Nuclear & Radiochemistry

Q1 (10)

(a) Postulate $-\dfrac{dN}{dt} \propto N \Rightarrow \dfrac{dN}{dt} = -\lambda N$ (1). Separate: $\int_{N_0}^{N}\frac{dN}{N} = -\lambda\int_0^t dt$ (1) $\Rightarrow \ln(N/N_0) = -\lambda t \Rightarrow N=N_0 e^{-\lambda t}$ (1).

(b) Half-life: $N_0/2 = N_0 e^{-\lambda t_{1/2}} \Rightarrow t_{1/2} = \dfrac{\ln 2}{\lambda}$ (1.5). Mean life $\tau = \dfrac{\int_0^\infty t\,\lambda N_0 e^{-\lambda t}dt}{\int_0^\infty \lambda N_0 e^{-\lambda t}dt} = \dfrac{1}{\lambda}$ (1.5). Hence $t_{1/2} = \dfrac{\ln 2}{\lambda} = \tau\ln 2$ (1).

(c) $A = -dN/dt = \lambda N_0 e^{-\lambda t} = \lambda N$ (1.5). Since $A \propto N$ with the same constant $\lambda$, both decay as $e^{-\lambda t}$, so they share $t_{1/2}=\ln2/\lambda$ (1.5).

Q2 (10)

(a) $\Delta m = Z\,m(^1\text{H}) + N\,m_n - m(^{56}\text{Fe})$ (1). Using $^1\text{H}$ atomic mass includes 1 electron; $Z$ of them supply the 26 electrons of the Fe atom, so electron masses cancel (binding energy of electrons neglected) (1).

(b) $Z=26$, $N=30$. $26\times1.007825 = 26.20345$; $30\times1.008665 = 30.25995$; sum $=56.46340$ (1). $\Delta m = 56.46340 - 55.934937 = 0.528463\ \text{u}$ (2).

(c) $BE = 0.528463 \times 931.5 = 492.3\ \text{MeV}$ (2). Per nucleon $= 492.3/56 = 8.79\ \text{MeV}$ (1).

(d) Near the BE/nucleon peak (~8.8 MeV, iron region), nucleons are most tightly bound; energy is released by fusing lighter nuclei up toward it and by splitting heavier nuclei down toward it (2).

Q3 (10)

(a) $A=\lambda N=A_0 e^{-\lambda t}$ (1) $\Rightarrow \lambda t = \ln(A_0/A)$ (1) $\Rightarrow t = \frac{1}{\lambda}\ln(A_0/A)=\frac{t_{1/2}}{\ln2}\ln(A_0/A)$ (1).

(b) $t = \dfrac{5730}{0.693}\ln\!\left(\dfrac{15.3}{9.2}\right)$. $\ln(1.6630)=0.5086$ (2). $t = 8268 \times 0.5086 \approx 4205\ \text{yr}$ (≈ 4200 yr) (2).

(c) Any two, e.g.: (i) atmospheric $^{14}\text{C}/^{12}\text{C}$ constant over time — if it was higher in past, true age understated; (ii) no contamination/exchange after death — modern carbon contamination makes sample look younger; (iii) sample stopped exchanging carbon at death (closed system). (3, 1.5 each for assumption + bias direction)

Q4 (10)

(a) (1 each)

• $\alpha$: $^A_ZX \to\ ^{A-4}_{Z-2}Y + ^4_2\text{He}$; $Z-2$, $N-2$, $A-4$.
• $\beta^-$: $^A_ZX \to\ ^{A}_{Z+1}Y + e^- + \bar\nu$; $Z+1$, $N-1$, $A$ same.
• $\beta^+$: $^A_ZX \to\ ^{A}_{Z-1}Y + e^+ + \nu$; $Z-1$, $N+1$, $A$ same.
• EC: $^A_ZX + e^- \to\ ^{A}_{Z-1}Y + \nu$; $Z-1$, $N+1$, $A$ same.

(b) Neutron-rich (above band): $\beta^-$ converts n→p, lowering N/Z (1). Proton-rich (below band): $\beta^+$/EC converts p→n, raising N/Z (1).

(c) $\Delta m = 209.982874 - (205.974465 + 4.002603) = 0.005806\ \text{u}$ (2). $Q = 0.005806\times931.5 = 5.41\ \text{MeV}$ (1). $Q>0$ ⇒ spontaneous (1).

Q5 (10)

(a) Moderator (e.g. water, graphite) slows fast fission neutrons to thermal energies to sustain fission in $^{235}$U (1). Control rods (B, Cd) absorb neutrons to regulate $k$ (1). Coolant removes heat / transfers it to generate steam (1). Thermal reactor: uses moderated slow neutrons, enriched/natural U; Fast reactor: no moderator, uses fast neutrons, can breed $^{239}$Pu (1).

(b) Critical mass = minimum fissile mass for a self-sustaining chain reaction (neutron production = losses) (1). $k$ = neutrons in one generation per neutron in previous: $k<1$ subcritical (dies out), $k=1$ critical (steady), $k>1$ supercritical (grows/explodes) (2).

(c) $^2_1\text{D} + ^3_1\text{T} \to\ ^4_2\text{He} + ^1_0n + 17.6\ \text{MeV}$ (1). Harder because: Coulomb barrier requires ~$10^8$ K plasma; confinement (magnetic/inertial) of hot plasma is extremely difficult; no self-sustaining critical-mass analogue at reachable densities (any two, 2).

Q6 (10)

(a) (4)

import math
def remaining(N0, half_life, t):
    lam = math.log(2) / half_life        # decay constant, per s
    N = N0 * math.exp(-lam * t)          # surviving nuclei
    A = lam * N                          # activity in Bq
    return N, A

(b) $t/t_{1/2} = 1800/600 = 3$ half-lives (1). $N = 10^{20}\times(1/2)^3 = 1.25\times10^{19}$ (1.5). $\lambda = 0.693/600 = 1.155\times10^{-3}\ \text{s}^{-1}$; $A = 1.155\times10^{-3}\times1.25\times10^{19} = 1.44\times10^{16}\ \text{Bq}$ (1.5).

(c) For $t\gg t_{1/2}$, $e^{-\lambda t}$ underflows to 0 losing precision; use math.exp in log-space or decimal/mpmath, or work with $\log N$ (2).

[
  {"claim":"Fe-56 mass defect and BE/nucleon", "code":"dm=26*1.007825+30*1.008665-55.934937; be=dm*931.5; bepn=be/56; result=(abs(dm-0.528463)<1e-4) and (abs(bepn-8.79)<0.05)"},
  {"claim":"Radiocarbon age ~4200 yr", "code":"import math; t=(5730/math.log(2))*math.log(15.3/9.2); result=abs(t-4205)<40"},
  {"claim":"Po-210 alpha Q-value 5.41 MeV", "code":"Q=(209.982874-(205.974465+4.002603))*931.5; result=abs(Q-5.41)<0.02"},
  {"claim":"Decay simulator N and A at t=1800s", "code":"import math; lam=math.log(2)/600; N0=1e20; t=1800; N=N0*math.exp(-lam*t); A=lam*N; result=(abs(N-1.25e19)<1e17) and (abs(A-1.44e16)/1.44e16<0.02)"}
]