The deuteron is bound by nuclear forces just as $\mathrm{H}$-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 of a Coulomb potential but with an effective charge $e\text{'}$

$F=\frac{1}{4\pi {\epsilon }_0}\frac{e^2}{r}$

Estimate the value of $\left(e^{\text{'}}/e\right)$, given that the binding energy of a deuteron is $2.2\mathrm{MeV}$

The activity $R$ of an unknown radioactive nuclide is measured at hourly intervals. The results found are tabulated as follows:

$\begin{array}{|lccccc|}\hline t\left(\mathrm{h}\right)& 0& 1& 2& 3& 4\\ R\left(\mathrm{MBq}\right)& 100& 35.36& 12.51& 4.42& 1.56\\ \hline\end{array}$

Plot the graph of $R$ versus $t$ and calculate half-life from the graph.

The activity $R$of an unknown radioactive nuclide is measured at hourly intervals. The results found are tabulated as follows:

$\begin{array}{|lccccc|}\hline t\left(\mathrm{h}\right)& 0& 1& 2& 3& 4\\ R\left(\mathrm{MBq}\right)& 100& 35.36& 12.51& 4.42& 1.56\\ \hline\end{array}$

Plot the graph of In  $\left(\frac{R}{R_0}\right)$versus $t$ and obtain the value of half-life from the graph.

Nuclei with magic no. of proton $Z=2,8,20,28,50,52$ and magic no. of neutrons $N=2,8,20,28,50,82\text{ and }126$ are found to be very stable,

 Verify this by calculating the proton separation energy $S_p$for ${}^{120}Sn(Z=50){\text{ and }}^{121}Sb=(Z=51)$. 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

${\mathrm{S}}_{\mathrm{p}}=\left(M_{Z-1}{,}_N+M_H-M_{Z,N}\right)c^2$

Given, ${}^{119}\mathrm{In}=118.9058\mathrm{u},{ }^{120}\mathrm{Sn}=119.902199\mathrm{u}{}^{121}\mathrm{Sb}=120.903824\mathrm{u}{,}^1\mathrm{H}=1.0078252\mathrm{u}$

Nuclei with magic no. of proton $Z=2,8,20,28,50,52$ and magic no. of neutrons $N=2,8,20,28,50,82\text{ and }126$ are found to be very stable,

 What does the existance of magic number indicate?