B M Sharma Solutions for Chapter: Nuclear Physics, Exercise 1: DPP

The $\alpha$-particles of energy $400 \mathrm{keV}$ are bombarded on nucleus of ${}_{82}\mathrm{Pb}$. In scattering of $\text{α}$-particles, its minimum distance from nucleus will be,

 In the nuclear reaction ${}_2\mathrm{He}^4{+}_z{X}^A{\to }_{z+2}{Y}^{A+3}+A$,  $A$ denotes,

For a nucleus to be stable, the correct relation between neutron number $N$ and proton number $Z$ is,

Two nucleons are at a separation of $1\times {10}^{-15} \mathrm{m}$. The net force(nuclear) between them is $F_1$ if both are neutrons, $F_2$ if both are protons and $F_3$ if one is a proton and other is a neutron. In such a case,

$M_n$ and $M_p$ represent mass of neutron and proton, respectively. If an element having atomic mass $M$ has $N$-neutrons and $Z$-protons, then the correct relation will be,

If $m, m_n$ and $m_p$ are the masses of ${}_{Z}X^A$ nucleus, neutron and proton, respectively, then,

The rest energy of an electron is,
 

Let $m_p$ be the mass of proton, $m_n$ the mass of neutron, $M_1$ the mass of ${}_{10}{}^{20}\mathrm{Ne}$ nucleus and $M_2$ the mass of ${}_{20}{}^{40}\mathrm{Ca}$ nucleus. Then,