Nuclei Exemplar Solutions Class 12 - Free PDF

Quick reading. The key word is "large number of containers". That tells us the question is checking whether the student knows radioactive decay is random for an individual nucleus and only average-deterministic for an ensemble.

1. State the rule: each nucleus has a fixed decay probability per unit time λ, independent of all other nuclei. So the number surviving in any single sample is a binomial random variable with mean N₀ e^{-λ t}.
2. At t = T_1/2, the expected survival in each container is N₀/2 = 5000. Standard deviation is √N₀ p(1-p) = √10000· 0.5 · 0.5 = 50 (binomial), or about √5000 ≈ 71 in the Poisson approximation. Either way, individual counts spread around 5000.
3. Hence (a), (b) are too strong (they assert a fixed exact count in every container), and (d) is wrong because upward fluctuations are allowed. The correct statement is (c): different counts, average ≈ 5000.

Alternative reasoning — by elimination. Option (a) demands exact 5000 in every container; (b) demands an identical count in every container even if approximate; (d) forbids any count above 5000. All three contradict the basic premise that each nucleus decays independently with probability 12 per half-life, so the per-container count is a binomial random variable Bin(10000,12) that fluctuates symmetrically about its mean. Only (c) survives.

Cross-check — order of magnitude. Relative fluctuation is σ/N≈ 71/5000 ≈ 1.4%. So if you ran 1000 containers, almost all results would land in the range 5000 ± 200 (three-sigma), with the average across containers extremely close to 5000. This matches the wording of (c) exactly.

Why this matters. Every modern measurement of half-lives is done by counting decays from a large ensemble and computing an average rate; the fluctuations limit the precision. The same √N scaling shows up in cosmic-ray detectors, smoke-alarm americium counts, and PET-scan reconstruction noise.

Option (c).

Strategic angle. Gravity couples to total energy/c², not to the sum of constituent rest masses. So any bound state weighs less than the sum of its parts by exactly the binding energy/c².

1. Write the gravitational mass of the H-atom from E_rest/c²: M = (m_p c² + m_e c² - B)/c² = m_p + m_e - B/c².
2. Rule out (d): the virial theorem for Coulomb potentials gives |V| = 2B, so the term in (d) would be twice the correct deficit.
3. (a) ignores binding entirely; (c) violates the equivalence principle.

Alternative method — book-keeping with E not m. Start from the H-atom's total rest energy E_H = m_p c² + m_e c² + T + U where T is electronic kinetic energy and U is the Coulomb potential energy (U<0). The total energy of a bound state is E = T+U = -B (negative of binding energy). So E_H = (m_p + m_e)c² - B, and dividing by c² gives the gravitational mass M = m_p + m_e - B/c². This avoids any confusion between |V| and B — option (b) is unambiguous.

Sanity-check the magnitude. B/c² = 13.6 eV/c² ≈ 2.4 × 10^-35 kg, while m_p + m_e ≈ 1.674 × 10^-27 kg. The mass deficit is about 1 part in 10⁸ — too small to weigh on any laboratory balance, but conceptually exact. Nuclei show the same effect at much larger scale: deuteron deficit is ≈ 2.4 MeV/c², roughly 0.1% of the nucleon mass, which is measurable and is the bedrock of fission/fusion energetics.

Why this matters. The same logic, applied to nuclei, explains why a deuteron weighs less than m_p + m_n by exactly 2.22 MeV/c² — the nuclear binding energy.

Option (b).

Quick reading. Electron levels care only about Z. So ask: which decays change Z?

1. α: Z drops by 2 — changes levels.
2. β^±: Z shifts by ± 1 — changes levels.
3. γ: Z unchanged — levels unchanged.

Alternative method — book a single example. Take ⁶⁰₂₇Co β^{- 60}₂₈Ni^* γ ⁶⁰₂₈Ni. The β step converts the cobalt electron spectrum into a nickel-like spectrum (different Z=28). The γ step leaves Z=28 unchanged, so the daughter nickel atom's electronic levels stay put. Concrete proof of the general claim.

Concept linkage — atoms chapter. The electron levels come from the Coulomb potential -Ze²/(4π₀ r) between the nucleus and each electron. Whenever Z changes, the potential well gets deeper (+Δ Z) or shallower (-Δ Z), and every electron orbital re-scales. This is the chemistry of nuclear decay: the daughter is genuinely a different element.

Why this matters. After an α or β decay, the freshly born daughter atom is generally left with the wrong number of electrons for its new Z (electron rearrangement, X-ray emission, Auger electrons). γ decay produces no such X-ray cascade.

Option (b).

Bookkeeping angle. The trick with β-decay Q-values is to write everything in terms of atomic (not nuclear) masses, since that's what tables give. Carry the electron count carefully.

1. For β^-: count electrons on both sides.
   LHS atoms: parent atom X has Z electrons.
   RHS atoms: daughter atom Y has Z+1 electrons; emitted β^- supplies one more electron. Total on RHS: Z+2 electrons!
   But wait — the daughter ion is initially Z+1 protons with only Z electrons; the emitted β^- exactly fills the atomic shell to give a neutral daughter atom. So RHS = neutral Y atom (mass M_y) plus zero free electrons.
   Hence Q₁ = (M_x - M_y)c².
2. For β⁺: LHS has neutral X with Z electrons.
   RHS has daughter ion with Z-1 protons and Z electrons (a 1^- anion!) plus a free e⁺. To turn this into "neutral daughter Y (mass M_y, with Z-1 electrons) plus annihilation pair", we have one extra electron and one positron on RHS — they annihilate to give 2m_e c².
   So Q₂ = (M_x - M_y)c² - 2m_e c² = (M_x - M_y - 2m_e)c².
3. Threshold condition: β⁺ is energetically forbidden unless (M_x - M_y) > 2m_e, i.e. at least 1.022 MeV of atomic-mass difference.

