Nuclear and Particle Physics: Binding, Moments, the Mass Formula, Nuclear Models, the Two-Nucleon Problem, Decays, Reactions, Detectors and the Quark Model

Section 10 of the GATE Physics paper, at the level of a standard textbook course: the properties of nuclei and the models that explain them, the ways nuclei decay and react, the instruments that accelerate and detect particles, and the classification of elementary particles by their conservation laws. It follows the syllabus: nuclear binding energy and electric and magnetic moments; the semi-empirical mass formula; the liquid-drop and shell models; the nuclear force and the two-nucleon problem; alpha decay, beta decay and electromagnetic transitions; Rutherford scattering, nuclear reactions and their conservation laws; fission and fusion; particle accelerators and detectors; photons, baryons, mesons and leptons; the quark model; and conservation laws, isospin, charge conjugation, parity and time-reversal invariance. Mass-formula coefficients are used only where a question states them; 1 u = 931.494 MeV/c² and e²/(4πε₀) = 1.44 MeV fm.

1. Nuclear size, binding energy and moments

Nuclear radii follow R = r₀A1/3 with r₀ ≈ 1.2 fm (3.6 fm for ²⁷Al), so the volume grows as A and the density, about 2.3 × 10¹⁷ kg/m³, is the same for all nuclei — the first sign that nuclear matter is like a liquid. The binding energy is B = [Zm_H + Nm_n − M(A, Z)]c²; for the deuteron, (1.007276 + 1.008665 − 2.013553) u × 931.494 MeV/u = 2.224 MeV (with the proton mass, since the deuteron mass is a nuclear mass). B/A rises steeply for light nuclei, peaks near ⁵⁶Fe–⁶²Ni at about 8.8 MeV, and falls slowly — so fusion of light nuclei and fission of heavy ones both release energy.

Nuclear magnetic moments are measured in nuclear magnetons μ_N = eħ/(2m_p), 1836 times smaller than μ_B. The proton has μ_p = +2.79 μ_N and the neutron μ_n = −1.91 μ_N; neither is the Dirac value (1 and 0), which shows the nucleons are composite. The electric quadrupole moment Q measures departure from spherical charge: Q > 0 for a prolate (cigar) nucleus, Q < 0 for oblate, and Q = 0 for spin 0 or ½. Nuclei far from closed shells are strongly deformed.

2. The semi-empirical mass formula, the liquid-drop and shell models

The liquid-drop model treats the nucleus as an incompressible charged drop, which gives the semi-empirical (Weizsäcker) mass formula B = a_vA − a_sA2/3 − a_cZ(Z − 1)/A1/3 − a_a(A − 2Z)²/A + δ. Each term has a reason: volume (saturation of short-range forces), surface (fewer neighbours at the surface), Coulomb repulsion, asymmetry (the Pauli principle favours N ≈ Z) and pairing (δ > 0 for even–even, 0 for odd A, < 0 for odd–odd). Setting ∂M/∂Z = 0 at fixed A gives the line of stability, which bends to N > Z for heavy nuclei as Coulomb repulsion grows. The same drop, deformed, explains fission: it happens when Coulomb energy wins over surface energy, roughly Z²/A ≳ 50 for spontaneous fission.

The shell model explains what the drop cannot: extra stability at the magic numbers 2, 8, 20, 28, 50, 82, 126, visible in binding energies, abundances and first excited states. Nucleons move independently in a mean potential; a harmonic-oscillator or Woods–Saxon well alone gives 2, 8, 20, 40, 70, 112, and the observed numbers need a strong spin–orbit term that pushes the j = l + ½ level of each high-l shell down into the shell below. Its prediction for an odd-A nucleus: the spin and parity are those of the last unpaired nucleon. ¹⁷O (one neutron beyond 16) has the neutron in 1d₅/₂, so 5/2⁺; ¹⁵N (a proton hole in 1p₁/₂) is 1/2⁻; parity is (−1)ˡ.

