Nuclear Properties & Structure I

(a) Confining an electron to nuclear dimensions ($\Delta x \sim 10^{-14}$ m) gives, by Heisenberg, $\Delta p \gtrsim \hbar/2\Delta x = 5.3 \times 10^{-21}$ kg m s⁻¹. It is relativistic, so $E \approx pc = 1.58 \times 10^{-12}\ \mathrm{J} \approx 10$ MeV — far above its 0.511 MeV rest energy. Equivalently by de Broglie: an electron with a typical β-decay energy (~1 MeV) has wavelength $\lambda = h/p \approx 870$ fm, ~100× the nuclear diameter — the wave simply does not fit. Yet electrons emitted in β-decay carry ≤ ~1–2 MeV. An electron needing ~10 MeV just to exist inside the nucleus contradicts every β-spectrum ever measured — so no electron resides in the nucleus; β-electrons are created at emission ($n \to p + e^- + \bar{\nu}_e$).

(b) Burning coal is chemistry: rearranging electron bonds, ~4 eV per CO₂ formed. Nuclear fuel must release nuclear binding energy (~200 MeV per fission, ~10⁷× more) via a self-sustaining chain reaction in fissile nuclides ($^{235}\mathrm{U}$, $^{239}\mathrm{Pu}$). Coal contains only ~1–3 ppm uranium (overwhelmingly non-fissile $^{238}\mathrm{U}$) — no fissile inventory, no chain reaction, no nuclear energy. (Its trace radioactivity is actually a pollution problem, not a fuel reserve.)

(a) Head-on ⟹ $\theta = 180^\circ$ in eq. (4): $d\sigma/d\Omega = (ke^2 Z_1Z_2/4E)^2 = (1.44 \times 2 \times 79/24)^2 = 89.9$ fm² $= 8.99 \times 10^{-25}$ cm². Nuclei per unit area: $nt = 19.8 \times 10^{-5} \times 6.022 \times 10^{23}/197 = 6.05 \times 10^{17}$ cm⁻². Fraction per unit area = $nt(d\sigma/d\Omega)/r^2 = 6.05 \times 10^{17} \times 8.99 \times 10^{-25}/12^2 = 3.8 \times 10^{-9}$ cm⁻². (Full steps in §4, Example 3.)

(b) The four decay series are 4n ($^{232}\mathrm{Th}$), 4n+2 ($^{238}\mathrm{U}$), 4n+3 ($^{235}\mathrm{U}$) and 4n+1 (neptunium, $^{237}\mathrm{Np}$). The first three have parents with half-lives comparable to Earth's age (10⁸–10¹⁰ yr) and survive in nature; $^{237}\mathrm{Np}$ has $t_{1/2} = 2.1 \times 10^6$ yr ≪ 4.5×10⁹ yr, so every primordial atom decayed away long ago. The series exists today only when made artificially (e.g. $^{238}\mathrm{U}(n,2n)^{237}\mathrm{U} \to ^{237}\mathrm{Np}$) — hence "artificial".

(a) Energetics. $^{238}\mathrm{U} \to ^{234}\mathrm{Th} + \alpha$: $Q = (238.050788 - 234.043601 - 4.002603) \times 931.5 = +4.27$ MeV — allowed, because the α-particle is extraordinarily tightly bound (28.3 MeV total, 7.07 MeV/nucleon), which pays for the emission. Knocking out a single proton or neutron costs its separation energy (~6–8 MeV) with no such payback ⟹ $Q < 0$, forbidden. The nucleus takes the decay path that releases energy.

(b)

The distinction is origin, not energy — the ranges overlap.

No. Uncertainty principle: $\Delta x \sim 10^{-14}$ m ⟹ $\Delta p \gtrsim 5 \times 10^{-21}$ kg m s⁻¹ ⟹ $E \approx pc \approx 10$ MeV — an order of magnitude above the largest β-decay energies (~1–3 MeV). A nuclear electron would also give nuclei the wrong spin/magnetic moment systematics. β-electrons are created at the instant of decay, not stored inside. (Full numbers in the 2020 card above.)

Pitchblende (uraninite, UO₂) holds $^{210}\mathrm{Po}$ ($t_{1/2} = 138$ d) as a daughter of the $^{238}\mathrm{U}$ series (via $^{210}\mathrm{Pb} \to ^{210}\mathrm{Bi} \to ^{210}\mathrm{Po}$), at only ~100 µg per ton of ore — the Curies' 1898 problem.

Wave nature: a ~1 MeV electron (typical β energy) has de Broglie wavelength $\lambda = h/p = hc/\sqrt{E^2 - m_e^2c^4} \approx 870$ fm — roughly a hundred nuclear diameters. A wave that large cannot be localised inside a ~10 fm nucleus. Particle nature: Heisenberg confinement to $\Delta x \sim 10^{-14}$ m demands $E \approx pc \approx 10$ MeV (computed in the 2020 card), contradicting measured β-spectra (endpoint energies ~0.5–3 MeV). Both natures agree: the electron is not a nuclear constituent.