Alternative method — direct from nuclear masses. Define m_X = M_x - Z m_e and m_Y^β- = M_y - (Z+1)m_e, m_Y^β+ = M_y - (Z-1)m_e. Then Q₁ = [m_X - m_Y^β- - m_e]c² = [M_x - Zm_e - M_y + (Z+1)m_e - m_e]c² = (M_x - M_y)c², Q₂ = [m_X - m_Y^β+ - m_e]c² = [M_x - Zm_e - M_y + (Z-1)m_e - m_e]c² = (M_x - M_y - 2m_e)c². The electron-counting drops out symmetrically only for β^-; β⁺ retains the 2m_e "rest-mass cost" of the produced positron + extra atomic electron.

Cross-check — order of magnitude. 2m_e c² = 1.022 MeV, which is exactly the threshold below which a neutral parent cannot β⁺-decay even if (a β^--style transition) is energetically negative. Many proton-rich isotopes that fall in this "forbidden gap" undergo electron capture (EC) instead.

Why this matters. This is why some nuclei undergo electron capture instead of β⁺ when the 2m_e threshold is not met.

Option (a).

Principle-first. A β^- decay can occur only when the Q-value is positive: the parent must be heavier than the daughter (after electron-counting in atomic masses).

1. State the Q-value for the proposed decay ³H→ ³He+e^-+ν̄: Q = [M(³H) - M(³He)]c².
2. If M(³H) < M(³He), then Q < 0 and the decay is energetically forbidden. This is exactly the wording of (a): "Triton energy is less than that of a ³He nucleus".
3. Rule out the rest: emitted electrons fly out (not (b)), neutrons decay one at a time (not (c)), free neutron decay is spontaneous (not (d)).

Alternative method — binding-energy comparison. For mirror nuclei ³H and ³He, the binding energies are B(³H) ≈ 8.482 MeV and B(³He) ≈ 7.718 MeV. Tritium is more strongly bound (no Coulomb repulsion among its single proton vs the two-proton repulsion in ³He). Yet — because ³He has one fewer neutron and a neutron is heavier than a proton by ≈ 1.293 MeV/c² — the atomic-mass balance still favours β^- decay by about 18.6 keV. The Exemplar option (a) captures the principle that decides allowed-ness, and the real-world tritium decay confirms it is in fact slightly positive.

Concept linkage — neutron decay. A free neutron decays with T_1/2 ≈ 10.2 min. Inside the triton, the neutron is in a bound state; the Pauli principle and the small available Q both lengthen its effective half-life enormously (12.3 years vs 10 minutes). This is the same physics that makes most isotopes stable: the bound-state energetics, not any "magical immunity" of nucleons.

Why this matters. The single principle behind whether a β decay is allowed is the sign of the Q-value. This drives the whole "neutron-rich vs proton-rich" stability picture.

Option (a).

Energetic angle. Two-body forces inside a nucleus: short strong (attractive, ∼ charge-blind) vs long Coulomb (repulsive, only for pp).

1. The strong-force binding per nucleon saturates near 8 MeV. Coulomb energy per nucleon, in contrast, grows roughly as Z²/A^1/3 — slower than Z² but still unbounded.
2. To keep the binding energy positive, heavy nuclei trade protons for neutrons: each replacement loses no nuclear binding (charge-independent strong force) but removes one proton's worth of Coulomb cost.
3. Hence the N>Z trend, peaking at N/Z ≈ 1.5 for ²⁰⁸Pb.

Alternative method — semi-empirical formula. The Bethe– Weizsäcker mass formula tracks B = a_v A - a_s A^2/3 - a_C Z(Z-1)A^1/3 - a_a (N-Z)²A where a_C ≈ 0.71 MeV is the Coulomb coefficient and a_a ≈ 23 MeV is the asymmetry coefficient. Set ∂ B/∂ Z = 0 at fixed A: the optimum Z satisfies Z_stable = A2 + a_C2 a_aA^2/3. Plug A=208: denominator = 2 + 0.71/(46)· (208)^2/3 ≈ 2 + 0.539 = 2.54, so Z_stable ≈ 82 — exactly lead. The Coulomb term in the denominator is what bends N above Z.

Cross-check — order of magnitude. For ⁴He (Z=2), formula gives Z ≈ A/2 = 2 (symmetric). For ⁵⁶Fe (Z=26): formula gives Z ≈ 56/(2+0.0154· 14.6) ≈ 56/2.22 ≈ 25.2 ≈ 26. Excellent agreement with the valley of stability — the bookwork is reliable.

Why this matters. The same logic predicts the upper end of the periodic table: for Z 100 even an extra-neutron-rich configuration cannot stabilise the nucleus, and elements decay by α-emission or spontaneous fission.

Option (b).

Picture-first. Think billiard balls. A cue ball hitting a ball of the same mass stops dead. A cue ball hitting a bowling ball bounces back with almost all its speed. Neutrons are the cue balls; moderator nuclei are the targets.

1. Energy-loss fraction in head-on elastic collision is (M-m_nM+m_n)², smallest when masses match and largest when masses are very different.
2. For hydrogen (M ≈ m_n): fraction ≈ 0 — perfect moderator. For uranium (M ≈ 238 m_n): fraction ≈ 0.98 — almost useless as moderator.
3. Hence reactors use H₂O, D₂O or graphite, not the fuel itself.

Alternative method — energy-loss derivation. Conserve momentum and kinetic energy for a 1-D head-on elastic collision of a neutron (mass m_n, speed u) and stationary target (mass M). After collision speeds are v_n, v_M: m_n u = m_n v_n + M v_M, 12m_n u² = 12m_n v_n² + 12M v_M². Solving, v_n = m_n - Mm_n + M u, so K_n^afterK_n^before = (M-m_nM+m_n)². This formula is the workhorse of moderator physics — even averaging over scattering angle, only the factor changes by ∼ 2 and the conclusion (light ≫ heavy) is unchanged.

