Nuclei — NEET Notes

Atomic masses & the composition of the nucleus

The atomic theory only became quantitative once chemists could weigh atoms. The unit they settled on, the unified atomic mass unit (symbol u, often written amu), is defined as one-twelfth the mass of a neutral carbon-12 atom in its ground state. Numerically 1 u = 1.6605 × 10⁻²⁷ kg. In these units a proton weighs 1.00727 u, a neutron 1.00866 u, and an electron a tiny 0.00055 u. The proton was identified by Rutherford in 1919; Chadwick discovered the neutron in 1932, completing the picture of the nucleus as a bound assembly of protons and neutrons collectively called nucleons.

Any nucleus is specified by two integers. The atomic number Z is the number of protons; the mass number A is the total number of nucleons. The number of neutrons follows automatically as N = A − Z. A nucleus is written ^A_ZX — for instance, ²³⁵₉₂U has 92 protons and 143 neutrons. Atoms of the same element with different mass numbers are isotopes (same Z, different A); nuclei with the same A but different Z are isobars; nuclei with the same N are isotones. NEET routinely tests these three terms in single-stem questions.

Size of the nucleus

Rutherford's alpha-scattering experiment was the first probe of the nucleus, but its dimensions could only be pinned down by later electron and neutron scattering. The result is remarkably simple: the radius of any nucleus follows

R = R₀ A^1/3

Nuclear radius — R₀ ≈ 1.2 fm = 1.2 × 10⁻¹⁵ m

where R₀ ≈ 1.2 fm. Two consequences fall out instantly. First, since the volume V = (4/3)πR³ ∝ A, every nucleon occupies the same volume regardless of which nucleus it sits in. Second, the nuclear density is a universal constant:

NEET 2022 tested the R ∝ A^(1/3) relation directly: when a parent of mass number 189 splits into daughters of mass numbers 125 and 64, the ratio of their radii is (125/64)^(1/3) = 5/4. The cube root keeps surfacing in stems involving fission fragments, so commit it to memory.

Mass-energy equivalence

The most consequential equation in twentieth-century physics is Einstein's 1905 result that mass and energy are interchangeable forms of the same quantity. Energy released in nuclear processes is always traceable to a small drop in total mass:

E = m c²

Einstein, 1905 — mass and energy are equivalent

Numerically, 1 u of mass corresponds to 931.5 MeV of energy. Every binding-energy calculation, every fission Q-value, every fusion yield in the chapter is just a careful bookkeeping of mass differences multiplied by this conversion. NEET 2020 turned this into a one-line calculation: the energy equivalent of 0.5 g of matter is 0.5 × 10⁻³ × (3 × 10⁸)² ≈ 4.5 × 10¹³ J. Quoted in TNT, that is over ten kilotons — illustrating why nuclear processes pack so much more energy than chemical ones, where mass changes are vanishingly small.

Binding energy per nucleon

If you weigh a nucleus carefully, it always weighs less than the sum of its free constituent protons and neutrons. The shortfall is the mass defect:

Δm = [Z m_p + (A − Z) m_n] − M_nucleus

Multiplied by c², this mass defect is the energy that would be required to disassemble the nucleus into separated free nucleons — and equivalently, the energy released when the same nucleons were assembled into the nucleus. This is the binding energy E_b. NCERT defines it both ways and either definition is fair game in NEET. Dividing by A gives the binding energy per nucleon:

E_bn = E_b / A

Plotted against mass number, E_bn rises sharply for light nuclei, peaks broadly around iron (A ≈ 56) at about 8.75 MeV per nucleon, and falls slowly for heavy nuclei. The shape of this curve is the single most important diagram in the chapter — it explains why heavy nuclei undergo fission (moving towards the peak) and why light nuclei undergo fusion (also moving towards the peak). Both processes release energy because both move from less-bound to more-bound configurations.

