Nuclear & Radiochemistry

Question 1

(a) ODEs (1): $\frac{dN_A}{dt}=-\lambda_A N_A,\qquad \frac{dN_B}{dt}=\lambda_A N_A - \lambda_B N_B.$ Solve first: $N_A = N_0 e^{-\lambda_A t}$ (1). Substitute: $\frac{dN_B}{dt}+\lambda_B N_B = \lambda_A N_0 e^{-\lambda_A t}.$ Integrating factor $e^{\lambda_B t}$ (1): $\frac{d}{dt}\!\left(N_B e^{\lambda_B t}\right)=\lambda_A N_0 e^{(\lambda_B-\lambda_A)t}.$ Integrate, apply $N_B(0)=0$ (1): $\boxed{N_B(t)=\frac{\lambda_A N_0}{\lambda_B-\lambda_A}\left(e^{-\lambda_A t}-e^{-\lambda_B t}\right)}$ (2).

(b) Set $dN_B/dt=0$: $-\lambda_A e^{-\lambda_A t}+\lambda_B e^{-\lambda_B t}=0$ (1) ⇒ $\lambda_B e^{-\lambda_B t}=\lambda_A e^{-\lambda_A t}$ (1) ⇒ $\ln(\lambda_B/\lambda_A)=(\lambda_B-\lambda_A)t$ (1) ⇒ $t_{\max}=\dfrac{\ln(\lambda_B/\lambda_A)}{\lambda_B-\lambda_A}$ (1).

(c) $\lambda=\ln2/5730$ yr$^{-1}$. Age $t=\dfrac{1}{\lambda}\ln\!\frac{A_0}{A}=\dfrac{5730}{\ln2}\ln\!\frac{15.30}{9.60}$ (2). $\ln(15.30/9.60)=\ln(1.59375)=0.46628$ (1). $t=\dfrac{5730}{0.6931}(0.46628)=8266.6\times0.46628\approx 3855$ yr (2). ≈ 3.9 × 10³ yr.

(d) Assumption: constant equilibrium $^{14}\text{C}/^{12}\text{C}$ in atmosphere. Real production varies with solar activity and geomagnetic field, and modern fossil-fuel dilution/ bomb spikes alter it (2). Hence calibration against tree-ring/varve records is needed (1). If past atmospheric $^{14}$C was higher, the sample started with more $^{14}$C than assumed; using the modern baseline underestimates initial activity, so the computed age is too young (raw age younger than true) (2).

Question 2

(a) $B/A$ = total nuclear binding energy divided by nucleon number; average energy needed to remove one nucleon (1). Curve rises steeply for light nuclei, peaks near $A\approx 56$ ($^{56}$Fe/$^{62}$Ni) (1). Moving toward the peak from either side increases $B/A$, so products are more tightly bound; the excess binding energy is released — fusing light nuclei (up the left slope) and splitting heavy nuclei (up the right slope) both climb toward the peak (2).

(b) $\Delta m = (m_D+m_T)-(m_{He}+m_n)$ $= (2.014102+3.016049)-(4.002602+1.008665)$ (2) $=5.030151-5.011267=0.018884\ \text{u}$ (2). $Q=0.018884\times931.494=17.59\ \text{MeV}$ (2). Q ≈ 17.6 MeV.

(c) Momentum conservation: $p_n=p_{He}$, so $E_n/E_{He}=m_{He}/m_n=4/1$? Careful: $E=p^2/2m$, so $E \propto 1/m$. With equal momenta $E_n/E_\alpha = m_\alpha/m_n = 4/1$ (2). Total KE $=Q=17.59$ MeV split so $E_n=\dfrac{m_\alpha}{m_\alpha+m_n}Q=\dfrac{4}{5}(17.59)=14.07\ \text{MeV}$ (2). Neutron carries ≈ 14.1 MeV. This high-energy neutron determines wall damage and is captured in a Li blanket ($^6\text{Li}+n\to{}^4\text{He}+\text{T}$) to breed tritium fuel (2).

(d) Using atomic masses, $Q=[4m(^1\text{H})-m(^4\text{He})-2m_e]c^2$. Atomic $^1$H includes 1 electron (4 total on LHS); atomic $^4$He includes 2 electrons; the reaction emits 2 positrons, so we must subtract $2m_e$ for the positron rest masses explicitly (2). $4(1.007825)=4.031300$; minus $4.002602 = 0.028698$; minus $2(0.000549)=0.001098$ ⇒ $\Delta m=0.027600\ \text{u}$ (2). $Q=0.027600\times931.494=25.71\ \text{MeV}$ (2). (The 2 positrons later annihilate adding $\approx2\times1.022$ MeV; net ~26.7 MeV.)

Question 3

(a)

import math
def activity(t, A0=7.4, thalf=6.0):
    lam = math.log(2)/thalf      # per hour
    return A0*math.exp(-lam*t)   # GBq

(3) At $t=9$: $A=7.4\cdot e^{-\ln2\cdot9/6}=7.4\cdot 2^{-1.5}=7.4/2.8284=2.616\ \text{GBq}$ (3). ≈ 2.6 GBq.

(b) $\lambda=\ln2/(6.0\times3600\ \text{s})=0.6931/21600=3.209\times10^{-5}\ \text{s}^{-1}$ (2). $A_0=\lambda N_0\Rightarrow N_0=A_0/\lambda=7.4\times10^{9}/3.209\times10^{-5}=2.31\times10^{14}$ atoms (2).

(c) $H = w_R \times D = 1\times0.20\ \text{mGy}=0.20\ \text{mSv}$ (1). Gy = absorbed energy per kg (J/kg), purely physical; Sv weights that by biological effectiveness ($w_R$) to reflect stochastic health risk (3).

(d) $0.01=e^{-\mu x}\Rightarrow x=\dfrac{\ln100}{\mu}=\dfrac{4.6052}{1.2}=3.84\ \text{cm}$ (4).

[
{"claim":"Radiocarbon age ~3855 yr","code":"t=5730/log(2)*log(15.30/9.60); result = abs(float(t)-3855)<15"},
{"claim":"D-T Q-value ~17.59 MeV","code":"dm=(2.014102+3.016049)-(4.002602+1.008665); Q=dm*931.494; result = abs(float(Q)-17.59)<0.05"},
{"claim":"Neutron KE = 14.07 MeV (4/5 of Q)","code":"dm=(2.014102+3.016049)-(4.002602+1.008665); Q=dm*931.494; En=Rational(4,5)*Q; result = abs(float(En)-14.07)<0.05"},
{"claim":"pp-chain Q ~25.71 MeV","code":"dm=4*1.007825-4.002602-2*0.000549; Q=dm*931.494; result = abs(float(Q)-25.71)<0.05"},
{"claim":"Tc-99m activity at 9h ~2.616 GBq","code":"A=7.4*2**Rational(-3,2); result = abs(float(A)-2.616)<0.01"},
{"claim":"Lead shield thickness for 1% ~3.84 cm","code":"x=log(100)/Rational(12,10); result = abs(float(x)-3.838)<0.01"}
]