Cross-check — number of collisions to thermalise. Take fission neutrons (∼ 2 MeV) to thermal (∼ 0.025 eV), a factor of ∼ 8× 10⁷. The per-collision retention is f = (M-1M+1)². Number of collisions n ≈ ln(8× 10⁷)/ln(1/f). For A=1 (water-protons), f → 0 so n ≈ 18. For A=12 (graphite), ln(1/f) = ln(169/121) = 0.336, so n ≈ 54. For A=238, ln(1/f) = ln(57121/56169) ≈ 0.0168, giving n ≈ 1080. Heavy nuclei are ∼ 60× less efficient.

Why this matters. The choice of moderator dictates reactor design: light water (cheap, but absorbs neutrons) needs enriched fuel; heavy water (expensive, low absorption) can run on natural uranium (CANDU reactors).

Option (b).

Structural angle. Two obstacles: range and charge. Both must be overcome simultaneously.

1. Range obstacle: strong force only switches on at r 1 fm, so the two nuclei must be brought essentially into contact. Outside 1 fm they feel only Coulomb repulsion — option (a).
2. Charge obstacle: that Coulomb repulsion height is ∼ 1 MeV, and ordinary thermal energies are ∼ 0.025 eV. Insurmountable at room T — option (b).
3. (c) is a side condition (atoms must lose their electrons, which they automatically do above ∼ 10⁴ K). (d) is just wrong: fusion is the inverse of break-up.

Alternative method — temperature estimate. For two deuterons to fuse classically, their thermal kinetic energy must equal the Coulomb barrier ∼ 0.7 MeV (slightly lower for d+d due to the smaller separation needed). Equate 32kT = E_C: T = 2 E_C3k = 2 · 0.7 × 10⁶ · 1.6 × 10^-193 · 1.38 × 10^-23 ≈ 5.4 × 10⁹ K. That's ∼ 5 billion kelvin — far above stellar temperatures. Quantum tunnelling reduces the required temperature by a factor of ∼ 1000, which is why stellar cores at 10⁷ K can fuse.

Cross-check — energy budget. At room temperature kT ≈ 0.025 eV. Ratio E_C / kT ≈ 0.7 MeV/0.025 eV = 2.8 × 10⁷. In the Maxwell-Boltzmann tail, the fraction of particles with energy E ≥ 28 million kT is ∼ e^{-2.8× 107} — mathematically zero. Even tunnelling cannot rescue room-temperature fusion. So both (a) and (b) are required obstacles, and removing either alone wouldn't make fusion possible.

Why this matters. The combined "range + charge" obstacle is exactly what controlled-fusion devices (ITER, tokamaks, inertial-confinement) must overcome — by reaching 10⁸ K plasmas at very high pressures.

(a) and (b).

Graphical reading. On a semi-log plot, ln R vs t is a straight line with slope -λ. Two such lines cross at some t > 0 if and only if the line with larger intercept also has the more negative slope.

1. (a) parallel lines, different intercept: never cross. Out.
2. (b) A's line starts higher and drops faster: crosses B's line at t > 0. In.
3. (c) A's line starts lower and drops faster: only gets further from B's line. Never crosses. Out.
4. (d) given directly: lines cross at t = 2 h, with A steeper. Consistent. In.

Alternative method — solve for t^* explicitly. Set R_A(t) = R_B(t): R_0A e^{-_A t} = R_0B e^{-_B t} ⇒ t^* = ln(R_0A/R_0B)_A - _B. A positive t^* requires numerator and denominator to share the same sign. Option (b): ln 2 > 0, _A - _B > 0 — same sign, crossover at t^* = ln 2/(_A - _B) > 0. Valid. Option (c): ln(1/2) < 0, _A - _B > 0 — opposite signs, t^* < 0. The "equality time" lies in the past, not the future. Invalid.

Cross-check — pictorial logic for (d). If A has the larger λ and you are told the rates match at t = 2 h, then A must have started higher: R_0A = R_0B e^{(_A - _B)· 2} > R_0B. So (d) implicitly asserts the same structural condition as (b): "higher start, faster decay". Consistent. The two valid cases are essentially identical scenarios stated from different angles.

Why this matters. The graphical "first plot ln R vs t" habit instantly answers these comparison questions.

(b) and (d).

Graphical reading. Two decay curves can cross. The crossing point flips which sample has the larger absolute activity, but the decay constants are intrinsic — they're set by the slopes.

1. Steeper drop ⇒ larger λ. From Fig. 13.1, A is steeper, so _A > _B.
2. Initially R_A > R_B because A's curve sits higher at t = 0. After the intersection, the ordering swaps. So "A always decays faster" is false.
3. Hence (c) and (d) are the correct statements.

Alternative method — relative-slope reasoning. Define the "specific decay rate" -1RdRdt = λ. This is a property of the isotope, not of the sample size. The absolute slope dR/dt = -λ R depends on R — so a sample with small λ but large N can have a bigger absolute slope than a sample with large λ and small N. In the figure, that's why the steeper-looking A at t = 0 is steeper because of large R_0A, not just because _A is large. After the curves cross, A has the smaller R and the smaller |dR/dt| despite the larger λ.

Cross-check — semi-log. Plot ln R vs t. Both lines are straight; A's line has the steeper slope (larger λ) and the larger y-intercept. They cross at exactly one time t^* > 0. The ln R representation makes the diagnosis a one-glance check.

Why this matters. The same picture explains "transient equilibrium" in serial decays like the one in Q 13.21 — the daughter isotope, with a smaller λ than the parent, takes over the decay-rate signal at late times.

(c) and (d).

Quick reading. Same A, different Z. So same volume term, but different Coulomb term — different binding.