NEET 2021 turned this into arithmetic. A nucleus of mass number 240 with binding energy per nucleon 7.6 MeV breaks into two fragments of mass number 120 each, with binding energy per nucleon 8.5 MeV. Total binding energy of products: 240 × 8.5 = 2040 MeV. Total binding energy of reactant: 240 × 7.6 = 1824 MeV. Gain in binding energy = 216 MeV — that is exactly the energy released. The arithmetic is mechanical once you know which way the inequality runs.

The nuclear force

Coulomb's law predicts that 92 positively charged protons should fly apart instantly. They do not, because at sub-femtometre distances a much stronger attractive force takes over — the strong nuclear force. It is the strongest force in nature (about 100 times stronger than the electromagnetic force at nuclear distances), but it is also extremely short-ranged. Five properties define it, all of which NCERT lists explicitly:

The fifth property is that the force is exchange-mediated — Yukawa proposed in 1935 that protons and neutrons exchange particles called mesons (later identified with pions), the way charged particles exchange virtual photons. NEET does not test the Yukawa mechanism directly, but it sometimes appears as a distractor option for "carrier of nuclear force." Saturation explains why E_bn is roughly constant for A between 30 and 170 — a nucleon deep inside a heavy nucleus has only a fixed number of immediate neighbours, no matter how big the whole nucleus gets.

Radioactivity — α, β and γ decay

Henri Becquerel stumbled onto radioactivity in 1896 when a uranium salt fogged a photographic plate without any external excitation. Marie and Pierre Curie soon isolated polonium and radium; Rutherford named the three emissions α, β and γ before anyone knew what they were. Today we know each in detail:

Three conservation laws control every decay equation: charge (sum of Z), nucleon number (sum of A), and mass-energy. Apply them mechanically and any decay product can be deduced. NEET 2022 used exactly this method: ²²₁₁Na → X + e⁺ + ν. Conservation of Z: 11 = Z + 1 ⇒ Z = 10 (neon). Conservation of A: 22 = A ⇒ A = 22. Therefore X = ²²₁₀Ne. NEET 2021 ran a similar drill across three sequential decays, asking which combination of β⁺, α and β⁻ reproduces the given Z changes.

Law of radioactive decay & half-life

Radioactive decay is the textbook example of a stochastic process at the level of a single atom that obeys a deterministic law in the bulk. Each unstable nucleus has a fixed probability per unit time of decaying — the decay constant λ. Because this probability does not depend on age, the rate at which nuclei disappear at any moment is proportional to how many are still present:

dN/dt = −λN

Integrating gives the law of radioactive decay:

N(t) = N₀ e^−λt

Exponential decay — the master equation of radioactivity

Two derived quantities matter for NEET. The half-life T₁/₂ is the time taken for N to fall to N₀/2; setting e^(−λt) = 1/2 gives

T₁/₂ = ln 2 / λ ≈ 0.693 / λ

The activity A is the number of disintegrations per second: A = λN. Its SI unit is the becquerel (1 Bq = 1 disintegration per second); the older unit is the curie (1 Ci = 3.7 × 10¹⁰ Bq). Because activity is proportional to N, it also decays as A(t) = A₀ e^(−λt), with the same half-life. Mean life τ = 1/λ ≈ 1.44 T₁/₂.

This is the workflow for NEET 2023 Q.2 (half-life 20 min, activity falls to 1/16 ⇒ t = 80 min), NEET 2018 (half-life 10 min, 540 of 600 nuclei disintegrated ⇒ 90 left ≈ N₀/8 ⇒ n ≈ 3 half-lives ≈ 30 min by official key, though variants exist), and NEET 2021 (T₁/₂ = 100 h, t = 150 h ⇒ A/A₀ = (1/2)^1.5 = 1/(2√2)). When the ratio is not a clean power of 1/2, fall back on the exponential directly.

