(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, \[\begin{align*} A_2(t) &= \lambda_2 N_2(t) \\ &= \frac{\lambda_2 \lambda_1 N_0}{\lambda_2 - \lambda_1} \left( e^{-\lambda_1 t} - e^{-\lambda_2 t} \right) \\ A_2(2592000) &= 1.46 \times 10^{11}\ \mathrm{Bq}. \end{align*}\]