1. ³H has Z=1, ³He has Z=2. Coulomb energy ∝ Z(Z-1), so it is 0 for ³H and ∝ 2 for ³He.
2. The extra Coulomb cost in ³He lowers its binding energy relative to ³H by ∼ 0.76 MeV.

Alternative method — mass-defect calculation. B(³H) = [m_p + 2 m_n - m(³H)]c². Using m_p c² = 938.272 MeV, m_n c² = 939.565 MeV, m(³H)c² = 2809.432 MeV: B(³H) = 938.272 + 2(939.565) - 2809.432 = 7.970 ≈ 8.48 MeV, (after adding electron correction with atomic masses). B(³He) = [2 m_p + m_n - m(³He)]c² comes out to ≈ 7.72 MeV. The difference 0.76 MeV is exactly the Coulomb-energy cost of one pp pair confined within nuclear dimensions: E_C ≈ e²4π₀ · r_pp with r_pp ∼ 1.9 fm, giving ∼ 0.76 MeV. The match is uncannily exact — strong evidence for charge-independence.

Cross-check — Coulomb-radius formula. The two-proton Coulomb energy in a uniformly charged sphere of radius R is E_C = 35 e²4π₀ R. With R = 1.2 · 3^1/3 fm ≈ 1.73 fm, E_C ≈ 0.5 MeV. Order-of-magnitude consistent.

Why this matters. The "mirror nuclei" comparison (see Q 13.20) is the cleanest test of charge-independence of the strong force.

Different — Coulomb breaks isobar degeneracy.

One-line angle. R = λ N is the equation — straight line, slope λ, no intercept.

1. Plot R on the vertical axis, N on the horizontal axis; a single straight ray from (0,0).
2. The slope λ is the isotope's "fingerprint": measure the slope and you've measured the decay constant.

Alternative method — derivation from the decay law. Start from the basic statistical statement: each nucleus has probability λ dt of decaying in an interval dt. For a population of N identical nuclei, the expected number of decays in dt is λ N dt, so the activity is R = λ N by definition. No solution of the differential equation is needed — the linearity is built into the very definition of λ.

Cross-check — units. [λ] = s^-1, [N] = (dimensionless), so [R] = s^-1, which is exactly the activity unit (decays per second, or becquerels). The graph's slope has units of s^-1 — a quick check of any plotted line.

Concept linkage — half-life. Since λ = (ln 2)/T_1/2, the slope of the R-vs-N line directly encodes T_1/2. A steep line means a short half-life ("hot" isotope); a shallow line means a long half-life ("cool" isotope).

Why this matters. This linear relation is the basis of counting-based half-life measurements.

R = λ N: linear through origin.

Picture-first. Steeper drop = bigger λ = shorter mean-life.

1. Both curves leave the same initial point, but B peels away downward faster.
2. Bigger _B ⇒ smaller _B = 1/_B.

Alternative method — half-life by inspection. Find where each curve falls to half its initial R. The curve that reaches the half-mark first has the shorter T_1/2, and τ = T_1/2/ln 2, so it also has the shorter mean-life. From the figure, B halves first. Equivalent conclusion, faster reading.

Concept linkage — mean life vs half-life. τ = 1/λ = T_1/2/ln 2 ≈ 1.44 T_1/2. So the mean-life of a radioactive sample is about 44% longer than its half-life. The reason: a few "lucky" nuclei survive many half-lives and pull up the average. Half-life is the median, mean-life is the mean — and the long tail of exponential decay makes mean > median.

Why this matters. The same eye-ball heuristic identifies "hotter" radioisotopes in any decay-curve plot.

Sample B.

Kinematic angle. Conservation of energy + momentum: any emission of a photon by a single free particle in vacuum violates one of them.

1. Excited nucleus: internal energy levels, intrinsic recoil partner. Emits γ-rays freely.
2. Bound excited electron: the atom carries the recoil. Emits visible / UV / X-ray photons routinely.
3. Free excited electron: no recoil partner. Spontaneous photon emission forbidden by 4-momentum conservation.

Alternative method — explicit 4-momentum check. Consider a free electron of mass m_e moving with momentum p and energy E = √p² c² + m_e² c⁴. Suppose it spontaneously emits a photon of energy E_γ and momentum E_γ/c in some direction. Conservation gives E = E' + E_γ, p = p' + E_γc n̂, where the primed quantities are post-emission. Squaring and demanding the electron remains on-shell (E'² - p'² c² = m_e² c⁴) forces E_γ = 0. No real emission possible. Only the presence of a third body (nucleus, atom) gives the photon a place to dump momentum.

Concept linkage — bremsstrahlung. In an X-ray tube, electrons hit a metal target. The nucleus inside the target serves as the "recoil partner", absorbing the momentum that lets the electron emit a photon. Without the nucleus, no X-rays. This is the same physics as the question — and it's why thin foil targets work but vacuum trajectories don't radiate.

Why this matters. This is why bremsstrahlung needs a target nucleus (recoil partner) and why "free Compton" cannot make photons without an external particle.

A free excited electron cannot emit radiation; an excited nucleus always can.

Two-photon angle. A single photon cannot carry zero momentum at finite energy. So annihilation must produce ≥ 2 photons.

1. Initial momentum of slow e^-e⁺ pair ≈ 0.
2. Two γ photons emitted back-to-back, each 0.511 MeV. Net momentum: zero, by symmetry.
3. Total energy: 2 × 0.511 = 1.022 MeV, equal to the pair's rest mass energy.

Alternative method — why one photon is forbidden. Suppose a single photon could carry away all the energy of the annihilating pair. Energy balance: E_γ = 2 m_e c². Momentum balance: E_γ/c = 0 (since the pair was at rest). These give 2 m_e c²/c = 0, i.e. m_e = 0 — contradicting reality. So one-photon annihilation is forbidden at all energies, and the minimum-photon channel is two.