Nuclear fission

In 1939 Otto Hahn and Fritz Strassmann found barium in a sample of uranium that had been bombarded with neutrons. Lise Meitner and Otto Frisch interpreted the result as fission: a heavy nucleus, having absorbed a slow neutron, becomes unstable and splits into two medium-mass fragments. The canonical example in NEET is uranium-235:

¹n + ²³⁵U → ¹⁴⁴Ba + ⁸⁹Kr + 3n + Q

Q ≈ 200 MeV per fission event

NEET 2020 Q.103 asked precisely this — given ²³⁵U + n → Kr (Z = 36) + 3n + ?, find the missing fragment. Mass numbers: 235 + 1 = 89 + 3 + A ⇒ A = 144. Atomic numbers: 92 + 0 = 36 + 0 + Z ⇒ Z = 56. Answer: ¹⁴⁴₅₆Ba. The energy released, about 200 MeV per fission, comes from the rise in binding energy per nucleon: about 7.6 MeV in U-235 climbing to about 8.5 MeV in the fragments, multiplied by 235 nucleons.

Chain reaction & nuclear reactors

The fission of one U-235 nucleus releases two or three neutrons. If on average more than one of those neutrons goes on to trigger another fission, the reaction sustains itself — a chain reaction. Three regimes are possible. If the multiplication factor k < 1, the chain dies out (sub-critical). If k = 1, the rate of fission is constant (critical) — this is how nuclear reactors operate. If k > 1, the rate doubles repeatedly in microseconds (super-critical) — this is what happens in a bomb.

A working reactor needs four ingredients, all of which NCERT names. Fuel — usually enriched uranium (a few percent ²³⁵U). Moderator — heavy water, graphite, or ordinary water — slows the fast neutrons emitted at fission down to thermal speeds, where they are far more easily absorbed by U-235. Control rods — cadmium or boron — absorb neutrons to keep k = 1 exactly. Coolant — water or liquid sodium — carries off heat to drive a conventional steam turbine. Remove the control rods and prevent moderation, and you have the principle of a fission bomb: an uncontrolled, super-critical mass releasing its full energy in microseconds.

Nuclear fusion — the energy of stars

The other route to the binding-energy peak runs from the light side. Two very small nuclei, brought close enough that the nuclear force can grab them, fuse into a single larger nucleus — and again the products are more tightly bound than the reactants. This is nuclear fusion. The barrier is the Coulomb repulsion of the two positive nuclei: only at temperatures of around 10⁷ K do hydrogen nuclei have enough kinetic energy to overcome it. That is what the core of the Sun supplies for free.

The dominant cycle in the Sun is the proton-proton chain, which can be summarised as

4 ¹H → ⁴He + 2 e⁺ + 2 ν + 26.7 MeV

Net p-p chain in the Sun's core

Four hydrogen nuclei, working through three intermediate steps involving deuterium and helium-3, end up as one helium-4 nucleus, two positrons, two neutrinos, gamma photons and 26.7 MeV. The positrons annihilate with electrons immediately. The Sun has been burning hydrogen this way for 4.5 billion years; it will continue for roughly another 5 billion before it runs out of core hydrogen and contracts into helium burning.

Two notes on terminology. A thermonuclear bomb (or hydrogen bomb) uses a fission bomb as a trigger to generate the temperatures required for fusion; the main yield comes from fusing isotopes of hydrogen (deuterium, tritium) to helium. Controlled thermonuclear fusion on Earth — the goal of ITER, tokamaks, and inertial confinement experiments — remains a long-standing engineering problem precisely because confining a 10⁸ K plasma for long enough to net-gain energy is exceptionally difficult. The Sun's secret is gravity: a billion-billion-billion kilograms of overburden hold the core together.

Per nucleon, fusion releases more energy than fission. The 26.7 MeV of a p-p cycle distributed over 4 nucleons is about 6.7 MeV/nucleon; the 200 MeV of a U-235 fission spread over 235 nucleons is about 0.85 MeV/nucleon. That is why hydrogen bombs are vastly more energetic per kilogram than fission bombs — and why fusion is the holy grail of clean energy.