Consider the chain of two decays

$${ }^{38} \mathrm{~S} \xrightarrow[2.48 \mathrm{~h}]{ }{ }^{38} \mathrm{Cl} \xrightarrow[0.62 \mathrm{~h}]{ }{ }^{38} \mathrm{Ar}$$

At time $t$, Let ${ }^{38} \mathrm{~S}$ have $N_1(t)$ active nuclei and ${ }^{38} \mathrm{Cl}$ have $N_2(t)$ active nuclei.

$$\begin{aligned} \frac{d N_1}{d t} & =-\lambda_1 N_1=\text { rate of formation of } \mathrm{Cl}^{38} \\ \text{Also,}\quad\frac{d N_2}{d t} & =-\lambda_1 N_2+\lambda_1 N_1 \\ \text{But}\quad N_1 & =N_0 e^{-\lambda_1 t} \\ \frac{d N_2}{d t} & =\lambda_1 N_0 e^{-\lambda_1 t}-\lambda_2 N_2\quad\text{.... (i)} \end{aligned}$$

Multiplying by $e^{\lambda z^t} d t$ and rearranging

$$e^{\lambda_2 t} d N_2+\lambda_2 N_2 e^{\lambda_2 t} d t=\lambda_1 N_0 e^{\left(\lambda_2-\lambda_1\right) t} d t$$

Integrating both sides

$$N_2 e^{\lambda_2 t}=\frac{N_0 \lambda_1}{\lambda_2-\lambda_1} e^{\left(\lambda_2-\lambda_1\right) t}+C$$

$$\begin{aligned} \therefore \quad N_2 e^{\lambda_2 t} & =\frac{N_0 \lambda_1}{\lambda_2-\lambda_1}\left(e^{\left(\lambda_2-\lambda_1\right) t}-1\right) \quad\text{.... (ii)}\\ N_2 & =\frac{N_0 \lambda_1}{\lambda_2-\lambda_1}\left(e^{-\lambda_1 t}-e^{-\lambda_2 t}\right) \end{aligned}$$

For maximum count, $\quad \frac{d N_2}{d t}=0$

$$\lambda_1 N_0 e^{-\lambda_1 t}-\lambda_2 N_2=0\quad$$ [From Eq. (i)]

$\Rightarrow \quad \frac{N_0}{N_2}=\frac{\lambda_2}{\lambda_1} e^{\lambda_1 t}\quad$ [From Eq. (ii)]

$$\begin{gathered} e^{\lambda_2 t}-\frac{\lambda_2}{\lambda_1} \cdot \frac{\lambda_1}{\left(\lambda_1-\lambda_1\right)} e^{\lambda_1 t}\left[e^{\left(\lambda_2-\lambda_1\right) t}-1\right]=0 \\ \text{or}\quad e^{\lambda_2 t}-\frac{\lambda_2}{\left(\lambda_2-\lambda_1\right)} e^{\lambda_2 t}+\frac{\lambda_2}{\left(\lambda_2-\lambda_1\right)} e^{\lambda_1 t}=0 \end{gathered}$$

$$\begin{aligned} 1-\frac{\lambda_2}{\left(\lambda_2-\lambda_1\right)}+\frac{\lambda_2}{\left(\lambda_2-\lambda_1\right)} e^{\left(\lambda_1-\lambda_2\right) t} & =0 \\ \frac{\lambda_2}{\left(\lambda_2-\lambda_1\right)} e^{\left(\lambda_1-\lambda_2\right) t} & =\frac{\lambda_2}{\left(\lambda_2-\lambda_1\right)}-1 \\ e^{\left(\lambda_1-\lambda_2\right) t} & =\frac{\lambda_1}{\lambda_2} \\ t & =\left(\log _e \frac{\lambda_1}{\lambda_2}\right) /\left(\lambda_1-\lambda_2\right) \\ & =\frac{\log _e\left(\frac{2.48}{0.62}\right)}{2.48-0.62} \\ & =\frac{\log _e 4}{1.86}=\frac{2.303 \times 2 \times 0.3010}{1.86} \quad\left(\because \lambda=\frac{0.693}{T_{1 / 2}}\right) \\ & =0.745 \mathrm{~S} \end{aligned}$$

Given binding energy $\quad B=2.2 \mathrm{~MeV}$

From the energy conservation law,

$$E-B=K_n+K_p=\frac{p_n^2}{2 m}+\frac{p_p^2}{2 m}\quad\text{.... (i)}$$

$$\begin{aligned} &\text { From conservation of momentum, }\\ &\begin{array}{ll} & p_n+p_p=\frac{E}{C} \quad\text{.... (ii)}\\ \text { As } & E=B, \text { Eq. (i) } p_n^2+p_p^2=0 \end{array} \end{aligned}$$

It only happen if $p_n=p_p=0$

So, the Eq. (ii) cannot satisfied and the process cannot take place.

