Given, $R=12 \mathrm{dis} / \mathrm{min}$ per $\mathrm{g}, R_0=16 \mathrm{dis} / \mathrm{min}$ per $\mathrm{g}, T_{1 / 2}=5760 \mathrm{yr}$ Let $t$ be the span of the tree.

According to radioactive decay law,

$$R=R_0 e^{-\lambda t} \text { or } \frac{R}{R_0}=e^{-\lambda t} \text { or } e^{\lambda t}=\frac{R_0}{R}$$

Taking log on both the sides

$$ \begin{aligned} &\begin{aligned} \lambda t \log _e e & =\log _e \frac{R_0}{R} \Rightarrow \lambda t=\left(\log _{10} \frac{16}{12}\right) \times 2.303 \\ t & =\frac{2.303(\log 4-\log 3)}{\lambda} \\ & =\frac{2.303(0.6020-4.771) \times 5760}{0.6931} \quad \left(\because \lambda=\frac{0.6931}{T_{1 / 2}}\right)\\ & =2391.20 \mathrm{~yr} \end{aligned}\\ \end{aligned}$$

Each particle (neutron and proton) present inside the nucleus is called a nucleon.

Let $\lambda$ be the wavelength $\lambda=10^{-15} \mathrm{~m}$

To detect separate parts inside a nucleon, the electron must have wavelength less than $10^{-15} \mathrm{~m}$.

We know that

$$\begin{aligned} \lambda & =\frac{h}{p} \text { and } K E=P E \quad\text{.... (i)}\\ \text{Energy}\quad & =\frac{h c}{\lambda}\quad\text{.... (ii)} \end{aligned}$$

$$\begin{aligned} &\text { From Eq. (i) and Eq. (ii), }\\ &\begin{aligned} \text { kinetic energy of electron } & =\mathrm{PE}=\frac{h \mathrm{c}}{\lambda}-\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{10^{-15} \times 1.6 \times 10^{-19}} \mathrm{eV} \\ \mathrm{KE} & =10^9 \mathrm{eV} \end{aligned} \end{aligned}$$

(a) According to question, a nuclide 1 is said to be mirror isobar of nuclide 2 , if $Z_1=N_2$ and $Z_2=N_1$.

Now in ${ }_{11} \mathrm{Na}^{23}, Z_1=11, N_1=23-11=12$

$\therefore$ Mirror isobar of ${ }_{11} \mathrm{Na}^{23}$ is ${ }_{12} \mathrm{Mg}^{23}$, for which $Z_2=12=N_1$ and $N_2=23-12=11=Z_1$

(b) $\mathrm{As}_{12}^{23} \mathrm{Mg}$ contains even number of protons (12) against ${ }_{11}^{23} \mathrm{Na}$ which has odd number of protons (11), therefore ${ }_{11}^{23} \mathrm{Mg}$ has greater binding energy than ${ }_{11} \mathrm{Na}^{23}$.

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

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