Before the neutrino hypothesis, the beta decay process was throught to be the transition.
$$n \rightarrow p+\bar{e}$$
If this was true, show that if the neutron was at rest, the proton and electron would emerge with fixed energies and calculate them. Experimentally, the electron energy was found to have a large range.
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, $$\begin{aligned}
E_p & =\left(m_p^2 c^4+p^2 c^2\right)^{\frac{1}{2}}=\sqrt{936^2+2^2} \\
& \simeq 936 \mathrm{MeV} \\
E_e & =\left(m_e^2 c^4+p^2 c^2\right)^{\frac{1}{2}}=\sqrt{(0.51)^2+2^2} \\
& \simeq 2.06 \mathrm{MeV}
\end{aligned}$$
The activity $R$ of an unknown radioactive nuclide is measured at hourly intervals. The result found are tabulated as follows
| $t(h)$ | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| $R(MB_q)$ | 100 | 35.36 | 12.51 | 4.42 | 1.56 |
(i) Plot the graph of $R$ versus $t$ and calculate half-life from the graph.
(ii) Plot the graph of $\operatorname{In}\left(\frac{R}{R_0}\right)$ versus $t$ and obtain the value of half-life from the graph.
In the table given below, we have listed values of $R\left(\mathrm{MB}_q\right)$ and $\ln \left(\frac{R}{R_0}\right)$.
| $t(h)$ | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| $R(MB_q)$ | 100 | 35.36 | 12.51 | 4.42 | 1.56 |
| $\frac{R}{R_0}$ | $-$ | $-1.04$ | $-2.08$ | $3.11$ | $-4.16$ |
(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 $t$.
$$\begin{aligned} & \text { Slope of this graph }=-\lambda \\ & \text { from the graph, } \\ & \lambda=-\left(\frac{-4.16-3.11}{1}\right) \Rightarrow=1.05 h^{-1} \\ & \text { Half-life } \\ & T_{1 / 2}=\frac{0.693}{\lambda}=\frac{0.693}{1.05}=0.66 \mathrm{~h} \\ & =39.6 \mathrm{~min} \approx 40 \mathrm{~min} \end{aligned}$$
Nuclei with magic number of proton $Z=2,8,20,28,50,52$ and magic number of neutrons $N=2,8,20,28,50,82$ and 126 are found to be very stable.
(i) Verify this by calculating the proton. separation energy $S_p$ for ${ }^{120} \operatorname{Sn}(Z=50)$ and ${ }^{121} \mathrm{Sb}(Z=51)$.
The proton separation energy for a nuclide is the minimum energy required to separate the least tightly bound proton from a nucleus of that nuclide. It is given by
$$\begin{aligned} S_p & =\left(M_{Z-1, N}+M_{\mathrm{H}}-M_{Z, N}\right) c^2 . \\ \text{Given,}\quad { }^{119} \mathrm{In} & =118.9058 \mathrm{u},{ }^{120} \mathrm{Sn}=199.902199 \mathrm{u}, \\ { }^{121} \mathrm{Sb} & =120.903824 \mathrm{u},{ }^1 \mathrm{H}=1.0078252 \mathrm{u} . \end{aligned}$$
(ii) What does the existence of magic number indicate?
(i) The proton separation energy is given by
$$\begin{aligned} S_{p S n} & =\left(M_{119.70}+M_H-M_{120,70}\right) c^2 \\ & =(118.9058+1.0078252-119.902199) c^2 \\ & =0.0114362 c^2 \\ \text{Similarly}\quad S_{p S p} & =\left(M_{120,70}+M_H-M_{121,70}\right) c^2 \\ & =(119.902199+1.0078252-120.903822) c^2 \\ & =0.0059912 c^2 \end{aligned}$$
Since, $S_{p S_n}>S_{p S b}$, Sn nucleus is more stable than Sb nucleus.
(ii) The existence of magic numbers indicates that the shell structure of nucleus similar to the shell structure of an atom. This also explains the peaks in binding energy/nucleon curve.