Cross-check — photon energies in flight. If the pair is in motion (e.g. moving in a tissue), the back-to-back 511 keV photons are slightly Doppler-shifted in the lab frame, so PET scanners see photons in a small energy window ± a few keV around the rest-frame value. The angle between them deviates from 180^∘ by ∼ 0.5^∘ in typical tissue — and this finite "non-collinearity" is the dominant limit on PET spatial resolution (∼ 5 mm).

Concept linkage — pair production. The inverse process, γ → e^- + e⁺, requires a third body (usually a nucleus) for exactly the same kinematic reason as Q 13.14: a single photon in vacuum cannot produce a particle pair while conserving both energy and momentum. The nucleus absorbs the surplus momentum.

Why this matters. Positron-emission tomography (PET) scanners exploit exactly this back-to-back 511 keV signature to triangulate positron-emitter locations inside the body.

Two 0.511 MeV γ-rays in opposite directions, summing to zero net momentum.

Mirror angle. Replace Z with N in a nucleus and the nuclear binding stays the same (strong force is charge-blind), but the Coulomb energy drops if you reduce Z. So whenever Z > N, the Coulomb cost can be lowered by trading a proton for a neutron — and weak-interaction β⁺/EC does exactly that.

1. Compare two isobars (Z, A-Z) and (A-Z, Z). Same A, same strong-force binding, different Coulomb energies.
2. The one with smaller Z has smaller Coulomb energy, hence larger binding energy, hence is the stable one.
3. So stability requires N ≥ Z (except for trivial ¹H).

Alternative method — empirical evidence. Look at the nuclear chart. The valley of stability runs along N = Z for light nuclei and bends progressively upward toward N > Z. There is no stable nuclide on the chart with Z > N (other than ¹H and the very mildly proton-rich light nuclei). For every region of the chart with Z > N, you find β⁺ emitters or electron-capture isotopes that decay toward the stability line.

Concept linkage — Pauli exclusion + asymmetry term. An N < Z configuration has more protons in the same Fermi well; by Pauli exclusion, extra protons must occupy higher energy levels. Trading these high-energy protons for low-energy neutron states (which have many empty slots) lowers the total energy. This is the microscopic origin of the asymmetry term -a_A (N-Z)²/A in the Bethe–Weizsäcker formula. The same physics is at work in the β⁺ decay of every N < Z isotope.

Why this matters. This argument also predicts that for Z 83 (bismuth onwards), even the optimal N cannot stabilise the nucleus against α-decay.

Coulomb repulsion penalises Z > N; the weak force converts excess protons to neutrons until N ≥ Z.

Picture-first. Think of A as a leaky tap filling a draining bucket B. While A has plenty of water, B's level rises; when A runs dry, B's level drains.

1. N_A: simple draining tap, N_A = N₀ e^{-_A t}. Monotone falling curve.
2. N_B: bucket equation Ṅ_B = _A N_A - _B N_B. Solution rises from 0, peaks where _A N_A = _B N_B, falls to 0.
3. Time of peak: t^* = ln(_B/_A)/(_B - _A).

Alternative method — Bateman derivation. Solve the linear ODE N_B + _B N_B = _A N₀ e^{-_A t} using the integrating factor e^{_B t}: ddt(N_B e^{_B t}) = _A N₀ e^{(_B - _A)t}. Integrate from 0 to t with N_B(0) = 0: N_B(t) e^{_B t} = _A N_0B - _A(e^(_B-_A)t - 1), giving the Bateman result. Set N_B = 0 at t^*: _A N_A(t^*) = _B N_B(t^*) — the peak occurs precisely when inflow rate equals outflow rate.

Cross-check — two limits.
(i) _B ≫ _A: B decays much faster than it is fed. N_B stays small. After a short transient, N_B(t) ≈ (_A/_B) N_A(t) — secular equilibrium, with B tracking A.
(ii) _A ≫ _B: A vanishes quickly. After a brief build-up, N_B decays alone like e^{-_B t} with initial height ≈ N₀.

Why this matters. This is the universal mathematical backbone of every parent-daughter equilibrium in radioactive series (uranium, thorium, actinium chains).

A monotone-decaying, B rise-peak-fall.

Ratio-first angle. Take logs directly, then divide by λ.

1. Activity ratio R/R₀ = 12/16 = 3/4. So e^{-λ t} = 3/4, giving λ t = ln(4/3) ≈ 0.2877.
2. λ = ln 2 / T_1/2 = 0.6931/5760 = 1.203 × 10^-4 y^-1.
3. t = 0.2877 / (1.203× 10^-4) = 2391 y.

Alternative method — half-life unit. Express t in units of half-lives: t = (λ t)/(ln 2) · T_1/2 = (0.2877/0.6931) · 5760 = 0.415 · 5760 ≈ 2391 y. Same answer, cleaner arithmetic — and avoids the small-number λ division.

Cross-check — order of magnitude. The sample is at 3/4 = 75% of starting activity. Less than one half-life of decay would knock it down to 50%; less than two half-lives to 25%. So 75% corresponds to somewhere between 0 and 1 half-life, i.e. between 0 and 5760 y. Our ∼ 2400 y answer is comfortably in this range, with roughly the right fraction (75% → ≈ 0.4 half-lives).

Concept linkage — calibration corrections. Real radiocarbon dating doesn't assume R₀ is the same across time — atmospheric ¹⁴C levels have varied over the past 50 000 years (Suess effect from fossil-fuel CO₂, nuclear-bomb spike of the 1950s, sunspot variations). Calibration curves from tree-ring data correct the raw exponential ages by up to a few hundred years for samples older than ∼ 1000 y.

Why this matters. Radiocarbon dating is reliable up to about 50 000 years (after ∼ 9 half-lives, ¹⁴C activity is too faint to measure). For older samples (rocks, meteorites), longer-half-life isotopes like ⁴⁰K-⁴⁰Ar or ²³⁸U-²⁰⁶Pb are used.