Prediction: the heavier isobar is unstable — $^{7}\mathrm{Be}$ (7.01693 u > 7.01600 u) decays to $^{7}\mathrm{Li}$. Disintegration energy: $Q = (7.01693 - 7.01600) \times 931.5 = 0.00093 \times 931.5 = 0.87$ MeV. The mode is K-electron capture ($Q < 1.022$ MeV forbids β⁺): $^{7}\mathrm{Be} + e^-_K \to ^{7}\mathrm{Li} + \nu_e$. Spectrum: no charged particle is emitted — EC is a two-body decay, so the neutrino is monoenergetic (invisible in practice); the observable spectrum is a single sharp γ-line at 0.48 MeV from the 89.6% branch to the excited $^{7}\mathrm{Li}^*$ state (10.4% goes to the ground state, neutrino carries the full 0.87 MeV). Sketch: one narrow peak at 0.48 MeV on the energy axis — not a continuous β-spectrum.

Same energetics as the 2020 card: α-emission has $Q = +4.27$ MeV because the α-particle's 28.3 MeV binding energy subsidises the decay; single-nucleon emission costs ~6–8 MeV separation energy with $Q < 0$. Nuclei decay only along energetically open channels.

Coal releases chemical energy (~4 eV per CO₂); nuclear fuel must release nuclear binding energy (~200 MeV per fission) through a sustained chain reaction in fissile material. Coal's ~ppm uranium is essentially all non-fissile $^{238}\mathrm{U}$ — no fissile inventory, no chain reaction. See the full 2020 answer.

Write β⁺ decay with atomic masses (what tables list). Nuclear decay: $m_{nuc}(Z,A) \to m_{nuc}(Z-1,A) + m_e + \cdots$. Atomic mass = nuclear mass + Zmₑ, so $Q = [M_{atom}(Z,A) - Zm_e - (M_{atom}(Z-1,A) - (Z-1)m_e) - m_e]c^2 = [M_{atom}(Z,A) - M_{atom}(Z-1,A) - 2m_e]c^2$. The 2mₑ = one positron mass + one electron mass (the daughter atom has one fewer electron than the parent atom). Hence β⁺ needs $\Delta M > 2m_e$, i.e. 1.022 MeV — while electron capture, $Q = [M_{atom}(Z,A) - M_{atom}(Z-1,A)]c^2$, pays no such tax and wins whenever both compete.

Each $^{235}\mathrm{U}$ fission liberates ~200 MeV (fragment kinetic energy ~168 MeV + neutrons + γ + decay heat ≈ 195 MeV recoverable; ~11 MeV escapes as neutrinos) — about 10⁷× a chemical bond. Crucially it also releases 2–3 new neutrons, which fission further $^{235}\mathrm{U}$ nuclei: a chain reaction. Hold the neutron multiplication factor at exactly k = 1 (critical) and the ~200 MeV per fission appears as steady heat — carried by coolant to raise steam and drive turbines. That controlled chain reaction is the nuclear reactor (see the 2024 diagram card).

Moderator = accelerator pedal (makes neutrons useful); control rods = brake pedal (eats the surplus neutrons). Coolant (H₂O, He, liquid Na) then carries the heat out.

Same machinery as the 2020 numerical, new numbers: $d\sigma/d\Omega = (1.44 \times 2 \times 79/(4 \times 4.5))^2 = (12.64)^2 = 159.8$ fm² $= 1.60 \times 10^{-24}$ cm²; $nt = 6.05 \times 10^{17}$ cm⁻² (same foil). Fraction per unit area = $6.05 \times 10^{17} \times 1.60 \times 10^{-24}/10^2 = 9.7 \times 10^{-9}$ cm⁻². Lower energy ⟹ larger cross section ($\propto 1/E^2$) ⟹ ~2.6× the 2020 value — check limiting cases like this in the exam.

K-electron capture: a proton-rich nucleus captures one of its own K-shell electrons (K-shell has the largest wavefunction overlap with the nucleus): $p + e^-_K \to n + \nu_e$. No β-particle is emitted; the K-vacancy is filled by an outer electron, giving characteristic X-rays (or Auger electrons). With atomic masses the electron masses cancel exactly, $Q_{EC} = [M_{atom}(Z,A) - M_{atom}(Z-1,A)]c^2$ — no threshold penalty. The 1.02 MeV statement: β⁺ decay must create a positron (mₑ) while the daughter atom also sheds one electron, costing 2mₑc² = 1.022 MeV (derived in the 2022 card). So a nucleus with $0 < \Delta Mc^2 < 1.022$ MeV can decay only by electron capture — e.g. $^{7}\mathrm{Be}$ (0.87 MeV). EC always competes with β⁺ and dominates at low Q.