3. The nuclear force and the two-nucleon problem

  • Properties: short range (about 1–2 fm), strongly attractive at intermediate distance with a repulsive core below about 0.5 fm, charge-independent (pp, nn and np forces are nearly equal), spin-dependent, with a non-central tensor component, and saturating (B/A roughly constant).
  • Yukawa’s meson theory: a force carried by a particle of mass m has range ħ/(mc); the pion (about 140 MeV/c²) gives about 1.4 fm, the range of the long-range attraction.
  • The deuteron, the only bound two-nucleon state: B = 2.224 MeV, J^π = 1⁺, isospin 0; no excited bound state, and no bound pp or nn (the spin-singlet force is too weak). Its magnetic moment (0.857 μ_N) is close to but not equal to μ_p + μ_n = 0.880 μ_N, and its quadrupole moment is non-zero: the ground state is mostly ³S₁ with about 4% ³D₁ admixture, which only a tensor force can produce.

4. Radioactive decay: α, β and γ

Decay is random with a constant probability per unit time λ: N = N₀e−λt, activity A = λN, half-life t½ = ln 2/λ and mean life τ = 1/λ = t½/ln 2 (14.43 min for t½ = 10 min). An activity falling from 8000 Bq to 1000 Bq, a factor of 2³, has passed three half-lives. In a chain parent → daughter with λ_P ≪ λ_D, the daughter reaches secular equilibrium, A_D = A_P.

  • α decay: (A, Z) → (A − 4, Z − 2) + α. Momentum sharing gives the α a kinetic energy T_α = Q(A − 4)/A — 4.78 MeV for ²²⁶Ra with Q = 4.87 MeV. The α tunnels through the Coulomb barrier (Gamow), so the half-life depends exponentially on Q−1/2: the Geiger–Nuttall law, which spans more than twenty orders of magnitude in half-life.
  • β decay: n → p + e⁻ + ν̄_e (β⁻), p → n + e⁺ + ν_e inside a nucleus (β⁺), and electron capture. The electron spectrum is continuous because three bodies share Q — the reason Pauli proposed the neutrino. Allowed transitions carry no orbital angular momentum: Fermi (ΔJ = 0, spins antiparallel) and Gamow–Teller (ΔJ = 0, ±1, not 0 → 0), both with no parity change. The weak interaction violates parity (Wu, ⁶⁰Co).
  • γ transitions between nuclear levels are classified by multipolarity, EL or ML, with parity change (−1)ᴸ for EL and (−1)L+1 for ML; rates fall steeply with L, so high-spin differences give long-lived isomers. Internal conversion ejects an atomic electron instead of a photon and is the only route for 0 → 0.

5. Rutherford scattering, nuclear reactions, fission and fusion

A point charge ze of kinetic energy E scattering from a nucleus Ze obeys dσ/dΩ = (zZe²/(4πε₀ · 4E))² / sin⁴(θ/2), and in a head-on collision it stops at the distance of closest approach d = zZe²/(4πε₀E): for a 5 MeV α on gold, d = 2 × 79 × 1.44/5 = 45.5 fm, far outside the nucleus, which is why Rutherford saw a point. Deviations at higher energy measure the nuclear radius.

  • Conservation in reactions a + A → b + B: charge, baryon number (hence nucleon number), energy, momentum, angular momentum and parity (the strong and electromagnetic interactions conserve parity).
  • Q-value Q = (m_a + m_A − m_b − m_B)c². An endoergic reaction (Q < 0) has a laboratory threshold E_th = −Q(1 + m_a/m_A) non-relativistically, because part of the projectile’s energy must go into the motion of the centre of mass. For ¹⁴N(α, p)¹⁷O, Q = −1.19 MeV and E_th = 1.19 × 18/14 = 1.53 MeV.
  • Compound nucleus: at low energy the projectile is absorbed, the energy is shared, and the nucleus "forgets" how it was formed before decaying — which gives sharp resonances.
  • Fission of ²³⁵U by a thermal neutron releases about 200 MeV, mostly as fragment kinetic energy, and 2–3 neutrons that sustain a chain reaction. Fusion of light nuclei needs temperatures of order 10⁸ K to overcome the Coulomb barrier by tunnelling; D + T → ⁴He + n releases 17.6 MeV, and the Sun burns hydrogen by the pp chain.