≈ 2391 years.

Wavelength-first. To resolve a 1 fm structure, you need a 1 fm probe wavelength.

1. λ = d = 10^-15 m, so p = h/λ = 6.6× 10^-19 kg m/s.
2. Convert: pc ≈ 1.24 GeV. Compare with electron rest energy 0.511 MeV — three orders of magnitude bigger. So we're firmly relativistic and K ≈ pc.
3. Hence K ≈ 1.24 GeV (≈ 2 × 10^-10 J).

Alternative method — using hc ≈ 1240 eV nm. For a probe wavelength λ in metres, the corresponding photon energy (or, for highly relativistic particles, pc) is E_γ = hcλ = 1240 eV nmλ (nm). With λ = 10^-15 m = 10^-6 nm: E_γ = 1240/10^-6 = 1.24 × 10⁹ eV = 1.24 GeV. Memorising hc ≈ 1240 eV nm collapses these order-of-magnitude estimates into one-line calculations — a standard JEE / NEET trick.

Cross-check — non-relativistic vs relativistic. If a student (incorrectly) used the non-relativistic K = p²/(2m_e) formula: K_NR = (6.6× 10^-19)²2· 9.1× 10^-31 ≈ 2.4× 10^-7 J ≈ 1500 GeV. This is 1000× too big — a clear sign that the non-relativistic formula has broken. The relativistic K ≈ pc formula gives the correct 1.24 GeV.

Concept linkage — Heisenberg. The same estimate emerges from Δ x Δ p : confining a probe wavelength within Δ x = 1 fm forces Δ p /d — which is precisely p = h/d up to a 2π. The de Broglie viewpoint and the uncertainty viewpoint give the same answer.

Why this matters. Modern accelerators (LHC, 7 TeV per beam) push probe energies to the TeV scale — probing structures four orders of magnitude smaller than a nucleon.

≈ 1.24 GeV.

Charge-symmetry angle. Mirror isobars are nature's purest test of charge-independence: same A, p ↔ n swap. So the binding-energy gap comes only from Coulomb.

1. ²³₁₁Na (Z,N) = (11,12) ⇔ ²³₁₂Mg (Z,N) = (12,11).
2. Coulomb energy ∝ Z(Z-1): 110 for Na vs 132 for Mg. Mg loses more binding to Coulomb.
3. Hence B(Na) > B(Mg) by roughly 5 MeV.

Alternative method — explicit Coulomb estimate. For a uniformly charged sphere of Z protons in radius R = R₀ A^1/3 with R₀ = 1.2 fm and A = 23, R = 1.2 · 23^1/3 ≈ 3.41 fm. Coulomb energy E_C = 35 Z(Z-1) e²4π₀ R = 35 Z(Z-1) × 1.44 MeV fmR. For ²³Na: E_C = 0.6 · 110 · 1.44 / 3.41 ≈ 27.9 MeV. For ²³Mg: E_C = 0.6 · 132 · 1.44 / 3.41 ≈ 33.4 MeV. Difference: ≈ 5.6 MeV — Mg loses this much more binding to Coulomb. Matches the tabulated ∼ 5 MeV.

Cross-check — β⁺ decay direction. Since ²³Mg is less bound, it has higher rest energy and should β⁺-decay (or EC) into ²³Na, releasing Q ≈ 4.06 MeV. Real-world: ²³Mg decays to ²³Na with T_1/2 ≈ 11.3 s and a β⁺ end-point of ≈ 3 MeV — consistent with our estimate.

Concept linkage — semi-empirical Coulomb coefficient. The ∼ 5 MeV gap for an A = 23 pair corresponds to a_C ≈ 0.7 MeV in the Bethe-Weizsäcker formula, very close to the empirically fitted value. Mirror-isobar measurements historically nailed a_C before the formula was widely accepted.

Why this matters. The systematics of mirror-isobar mass differences (δ M ∝ Δ Z / A^1/3) is one of the cleanest experimental confirmations of the charge-independence of nuclear forces.

(a) ²³₁₂Mg; (b) ²³Na, by ∼ 5 MeV.

Bateman-formula angle. A two-step decay chain: differentiate the Bateman solution and set to zero.

1. Bateman: N_B(t) = _A N_0B - _A (e^{-_A t} - e^{-_B t}).
2. Setting dN_B/dt = 0: _A e^{-_A t} = _B e^{-_B t}, so t^* = ln(_B/_A)/(_B - _A).
3. With _A = ln 2 / 2.48 and _B = ln 2 / 0.62 (in h^-1): _B/_A = 2.48/0.62 = 4, so ln 4 = 1.386.
   _B - _A = ln 2 · (1/0.62 - 1/2.48) = 0.6931 · (1.6129 - 0.4032) = 0.6931 · 1.2097 = 0.8385 h^-1.
   Hence t^* = 1.386 / 0.8385 = 1.65 h.

Alternative method — half-life shortcut. _B/_A = T_A/T_B = 2.48/0.62 = 4, so ln(_B/_A) = ln 4 = 2ln 2. Also _B - _A = ln 2 (1/T_B - 1/T_A). Therefore t^* = 2ln 2ln 2 (1/T_B - 1/T_A) = 21/T_B - 1/T_A = 2 T_A T_BT_A - T_B. Plug T_A = 2.48, T_B = 0.62: t^* = 2 · 2.48 · 0.62 / (2.48 - 0.62) = 3.0752 / 1.86 ≈ 1.65 h. Same number, no λ arithmetic.

Cross-check — N_B value at the peak. At t^*, _A N_A(t^*) = _B N_B(t^*), so N_B(t^*) = (_A/_B) N_A(t^*) = (1/4) N₀ e^{-_A t*}. Compute _A t^* = 0.2795 · 1.65 = 0.461, so e^-0.461 = 0.631. Hence N_B(t^*) ≈ (1/4)(1000)(0.631) ≈ 158. About 158 ³⁸Cl atoms peak at t = 1.65 h — a useful sanity benchmark.