Particle-in-a-box: confining the electron's position to $\Delta x \sim 10^{-14}$ m makes its momentum uncertain by $\Delta p \gtrsim \hbar/2\Delta x \approx 5.3 \times 10^{-21}$ kg m s⁻¹, i.e. a kinetic energy $E \approx pc \approx 10$ MeV. No β-decay releases anywhere near that (endpoints are ~0.5–3 MeV), and a 10 MeV electron would have been seen long ago. Hence the electron cannot pre-exist inside the nucleus.

Proof: $Q = \tfrac{1}{e}\int(3z^2 - r^2)\rho\,d\tau$. Spherical symmetry ⟹ $\langle x^2\rangle = \langle y^2\rangle = \langle z^2\rangle = \langle r^2\rangle/3$, so $3\langle z^2\rangle - \langle r^2\rangle = 0$ and the integral vanishes — Q = 0 exactly. Non-zero Q appears when the nucleus is permanently deformed: for a uniform ellipsoid, $Q = \tfrac{2}{5}Z(a^2-b^2)$ — prolate (cigar, $a > b$) gives Q > 0, oblate (disc, $a < b$) gives Q < 0. Measured Q values (e.g. large positive Q in the rare-earth region) are direct evidence of deformation. (Full derivation: §3, D5.)

Origin, not energy: X-rays come from electron rearrangements (inner-shell vacancies, bremsstrahlung — keV range); γ-rays come from nuclear de-excitation (keV–MeV). Same electromagnetic radiation, different birthplace. See the 2020 comparison table.

Eight points, one line each: short-ranged (~1–2 fm, Yukawa $e^{-\mu r}/r$); strongly attractive (~50× Coulomb at 1 fm); saturating (nearest neighbours only ⟹ constant density, $B \propto A$); charge-independent (p–p ≈ n–n ≈ p–n); spin-dependent (triplet binds the deuteron, singlet doesn't); repulsive core below ~0.5 fm; tensor (non-central) component; exchange character — meson (pion) exchange, Yukawa 1935. (Expanded in §2.11.)

Fission is the splitting of a heavy nucleus into two lighter fragments (plus neutrons and γ-rays), e.g. $^{235}\mathrm{U} + n \to ^{236}\mathrm{U}^* \to ^{141}\mathrm{Ba} + ^{92}\mathrm{Kr} + 3n$. It is exoergic (~200 MeV) because the fragments sit higher on the B/A curve than the parent (§2.7). Discovered by Hahn–Strassmann (1938), explained by Meitner–Frisch with the liquid-drop picture — the drop deforms, necks, and snaps.

The ~200 MeV per fission appears mostly as kinetic energy of the fragments (~168 MeV), which thermalises in the fuel as heat; add neutron (~5 MeV) and prompt-γ (~7 MeV) heating plus decay heat (~15 MeV from the fragments' β/γ decay). A coolant carries this heat to a heat exchanger → steam → turbine → electricity. The reaction sustains itself because each fission emits 2–3 neutrons (chain reaction, held at k = 1 by control rods). ~11 MeV per fission escapes as neutrinos — unrecoverable.

The only stable odd-odd nuclei in nature are $^{2}\mathrm{H}$, $^{6}\mathrm{Li}$, $^{10}\mathrm{B}$ and $^{14}\mathrm{N}$. For even A the pairing term splits isobars into two mass parabolas (Fig. 2, §5): even-even on the lower, odd-odd on the upper, separated by $2\delta$. An odd-odd nucleus therefore always has an even-even isobaric neighbour of lower mass on one side or the other, and β-decays (β⁻ or β⁺/EC) into it. Beyond A = 14 the Coulomb-plus-asymmetry parabolas are steep enough that this is unavoidable — every heavier odd-odd isobar is stranded on the upper parabola with a downhill path. $^{14}\mathrm{N}$ survives only because at A = 14 the parabolas are shallow and shell effects intervene.

The neptunium series' longest-lived member, $^{237}\mathrm{Np}$ ($t_{1/2} = 2.1 \times 10^6$ yr), is far too short-lived to have survived since Earth's formation — unlike the $^{238}\mathrm{U}$, $^{235}\mathrm{U}$ and $^{232}\mathrm{Th}$ series parents. It exists today only when synthesised artificially. See the full 2020 answer.

Stability ⟺ lowest mass. Atomic masses: $^{40}\mathrm{Ar} = 39.962383$ u, $^{40}\mathrm{Ca} = 39.962591$ u, $^{40}\mathrm{K} = 39.963998$ u.

