to energy

E = mc²,

where c = 3 × 10⁸ m s⁻¹. The conversion factor c² = 9 × 10¹⁶ m² s⁻² is enormous, which is why ordinary chemistry never notices the mass change. NCERT Example 13.2 makes the point: 1 g of any substance is equivalent to E = 10⁻³ × 9 × 10¹⁶ = 9 × 10¹³ J, a colossal amount of energy.

The consequence for physics is that the law of conservation of energy now includes mass. In a nuclear reaction the total mass of the products can differ from the total mass of the reactants, and the difference appears as kinetic energy, radiation, or is supplied from the kinetic energy of the incoming particles. So we speak of conservation of mass-energy, not of mass alone.

The number you will use constantly. The energy equivalent of one atomic mass unit is

E = 1.6605 × 10⁻²⁷ × (2.9979 × 10⁸)² = 1.4924 × 10⁻¹⁰ J = 1.4924 × 10⁻¹⁰ / 1.602 × 10⁻¹⁹ eV = 931.5 × 10⁶ eV.

So 1 u = 931.5 MeV/c². Nuclear masses are therefore often quoted in MeV/c², and a mass difference Δm in u converts to energy simply as Δm × 931.5 MeV. Every binding-energy and Q-value numerical in this chapter is a mass bookkeeping exercise followed by this one multiplication.

Mass defect. The measured mass of a nucleus is always less than the sum of the masses of its free protons and neutrons. For oxygen-16: 8 protons + 8 neutrons = 8 × 1.00727 + 8 × 1.00866 = 16.12744 u, while the mass of the ¹⁶O nucleus (atomic mass 15.99493 u minus 8 electrons) is 15.99053 u. The difference, ΔM = 0.13691 u, is the mass defect. In general

ΔM = [Z m_p + (A − Z) m_n] − M.

Binding energy. By E = mc², the mass defect is the energy that was released when the nucleons came together, and equally the energy you must supply to pull the nucleus apart into free nucleons: E_b = ΔM c². For ¹⁶O, E_b = 0.13691 × 931.5 = 127.5 MeV. A more useful measure is the binding energy per nucleon, E_bn = E_b/A, the average energy needed to remove one nucleon; for ¹⁶O it is 127.5/16 = 7.97 MeV. The higher E_bn, the more stable the nucleus.

The curve. Plot E_bn against A for all stable nuclei and you get the most examined graph of the chapter. Its features: (i) E_bn is practically constant, close to 8 MeV, for 30 < A < 170, with a broad maximum of about 8.75 MeV at A = 56 (iron) and a value of 7.6 MeV at A = 238 (uranium); (ii) it is lower for light nuclei (A < 30) and falls slowly for heavy nuclei (A > 170); (iii) light nuclei such as ⁴He show small sharp peaks above the trend.

Conclusions the examiner wants. First, the force between nucleons is attractive and strong enough to bind by several MeV per nucleon. Second, the constancy of E_bn over the middle range shows the nuclear force is short-ranged and saturates: a nucleon interacts only with its few nearest neighbours, so adding more nucleons does not raise the energy per nucleon. For heavy nuclei the Coulomb repulsion, which is long-ranged and acts between all proton pairs, slowly lowers E_bn. Third, a heavy nucleus of A = 240 (E_bn ≈ 7.6 MeV) that splits into two fragments of A = 120 (E_bn ≈ 8.5 MeV) moves to more tightly bound products, so energy is released: this is fission. Fourth, two very light nuclei (A ≤ 10) that join to form a heavier one move up the steep left side of the curve, so energy is again released: this is fusion. In both cases the products are more tightly bound than the reactants and the difference in binding energy is released.

Protons packed into a few femtometres repel each other with Coulomb forces far larger than anything in chemistry, and the gravitational attraction between nucleons is hopelessly weak. A binding energy of about 8 MeV per nucleon needs a new force, the strong nuclear force. Its properties are read off the experiments and from the binding-energy curve:

1. Strongest force known. It is much stronger than the Coulomb force and about 10³⁸ times the gravitational force between two protons; otherwise no nucleus with more than one proton could exist.
2. Short range and saturating. It falls to practically zero for separations beyond a few femtometres. Each nucleon therefore interacts only with its near neighbours, which is why E_bn is roughly constant from A = 30 to 170. The Coulomb force, in contrast, is long-ranged and is felt between every pair of protons.
3. Attractive, then repulsive. The potential energy of a pair of nucleons plotted against their separation r has a minimum at r₀ ≈ 0.8 fm. For r > r₀ the force is attractive; for r < r₀ it is strongly repulsive, which stops the nucleus collapsing. This graph is a regular 2- or 3-mark question.
4. Charge independent. The n–n, p–p and n–p nuclear forces are very nearly equal; the force does not depend on electric charge.
5. No simple law. Unlike Coulomb's inverse-square law there is no compact mathematical form; the force is described from experiment. It also depends on the relative spins of the nucleons, a point beyond this course.

When asked to 'draw the plot of potential energy of a pair of nucleons as a function of their separation', sketch a curve that is large and positive at very small r, crosses zero, drops to a minimum at about 0.8 fm, then rises back to zero and stays there beyond a few fm. Mark the attractive region (r > r₀) and the repulsive region (r < r₀), and write two features beside it: short range and charge independence.