6. Particle accelerators and detectors

  • Cyclotron: a particle in B circles at f = qB/(2πm), independent of radius, so a fixed-frequency field across the dees accelerates it every half-turn — 15.2 MHz for protons in 1 T. The energy is limited by the relativistic mass increase, which the synchrocyclotron (falling frequency) and synchrotron (rising field, fixed orbit) overcome. Linacs accelerate in a straight line; colliders give the highest centre-of-mass energy.
  • Gas detectors: an ionisation chamber collects the primary ions; a proportional counter multiplies them in proportion to the deposited energy; a Geiger–Müller counter discharges fully, so its pulse size is independent of energy — it counts but cannot measure energy.
  • Scintillation detectors (NaI(Tl)) convert energy to light read by a photomultiplier; semiconductor detectors (HPGe, cooled) create electron–hole pairs at about 3 eV each, and so resolve γ-ray energies far better than NaI.
  • Cherenkov detectors: a charged particle faster than light in the medium (v > c/n) emits a cone of light at cos θ = 1/(nβ) — 48.2° for β → 1 in n = 1.5 — which gives a velocity threshold. Cloud and bubble chambers show tracks; calorimeters measure total energy.

7. Elementary particles, the quark model and conservation laws

The particle families
FamilyMembersContent and interactions
Leptons (spin ½)e, μ, τ and ν_e, ν_μ, ν_τpoint-like; weak (and EM if charged); lepton numbers Lₑ, L_μ, L_τ
Baryons (half-integer spin)p (uud), n (udd), Λ (uds), Σ, Ξ, Ω⁻ (sss)three quarks; B = 1
Mesons (integer spin)π⁺ (ud̄), π⁰, K⁺ (us̄), K⁰ (ds̄)quark–antiquark; B = 0
Gauge bosonsγ, W±, Z⁰, gluonscarry the EM, weak and strong forces

Quarks carry charges +2/3 (u, c, t) and −1/3 (d, s, b), baryon number 1/3 and colour; hadrons are colour singlets. The Gell-Mann–Nishijima relation Q = I₃ + Y/2, with hypercharge Y = B + S, organises them into isospin multiplets: (p, n) is an I = ½ doublet, (π⁺, π⁰, π⁻) an I = 1 triplet. Isospin symmetry — u and d nearly degenerate — is why nuclear forces are charge-independent; it is conserved by the strong interaction and broken by electromagnetism.

What each interaction conserves
QuantityStrongEMWeak
Charge, baryon number, lepton numbersyesyesyes
Strangeness (and other flavours)yesyesno (ΔS = 0, ±1)
Isospin Iyesnono
Parity P, charge conjugation Cyesyesno (maximally violated)
CPyesyesviolated slightly (K⁰, B⁰)

C swaps particles and antiparticles, P inverts space and T reverses time. The weak interaction violates P and C separately (neutrinos are only left-handed, antineutrinos right-handed), CP is violated slightly, and the combination CPT is conserved by every local Lorentz-invariant field theory, which makes particle and antiparticle masses and lifetimes equal. Strange particles are made in pairs by the strong interaction (π⁻ + p → K⁰ + Λ, "associated production") and decay slowly by the weak one (Λ → p + π⁻).

Key takeaways

  • R = 1.2A1/3 fm, constant density; B/A peaks near 8.8 MeV at Fe–Ni; μ_N = eħ/2m_p; Q > 0 prolate.
  • SEMF: volume, surface, Coulomb, asymmetry, pairing. Shell model: magic numbers 2, 8, 20, 28, 50, 82, 126 need spin–orbit; odd-A J^π from the last nucleon.
  • The deuteron’s quadrupole moment proves a tensor force; Yukawa range ħ/(m_πc) ≈ 1.4 fm.
  • τ = t½/ln 2; T_α = Q(A − 4)/A; β spectra are continuous; E_th = −Q(1 + m_a/m_A); d_min = zZe²/(4πε₀E).
  • Q = I₃ + (B + S)/2; strong conserves S and I, weak breaks S, P and C; CPT always holds.