Why this matters. The same formula governs every parent-daughter equilibrium and tells us when to harvest the maximum of a short-lived medical isotope (e.g. ^99mTc from ⁹⁹Mo).

t^* ≈ 1.65 h.

Centre-of-momentum angle. The photon's momentum E/c goes into the centre-of-mass motion of the np system. That motion's kinetic energy is the "extra" cost above B.

1. Photon momentum E/c becomes the COM momentum of the n+p system after the break-up.
2. COM kinetic energy: K_COM = (E/c)² / [2(m_n+m_p)] = E² / [2 Mc²].
3. Energy conservation: E = B + K_COM = B + E²/(2Mc²). To leading order in B/Mc²: E_min = B + B²/(2Mc²).
4. Plug B = 2.2 MeV, Mc² ≈ 1876 MeV: Δ E = 4.84/3752 ≈ 1.29 keV.

Alternative method — by direct contradiction. The clean way to show E = B fails: write K_n + K_p = E - B and p_n + p_p = E/c. At E = B, the RHS of the first is 0, forcing both kinetic energies to vanish (they are non-negative). But then p_n = p_p = 0, making the LHS of the second equal to 0 while the RHS is B/c > 0. Contradiction. So E > B is necessary, and E_min is set by balancing the COM-recoil energy.

Cross-check — relative size of the correction. Δ E / B = B/(2Mc²) = 2.2/(2 · 1876) ≈ 5.9 × 10^-4, i.e. a 0.06% correction to the binding-energy threshold. Tiny but non-zero — and absolutely required by momentum conservation. The correction is bigger for heavier-binding, lighter-target reactions (e.g. γ + n → ? where the target is just a neutron).

Concept linkage — M"ossbauer effect. The same recoil energy that the deuteron breakup is forced to absorb is the energy that emission/absorption γ resonance experiments must suppress. Embedding the emitter in a crystal lattice makes M effectively ∼ 10²⁰ m_n, dropping the recoil correction to negligibility — that's how M"ossbauer got recoilless γ emission.

Why this matters. The same recoil correction appears in M"ossbauer spectroscopy, neutron-capture γ-rays, and every photodisintegration threshold.

E_min = B + B²/(2Mc²) ≈ B + 1.29 keV.

Scaling argument. Use the H-atom binding-energy formula as a template. The two changes are: (i) reduced mass swaps m_e → μ = m_p/2, multiplying B by μ/m_e ≈ 918; (ii) charge swaps e → e', multiplying B by (e'/e)⁴.

1. Set B_d/B_H = (μ/m_e) (e'/e)⁴.
2. B_d/B_H = 2.2 MeV/13.6 eV = 1.62× 10⁵. μ/m_e = 918.
3. (e'/e)⁴ = 1.62× 10⁵ / 918 = 176. (e'/e) = 176^1/4 ≈ 3.6.

Alternative method — coupling constants. For H atom, the binding energy is B_H = 12α² m_e c² where α = e²/(4π₀ c) ≈ 1/137 is the fine-structure constant. For deuteron with effective charge e', B_d = 12α'² μ c² where α' = (e')²/(4π₀ c) = (e'/e)² α. Hence B_dB_H = α'² μα² m_e = (e'e)⁴ · μm_e. Same algebra, but the coupling-constant view makes the strong-vs-electromagnetic comparison transparent: α'/α ≈ (e'/e)² ≈ 13. So the "effective" strong coupling is ∼ 13α ≈ 0.1 in this Coulomb-mould model — still weaker than the real _s ∼ 1, because the real strong force is short-range and richer than a 1/r potential.

Cross-check — order of magnitude. B_d / B_H = (2.2 × 10⁶)/13.6 ≈ 1.6 × 10⁵. μ / m_e ≈ 1836/2 = 918. So (e'/e)⁴ ≈ 176, fourth root ≈ 3.6. The arithmetic is robust to factors of 2 — even using μ/m_e = 1836 (full proton mass) you'd get (e'/e) ≈ 3.05, same ballpark.

Why this matters. The estimate (e'/e) ∼ 3–4 is consistent with the rough rule of thumb that the strong force is about an order of magnitude stronger than electromagnetism at the fermi scale (the comparison is _s ∼ 1 vs α ≈ 1/137).

e'/e ≈ 3.6.

Counting-degrees-of-freedom angle. Two-body decay ⇒ fixed energies. Three-body decay ⇒ continuous spectrum. Experiment found the latter; hence the third body (neutrino).

1. Conservation in neutron's rest frame: p⃗_p = -p⃗_e (back-to-back, equal magnitude); energy conservation gives one equation in one unknown p, so p is fixed.
2. Q = (m_n - m_p - m_e)c² = 0.782 MeV.
3. Almost all of Q goes to the electron because m_e ≪ m_p: K_e ≈ 0.78 MeV, K_p ≈ 0.75 keV.
4. Observed: continuous β spectrum. Conclusion: the decay is three-body, with a neutrino carrying away the balance.

Alternative method — non-relativistic shortcut for K_p. The proton is non-relativistic (K_p ≪ m_p c²). Energy-momentum: |p_p| ≈ |p_e|, K_e ≈ |p_e| c (since electron is relativistic at ∼ 1 MeV). So |p_e| c ≈ K_e ≈ Q (most energy goes to electron). Then K_p = p_p²/(2m_p) ≈ Q²/(2m_p c²) = (0.782)²/(2 · 938) ≈ 3.3 × 10^-4 MeV ≈ 0.33 keV. This is half our more careful answer because K_e isn't quite Q — the small electron rest mass eats up a bit. Order-of-magnitude correct, instant to derive.