Let $E=B+X$, where $X<

Put value of $p_{\mathrm{n}}$ from Eq. (ii) in Eq. (i), we get

$$\begin{aligned} X & =\frac{\left(\frac{E}{c}-p_{\mathrm{p}}\right)}{2 m}+\frac{p_{\mathrm{p}}^2}{2 m} \\ \text { or } \quad 2 p_{\mathrm{p}}^2-\frac{2 E p_{\mathrm{p}}}{c}+\frac{E^2}{c^2}-2 m X & =0 \end{aligned}$$

Using the formula of quadratic equation, we get

$$p_p=\frac{\frac{2 E}{c} \pm \sqrt{\frac{4 E^2}{c^2}-8\left(\frac{E^2}{c^2}-2 m X\right)}}{4}$$

For the real value $p_p$, the discriminant is positive

$$\begin{aligned} \frac{4 E^2}{c^2} & =8\left(\frac{E^2}{c^2}-2 m X\right) \\ 16 m X & =\frac{4 E^2}{c^2} \\ X & =\frac{E^2}{4 m c^2} \approx \frac{B^2}{4 m c^2} \quad[\therefore X<< B \Rightarrow E \cong B] \end{aligned}$$

The binding energy is H -atom

$$E=\frac{m e^4}{\pi \varepsilon_0^2 h^2}=13.6 \mathrm{eV}\quad\text{.... (i)}$$

If proton and neutron had charge $e^{\prime}$ each and were governed by the same electrostatic force, then in the above equation we would need to replace electronic mass $m$ by the reduced mass $m^{\prime}$ of proton-neutron and the electronic charge e by $e^{\prime}$.

$$\begin{aligned} m^{\prime} & =\frac{M \times N}{M+N}=\frac{M}{2} \\ & =\frac{1836 \mathrm{~m}}{2}=918 \mathrm{~m} \end{aligned}$$

$$\begin{aligned} &\text { Here, } M \text { represents mass of a neutron/proton }\\ &\therefore \quad \text { Binding energy }=\frac{918 m\left(e^{\prime}\right)^4}{8 \varepsilon_0^2 h^2}=2.2 \mathrm{MeV}\quad\text{.... (ii)} \end{aligned}$$

$$\begin{aligned} &\text { Dividing Eqs. (ii) and (i), we get }\\ &\begin{aligned} 918\left(\frac{e^{\prime}}{e}\right)^4 & =\frac{2.2 \mathrm{MeV}}{13.6 \mathrm{eV}}=\frac{2.2 \times 10^6}{13.6} \\ \left(\frac{e^{\prime}}{e}\right)^4 & =\frac{2.2 \times 10^6}{13.6 \times 918}=176.21 \\ \frac{e^{\prime}}{e} & =(176.21)^{1 / 4}=3.64 \end{aligned} \end{aligned}$$

Before $\beta$-decay, neutron is at rest. Hence, $E_n=m_n c^2, p_n=0$

$$\begin{aligned} \mathbf{p}_n & =\mathbf{p}_\rho+\mathbf{p}_e \\ \text{Or}\quad\mathbf{p}_\rho+\mathbf{p}_e & =0 \Rightarrow\left|\mathbf{p}_\rho\right|=\left|\mathbf{p}_{\mathrm{e}}\right|=\mathbf{p} \\ \text{Also}\quad E_p & =\left(m_\rho^2 c^4+p_\rho^2 c^2\right)^{\frac{1}{2}} \\ E_e & =\left(m_e^2 c^4+p_\rho^2 c^2\right)^{\frac{1}{2}} \\ & =\left(m_e^2 c^4+p_e^2 c^2\right)^{\frac{1}{2}} \end{aligned}$$

From conservation of energy,

$$\begin{aligned} &\left(m_p^2 c^4+p^2 c^2\right)^{\frac{1}{2}}+=\left(m_e^2 c^4+p^2 c^2\right)^{\frac{1}{2}}=m_n c^2 \\ & m_p c^2 \approx 936 \mathrm{MeV}, m_n c^2 \approx 938 \mathrm{MeV}, m_e c^2=0.51 \mathrm{MeV} \end{aligned}$$

Since, the energy difference between $n$ and $p$ is small, $p c$ will be small, $p c<

$\Rightarrow \quad m_p c^2+\frac{p^2 c^2}{2 m_p^2 c^4} \simeq m_n c^2-p c$

To first order $\quad p c \approx m_n c^2-m_p c^2=938 \mathrm{MeV}-936 \mathrm{MeV}=2 \mathrm{MeV}$

This gives the momentum of proton or neutron. Then,

$$\begin{aligned} E_p & =\left(m_p^2 c^4+p^2 c^2\right)^{\frac{1}{2}}=\sqrt{936^2+2^2} \\ & \simeq 936 \mathrm{MeV} \\ E_e & =\left(m_e^2 c^4+p^2 c^2\right)^{\frac{1}{2}}=\sqrt{(0.51)^2+2^2} \\ & \simeq 2.06 \mathrm{MeV} \end{aligned}$$

In the table given below, we have listed values of $R\left(\mathrm{MB}_q\right)$ and $\ln \left(\frac{R}{R_0}\right)$.

(i) When we plot the graph of $R$ versus $t$, we obtain an exponential curve as shown.

From the graph we can say that activity $R$ reduces to $50 \%$ in $t=O B \approx 40 \mathrm{~min}$

So, $t_{1 / 2} \approx 40 \mathrm{~min}$.

(ii) The adjacent figure shows the graph of $\ln \left(R / R_0\right)$ versus $t$.

$$\begin{aligned} & \text { Slope of this graph }=-\lambda \\ & \text { from the graph, } \\ & \lambda=-\left(\frac{-4.16-3.11}{1}\right) \Rightarrow=1.05 h^{-1} \\ & \text { Half-life } \\ & T_{1 / 2}=\frac{0.693}{\lambda}=\frac{0.693}{1.05}=0.66 \mathrm{~h} \\ & =39.6 \mathrm{~min} \approx 40 \mathrm{~min} \end{aligned}$$