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. The activity of a sample falls from 8000 Bq to 1000 Bq in 12 days. Its half-life, in days, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 4

    8000/1000 = 8 = 2³, so three half-lives have passed: t½ = 12/3 = 4 days. Dividing 12 by the ratio 8 gives 1.5 days, and reading the fall as linear gives no half-life at all.
  2. A nuclide has a half-life of 10.0 min. Its mean life, in minutes, to two decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 14.43

    τ = 1/λ = t½/ln 2 = 10.0/0.6931 = 14.43 min. The mean life is longer than the half-life; multiplying by ln 2 instead gives 6.93 min.
  3. With m_p = 1.007276 u, m_n = 1.008665 u, the deuteron mass 2.013553 u and 1 u = 931.494 MeV/c², the deuteron binding energy, in MeV, to two decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 2.22

    Δm = 1.007276 + 1.008665 − 2.013553 = 0.002388 u, and B = 0.002388 × 931.494 = 2.224 MeV → 2.22. The masses given are nuclear, so no electron masses enter; mixing an atomic hydrogen mass with the nuclear deuteron mass would add 0.511 MeV.
  4. Use B = a_vA − a_sA2/3 − a_cZ(Z − 1)/A1/3 − a_a(A − 2Z)²/A + δ with a_v = 15.8, a_s = 18.3, a_c = 0.714, a_a = 23.2 MeV and δ = +12/√A MeV (even–even). For ⁵⁶Fe (Z = 26), the binding energy per nucleon, in MeV, to two decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 8.76

    Volume 884.8; surface 18.3 × 14.64 = 268.0; Coulomb 0.714 × 26 × 25/3.826 = 121.3; asymmetry 23.2 × 16/56 = 6.6; pairing 12/7.48 = 1.6. B = 884.8 − 268.0 − 121.3 − 6.6 + 1.6 = 490.6 MeV, and B/A = 8.76 MeV. Using Z² in the Coulomb term lowers it to about 8.67.
  5. Taking r₀ = 1.2 fm, the radius of the ²⁷Al nucleus, in fm, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 3.6

    R = r₀A1/3 = 1.2 × 271/3 = 1.2 × 3 = 3.6 fm. Using Z = 13 in place of A gives 2.8 fm; the radius depends on the nucleon number because nuclear density is constant.
  6. ²²⁶Ra decays by α emission with Q = 4.87 MeV. The kinetic energy of the α particle, in MeV, to two decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 4.78

    Momentum balance with the recoiling ²²²Rn gives T_α = Q(A − 4)/A = 4.87 × 222/226 = 4.78 MeV; the daughter takes the remaining 0.09 MeV. Taking T_α = Q ignores the recoil.
  7. A 5.0 MeV α particle approaches a gold nucleus (Z = 79) head-on. With e²/(4πε₀) = 1.44 MeV fm, the distance of closest approach, in fm, to one decimal place, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 45.5

    At closest approach all the kinetic energy is Coulomb potential: d = zZe²/(4πε₀E) = 2 × 79 × 1.44/5.0 = 45.5 fm. Forgetting the α’s charge z = 2 gives 22.8 fm.
  8. The reaction ¹⁴N(α, p)¹⁷O has Q = −1.19 MeV. Taking the masses of α and ¹⁴N as 4 u and 14 u, the non-relativistic laboratory threshold energy of the α, in MeV, to two decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 1.53

    E_th = −Q(1 + m_a/m_A) = 1.19 × (1 + 4/14) = 1.19 × 1.2857 = 1.53 MeV. |Q| = 1.19 MeV is only the energy needed in the centre-of-mass frame; the extra 0.34 MeV goes into moving the centre of mass.
  9. A proton (m = 1.6726 × 10⁻²⁷ kg, e = 1.602 × 10⁻¹⁹ C) circulates in a cyclotron with B = 1.0 T. The cyclotron frequency, in MHz, to one decimal place, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 15.2

    f = eB/(2πm) = 1.602 × 10⁻¹⁹ × 1.0/(2π × 1.6726 × 10⁻²⁷) = 1.524 × 10⁷ Hz = 15.2 MHz. eB/m = 95.8 × 10⁶ rad/s is the angular frequency, not f.
  10. A particle with β ≈ 1 passes through a radiator of refractive index 1.5. The Cherenkov angle, in degrees, to one decimal place, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 48.2

    cos θ = 1/(nβ) = 1/1.5 = 0.667, so θ = 48.2°. The radiation needs β > 1/n = 0.667; sin θ = 1/n (a critical-angle formula) gives 41.8°.
  11. Which set lists only nuclear magic numbers?

