7th April 2021

# (a)

By conservation of energy-mass, the total energy released is \begin{align*} E &= \left[M({}^{210}_{\phantom{2}84}\mathsf{Po}) - M({}^{206}_{\phantom{2}82}\mathsf{Pb}) - M({}^{4}_{2}\mathsf{He})\right]c^2 \\ &= (209.9828736 - 205.9744653 - 4.002603254)uc^2 \\ &= 0.005805046 uc^2 \\ &= 5.407 \ \mathrm{MeV} \end{align*}

For recoil, let $$M = M({}^{206}_{\phantom{2}82}\mathsf{Pb})$$ and $$m = M({}^{4}_{2}\mathsf{He})$$, \begin{align*} \frac{M({}^{206}_{\phantom{2}82}\mathsf{Pb})) V^2}2 &= \frac{M({}^{4}_{2}\mathsf{He}) E}{M({}^{206}_{\phantom{2}82}\mathsf{Pb}) + M({}^{4}_{2}\mathsf{He})} \\ V^2 &= \frac{2 m E}{M(M + m)} \\ &= 2 \frac{4.002603254}{205.9744653} \frac{0.005805046 c^2}{205.9744653 + 4.002603254} \\ V &= 3.108 \times 10^5\ \mathrm{m}\,\mathrm{s}^{-1} \end{align*} the fraction of total energy is given by $\frac{m}{M + m} = 0.019062097.$

# (b)

For $${}^{232}_{\phantom{2}90}\mathsf{Th}$$ to decay into $${}^{208}_{\phantom{2}82}\mathsf{Pb}$$, it loses $$24$$ nucleons, so there are $$6$$ $$\alpha$$-decays.

If $${}^{232}_{\phantom{2}90}\mathsf{Th}$$ undergoes six $$\alpha$$-decays, its atomic number would have been $$78$$, so there are $$4$$ $$\beta$$-decays.

# (c)

Let $$N_1(t)$$ denote number of $${}^{210}_{\phantom{2}83}\mathsf{Bi}$$ atoms at time $$t$$. Assuming one month is 30 days, we have $$2592000$$ seconds in a month.

Initially, there are $N_0 = N_1(0) = \frac{10^{-6}}{209.9841204 u} = 2.86790294 \times 10^{18}$ many $${}^{210}_{\phantom{2}83}\mathsf{Bi}$$ atoms.

So the number of $${}^{210}_{\phantom{2}83}\mathsf{Bi}$$ beta decays per second after one month is given by \begin{align*} N_1(t) &= N_0 e^{-\lambda_1 t} \\ A_1(t) &= \lambda N_1(t) \\ &= \lambda N_0 e^{-\lambda_1 t} \\ A_1(2592000) &= \lambda_1 N_0 e^{-\lambda_1 2592000} \\ &= 7.25\times 10^{10} \ \mathrm{Bq}. \end{align*}

Let $$N_2(t)$$ denote number of $${}^{210}_{\phantom{2}84}\mathsf{Po}$$ atoms at time $$t$$. $$N_2(t)$$ is governed by the differential equation $\frac{dN_2(t)}{dt} = \lambda_1 N_1(t) - \lambda_2 N_2(t)$ and we know that $$N_2(0) = 0$$.

Solving the DE, \begin{align*} \frac{dN_2(t)}{dt} &= \lambda_1 N_0 e^{-\lambda_1 t} - \lambda_2 N_2(t) \\ e^{\lambda_2 t} \frac{dN_2(t)}{dt} + e^{\lambda_2 t} \lambda_2 N_2(t) &= \lambda_1 N_0 e^{(\lambda_2-\lambda_1) t} \\ \frac{d}{dt}\left[ e^{\lambda_2 t} N_2(t) \right] &= \lambda_1 N_0 e^{(\lambda_2-\lambda_1) t} \\ e^{\lambda_2 t} N_2(t) &= \frac{\lambda_1 N_0}{\lambda_2 - \lambda_1} e^{(\lambda_2-\lambda_1) t} + C \end{align*} substituting $$N_2(0) = 0$$ to find $$C$$, we have $C = -\frac{\lambda_1 N_0}{\lambda_2 - \lambda_1}$ and we have the expression \begin{align*} e^{\lambda_2 t} N_2(t) &= \frac{\lambda_1 N_0}{\lambda_2 - \lambda_1} \left( e^{(\lambda_2-\lambda_1) t} - 1 \right) \\ N_2(t) &= \frac{\lambda_1 N_0}{\lambda_2 - \lambda_1} \left( e^{-\lambda_1 t} - e^{-\lambda_2 t} \right) \end{align*} now we can find the number of $${}^{210}_{\phantom{2}84}\mathsf{Po}$$ alpha decays per second after one month, \begin{align*} A_2(t) &= \lambda_2 N_2(t) \\ &= \frac{\lambda_2 \lambda_1 N_0}{\lambda_2 - \lambda_1} \left( e^{-\lambda_1 t} - e^{-\lambda_2 t} \right) \\ A_2(2592000) &= 1.46 \times 10^{11}\ \mathrm{Bq}. \end{align*}