The binding-energy curve says that a heavy nucleus can lower its energy by splitting, and a light pair can lower theirs by joining. Both processes release the binding-energy difference, and both are used as energy sources. Chemical reactions release a few eV per atom; nuclear reactions release a few MeV per nucleon, about a million times more per kilogram.

Fission was discovered in 1939 by Hahn and Strassmann. When a slow (thermal) neutron is absorbed by uranium-235 the compound nucleus ²³⁶U is highly excited and splits into two medium-mass fragments plus two or three fresh neutrons. One common channel:

¹₀n + ²³⁵₉₂U → ²³⁶₉₂U → ¹⁴⁴₅₆Ba + ⁸⁹₃₆Kr + 3 ¹₀n

Other channels give ¹³³Sb + ⁹⁹Nb + 4n, or ¹⁴⁰Xe + ⁹⁴Sr + 2n. The fragments are themselves unstable and undergo further decays. Check the bookkeeping in any equation you write: Z and A must balance on both sides (92 = 56 + 36; 236 = 144 + 89 + 3).

Energy released. The parent has E_bn ≈ 7.6 MeV, the fragments about 8.5 MeV, so roughly 0.9 MeV per nucleon × 236 nucleons ≈ 200 MeV is released per fission. NCERT quotes Q ≈ 200 MeV. It appears first as kinetic energy of the fragments and neutrons and is then transferred to the surrounding matter as heat. One kilogram of ²³⁵U contains (1000/235) × 6.022 × 10²³ ≈ 2.56 × 10²⁴ nuclei, so complete fission of 1 kg gives 2.56 × 10²⁴ × 200 × 1.6 × 10⁻¹³ J ≈ 8.2 × 10¹³ J.

Chain reaction. The two or three neutrons released can trigger further fissions. If on average more than one of them does, the number of fissions grows geometrically: uncontrolled, this is an atom bomb; controlled so that exactly one neutron per fission causes the next fission, it is a nuclear reactor, the source of nuclear electricity. The reactor's design (moderator, control rods, coolant) is no longer in the CBSE syllabus, but the one-line idea of the chain reaction still appears in case-based questions.

Fusion is the joining of two light nuclei into a heavier one. Because the product sits higher on the binding-energy curve, energy is released. Three deuterium reactions from NCERT:

¹₁H + ¹₁H → ²₁H + e⁺ + ν + 0.42 MeV
²₁H + ²₁H → ³₂He + n + 3.27 MeV
²₁H + ²₁H → ³₁H + ¹₁H + 4.03 MeV

Why it is hard. Two nuclei must come within a few fm for the nuclear force to act, but until then they repel by the Coulomb force. The height of the barrier for two protons is about 400 keV, and higher for nuclei of larger charge. A gas whose particles have this much thermal energy on average, using (3/2)kT ≈ 400 keV, would be at about 3 × 10⁹ K. Fusion is therefore called a thermonuclear reaction: it needs extreme temperature and pressure, which exist only in the interiors of stars. In practice the required temperature is lower than the crude estimate, because in a gas at a given temperature a fraction of particles has energies well above the average (the Maxwellian tail), and those are the ones that fuse. The Sun's core, at about 1.5 × 10⁷ K, is hot enough.

The proton-proton chain. The Sun's energy comes from converting hydrogen to helium in these steps:

¹₁H + ¹₁H → ²₁H + e⁺ + ν + 0.42 MeV
e⁺ + e⁻ → γ + γ + 1.02 MeV
²₁H + ¹₁H → ³₂He + γ + 5.49 MeV
³₂He + ³₂He → ⁴₂He + ¹₁H + ¹₁H + 12.86 MeV

The first three steps occur twice for every fourth step. Adding everything up, the net reaction is

4 ¹₁H + 2 e⁻ → ⁴₂He + 2ν + 6γ + 26.7 MeV.

Four hydrogen atoms become one helium atom with the release of 26.7 MeV. Helium is not further burnt while hydrogen lasts; the Sun has burnt hydrogen for about 5 × 10⁹ years and has enough for about the same again. When the hydrogen runs out the core contracts, heats up, and helium begins to fuse into heavier elements, but that stage is beyond the syllabus.

Fission reactors work; fusion reactors do not yet exist commercially, and the reason is the temperature. The natural fusion process in the Sun is slow (the first proton-proton step has a very low probability), so a reactor would use faster reactions with deuterium and tritium, such as

²₁H + ³₁H → ⁴₂He + n + 17.59 MeV,

together with the deuterium-deuterium reactions releasing 3.27 MeV and 4.03 MeV. The fuel is attractive: deuterium is present in ordinary water, and the products are not long-lived radioactive waste.

To make the reaction go at a useful rate the fuel must be heated to about 10⁸ K. At that temperature the gas is fully ionised, a plasma of bare nuclei and electrons, and no material container can hold it; contact with any wall would cool the plasma and destroy the wall. The central engineering problem is therefore confinement: holding a plasma at 10⁸ K, at sufficient density, for long enough that fusion releases more energy than was spent heating it. Magnetic confinement (the tokamak) uses strong magnetic fields to keep charged particles away from the walls; inertial confinement compresses a fuel pellet with intense laser pulses. Large international projects are working on both, and the chapter notes that a practical fusion reactor is still some way off.

What the board can ask from this short section: the reaction and its Q value, why the temperature must be so high, why a plasma cannot be held in an ordinary vessel, and one advantage of fusion over fission as an energy source (abundant fuel, no long-lived radioactive fragments). Keep the answers to two or three lines each.