    1. 2, 8, 20, 28, 50, 82, 126
    2. 2, 8, 18, 32, 50, 72
    3. 2, 8, 20, 40, 70, 112
    4. 2, 10, 18, 36, 54, 86
    Show answer

    Answer: A — 2, 8, 20, 28, 50, 82, 126

    The observed nuclear closures are 2, 8, 20, 28, 50, 82, 126. 2, 8, 20, 40, 70, 112 is what a harmonic well gives without spin–orbit coupling; 2, 10, 18, 36, 54, 86 are the atomic noble-gas numbers; 2, 8, 18, 32 are 2n².
  12. The shell model predicts the ground-state spin and parity of ¹⁷O (8 protons, 9 neutrons) to be:

    1. 5/2⁺
    2. 1/2⁻
    3. 3/2⁻
    4. 0⁺
    Show answer

    Answer: A — 5/2⁺

    Eight protons and eight neutrons close the 1p shell; the ninth neutron enters 1d₅/₂ (l = 2, even parity), so J^π = 5/2⁺. 1/2⁻ is the p₁/₂ hole of ¹⁵O or ¹⁵N, and 0⁺ belongs to even–even nuclei such as ¹⁶O.
  13. The non-zero electric quadrupole moment of the deuteron shows that:

    1. The nuclear force has a tensor (non-central) component
    2. The nuclear force is charge-dependent
    3. The deuteron has spin 0
    4. The deuteron has an excited bound state
    Show answer

    Answer: A — The nuclear force has a tensor (non-central) component

    A pure ³S₁ state is spherical and has Q = 0; a quadrupole moment needs the ~4% ³D₁ admixture, which a central force cannot mix in and a tensor force can. A spin-0 state could have no quadrupole moment at all, and the deuteron has no bound excited state.
  14. Which of the following statements about conservation laws are correct?

    1. The strong interaction conserves strangeness
    2. The strong interaction conserves isospin
    3. The weak interaction conserves parity
    4. All known interactions conserve baryon number
    Show answer

    Answer: A — The strong interaction conserves strangeness; B — The strong interaction conserves isospin; D — All known interactions conserve baryon number

    Strangeness and isospin are strong-interaction symmetries (associated production, charge independence). Baryon number is conserved by all three. The weak interaction violates parity maximally, as the ⁶⁰Co experiment showed.
  15. Which of these processes violates lepton-number conservation?

    1. n → p + e⁻ + ν_e
    2. n → p + e⁻ + ν̄_e
    3. μ⁻ → e⁻ + ν̄_e + ν_μ
    4. π⁺ → μ⁺ + ν_μ
    Show answer

    Answer: A — n → p + e⁻ + ν_e

    In n → p + e⁻ + ν_e the electron (Lₑ = +1) and the neutrino (Lₑ = +1) give Lₑ = 2 from 0 — forbidden. With the antineutrino the total is 0, which is ordinary β decay. The muon and pion decays balance Lₑ and L_μ separately.
  16. Using Q = I₃ + (B + S)/2, a baryon with I₃ = 0 and strangeness S = −1 has charge:

    1. 0 — for example the Λ (uds)
    2. +1
    3. −1
    4. +½
    Show answer

    Answer: A — 0 — for example the Λ (uds)

    Q = 0 + (1 − 1)/2 = 0: the Λ (and Σ⁰), with quark content uds, has charge 2/3 − 1/3 − 1/3 = 0. Forgetting the baryon number gives −½, which is not a possible charge.
  17. Which statements about radiation detectors are correct?

    1. Cherenkov light is emitted only if the particle’s speed exceeds c/n
    2. The pulse height of a Geiger–Müller counter does not depend on the energy deposited
    3. An HPGe detector resolves γ-ray energies better than a NaI(Tl) scintillator
    4. A scintillation detector needs no photodetector
    Show answer

    Answer: A — Cherenkov light is emitted only if the particle’s speed exceeds c/n; B — The pulse height of a Geiger–Müller counter does not depend on the energy deposited; C — An HPGe detector resolves γ-ray energies better than a NaI(Tl) scintillator

    Cherenkov emission has the threshold β > 1/n. A GM tube discharges fully whatever starts it, so it only counts. HPGe needs about 3 eV per charge pair, NaI far more per photoelectron, so germanium’s statistics are much better. A scintillator’s light must be converted by a photomultiplier or photodiode.