Deuteron is a bound state of a neutron and a proton with a binding energy $B=2.2 \mathrm{MeV}$. A $\gamma$-ray of energy $E$ is aimed at a deuteron nucleus to try to break it into a (neutron + proton) such that the $n$ and $p$ move in the direction of the incident $\gamma$-ray. If $E=B$, show that this cannot happen. Hence, calculate how much bigger than $B$ must be $E$ be for such a process to happen.
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 deuteron is bound by nuclear forces just as H -atom is made up of $p$ and $e$ bound by electrostatic forces. If we consider the force between neutron and proton in deuteron as given in the form a coulomb potential but with an effective charge $e^{\prime}$
$$F=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{e^{\prime 2}}{r}$$
estimate the value of $\left(e^{\prime} / e\right)$ given that the binding energy of a deuteron is 2.2 MeV .
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 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.