(i) A = 40 is even: ⁴⁰Ar and ⁴⁰Ca are even-even (lower parabola — both stable), ⁴⁰K is odd-odd (upper parabola — unstable, β⁻ to Ca and EC to Ar, t_½ = 1.25×10⁹ yr). Order: ⁴⁰Ar ≥ ⁴⁰Ca ≫ ⁴⁰K — Ar is the absolute minimum by 0.0002 u (≈ 0.19 MeV), a subtlety; the exam point is that the even-even pair both beat the odd-odd member. (Naive parabola minimum $Z_0 = 40/(2+0.015 \times 40^{2/3}) = 18.4$ also favours Ar; Ca's near-degeneracy reflects its Z = 20 magic number.)

(ii) $^{6}\mathrm{Li} = 6.015123$ u < $^{6}\mathrm{He} = 6.018889$ u < $^{6}\mathrm{Be} = 6.019726$ u ⟹ ⁶Li > ⁶He > ⁶Be. Only ⁶Li is stable: ⁶He β⁻-decays (t_½ ≈ 0.8 s) and ⁶Be is proton-unbound (falls apart to α + 2p essentially instantly). ⁶Li is the odd-odd exception at light A (like ²H, ¹⁰B, ¹⁴N), where the parabolas are shallow; the "no stable odd-odd" rule bites only past ¹⁴N.

Head-on: $d_{min} = kZ_1Z_2e^2/E = 1.44 \times 2 \times 79/5.5 = 41.4$ fm $= 4.14 \times 10^{-14}$ m — about 6× the touching distance (8.9 fm), so the scattering is purely Coulombic. (Full steps in §4, Example 4.)

An electron confined to $\Delta x \sim 10^{-14}$ m has $\Delta p \gtrsim \hbar/2\Delta x$, i.e. $E \approx pc \approx 10$ MeV — or in wave language, a β-decay electron's de Broglie wavelength (~870 fm at 1 MeV) dwarfs the nucleus. Both contradict measured β-energies (≤ ~3 MeV). No resident electron; β-particles are created at decay. (Full numbers in the 2020 card.)

Using atomic masses, $Q_{\beta^+} = [M_{atom}(Z,A) - M_{atom}(Z-1,A) - 2m_e]c^2$: one mₑ for the created positron, one for the electron the daughter atom sheds. Positron emission needs the atomic-mass difference to exceed 2mₑc² = 1.022 MeV. (Full derivation in the 2022 card.)

A primordial nuclide is one that has existed since before Earth's formation — i.e. with a half-life comparable to the age of the Earth (~10⁹ yr), so a measurable fraction survives today. Examples: $^{238}\mathrm{U}$ ($t_{1/2} = 4.5 \times 10^9$ yr), $^{235}\mathrm{U}$ (7.0×10⁸ yr), $^{232}\mathrm{Th}$ (1.4×10¹⁰ yr), $^{40}\mathrm{K}$ (1.25×10⁹ yr). (This is also why the 4n+1 series is "artificial" — its parent is not primordial.)

$R = R_0A^{1/3}$, $R_0 \approx 1.2$ fm. Consequence: volume ∝ A, density constant (~10¹⁷ kg m⁻³).

The Coulomb term $-a_cZ(Z-1)/A^{1/3}$: the only long-range, non-saturating term — it grows as $Z^2$ while cohesion grows only as A.

One — the pairing term vanishes for all odd-A nuclei, so there is no even-even/odd-odd split. Only the minimum-Z isobar is stable.

Q > 0: exoergic — proceeds spontaneously, excess appears as kinetic energy. Q < 0: endoergic — needs projectile energy above threshold $E_{thr} = -Q(1+m/M)$.

$1\ \mathrm{barn} = 10^{-28}\ \mathrm{m^2} = 10^{-24}\ \mathrm{cm^2}$ — the unit of nuclear cross section.

The decay of a compound nucleus depends only on its energy, angular momentum and parity — not on how it was formed.

They give the magnetic moment of an odd-A nucleus from its last nucleon (two lines: $j = l \pm 1/2$, separate for odd-p/odd-n). Real moments falling between the lines signal configuration mixing beyond the single-particle picture.

Prolate (cigar-shaped) — Q > 0 means elongation along the spin axis; Q < 0 would be oblate.

Neutron capture on the odd-N $^{235}\mathrm{U}$ makes even-even $^{236}\mathrm{U}^*$ — the extra pairing gain puts its excitation above the fission barrier; capture on even-even $^{238}\mathrm{U}$ makes odd-N $^{239}\mathrm{U}^*$ with excitation below the barrier, so the neutron must bring extra kinetic energy.

Z²/A measures Coulomb disruption vs surface cohesion; the fission barrier vanishes at $Z^2/A \approx 48$ — beyond it, spontaneous fission.