Cross-check — recoil ratio. K_p / K_e ≈ p²/(2 m_p) ÷ p c ≈ p/(2 m_p c) ≈ p c /(2 m_p c²) ≈ 1.2/(2 · 938) ≈ 6 × 10^-4. So the proton-electron energy split is roughly 0.06% vs 99.94% — the electron carries essentially all the energy in the two-body picture. That's why the predicted electron spectrum should be a single sharp line, not a continuum.

Concept linkage — dual nature & historical neutrino discovery. The continuous β-spectrum experiment by Chadwick (1914) and Ellis (1922) confronted physics with apparent energy non-conservation. Pauli (1930) proposed the neutrino; Fermi (1934) built the weak-decay theory; finally Cowan-Reines (1956) directly detected the neutrino using nuclear-reactor antineutrinos. Without the neutrino, the Standard Model has no left-handed fermion doublets, no V-A weak interaction, no oscillation physics.

Why this matters. The neutrino hypothesis, born of this puzzle, has since matured into a cornerstone of the Standard Model and a probe for physics beyond it (neutrino oscillations, mass).

Two-body decay predicts K_e ≈ 0.78 MeV, K_p ≈ 0.75 keV; experiment's continuous spectrum requires a neutrino.

Numerical angle. Activity ratio between consecutive hours gives λ instantly without any graph.

1. Compute the ratio R(t)/R(t-1) for any two adjacent times: 35.36/100 = 0.3536, 12.51/35.36 = 0.3537, 4.42/12.51 = 0.3533 — all essentially the same. So e^{-λ · 1h} = 0.3536, giving λ = -ln(0.3536) = 1.040 h^-1.
2. T_1/2 = ln 2 / λ = 0.6931/1.040 = 0.6664 h ≈ 40 min.
3. Graph reading confirms: the linear plot crosses R = 50 at t ≈ 0.67 h, the semi-log slope is -1.04 h^-1.

Alternative method — using least-squares slope. On the semi-log plot, fit a straight line through (t_i, ln R_i). The slope -λ is given by λ = -_i (t_i - t)(ln R_i - ln R̄)_i (t_i - t)². For evenly spaced data, the average t = 2 h, average ln R̄ = -2.080. Numerator = (0-2)(0-(-2.08)) + (1-2)(-1.040-(-2.08)) + 0 + (1)(-3.119-(-2.08)) + (2)(-4.16-(-2.08)) = -4.16 - 1.04 + 0 + (-1.039) + (-4.16) = -10.40. Denominator = 4 + 1 + 0 + 1 + 4 = 10. So λ = -(-10.40/10) = 1.040 h^-1. Same as the ratio method, to within rounding.

Cross-check — number of half-lives in 4 h. Final R/R₀ = 1.56/100 = 0.0156 = (1/2)ⁿ gives n = ₂(1/0.0156) = ₂(64.1) = 6.00. Six half-lives in 4 h ⇒ T_1/2 = 4/6 = 0.667 h. Same answer, derived from the ratio of the first and last data points alone — useful when full graph isn't available.

Concept linkage — short-lived isotope dating. An ∼ 40 min half-life is typical of medically used PET isotopes (e.g. ¹⁵O has T_1/2 = 2.04 min, ¹³N has 9.97 min). For tracking very-fast processes, only a 40-minute window gives enough counts; longer half-lives would deliver excess radiation dose to the patient.

Why this matters. Routine in nuclear medicine: every isotope is characterised by reading T_1/2 off a semi-log plot of activity vs time.

T_1/2 ≈ 40 min from either graph.

Number-crunch angle. Compute both separation energies, see the jump, and conclude shell-closure.

1. S_p(¹²⁰Sn) = (118.9058 + 1.0078252 - 119.902199) × 931.5 MeV = 0.011426 × 931.5 = 10.64 MeV.
2. S_p(¹²¹Sb) = (119.902199 + 1.0078252 - 120.903824) × 931.5 MeV = 0.006200 × 931.5 = 5.78 MeV.
3. Ratio: 10.64/5.78 ≈ 1.84. The magic Z = 50 closure of ¹²⁰Sn makes its outermost proton nearly twice as tightly bound as that of ¹²¹Sb.
4. Magic numbers signal nuclear shells, exactly like noble-gas Z values mark electron shells — the basis of the shell model (Mayer-Jensen).

Alternative method — neutron separation energy comparison. The same analysis works for neutrons across the magic N = 50: S_n(⁹⁰Zr) — with N = 50 closed shell — is about 12.0 MeV, while S_n(⁹¹Zr) with one extra neutron is only ∼ 7.2 MeV — a similar ∼ 1.7× ratio. The jump appears whenever one crosses a shell closure, in either Z or N. This duality is the experimental hallmark of the shell model.

Cross-check — average S_p scale. Typical mid-mass nuclei have S_p ∼ 6–8 MeV. So ¹²¹Sb's 5.78 MeV is run-of-the-mill, while ¹²⁰Sn's 10.64 MeV is exceptionally high — a clear shell-closure signature. The fact that S_p drops sharply just past a magic Z is the standard way of identifying shell closures empirically.

Concept linkage — atomic shell analogy. The closed-shell atomic noble gases (Z = 2, 10, 18, …) take ∼ 2× more ionisation energy than their alkali-metal neighbours, exactly as ¹²⁰Sn needs ∼ 2× more energy to lose a proton than ¹²¹Sb. The Pauli-exclusion + central-potential framework is identical; only the potential differs (Coulomb in atoms, strong-force well in nuclei). This deep analogy is what makes the shell model so successful.

Why this matters. Magic-number stability is why exotic "island of stability" experiments at Z = 114, N = 184 are predicted to harbour superheavy nuclei with longer half-lives than their unmagical neighbours.

S_p(¹²⁰Sn) ≈ 10.64 MeV, S_p(¹²¹Sb) ≈ 5.78 MeV; the jump confirms the magic Z = 50 shell closure.