$1 \text{ a.m.u.}$ is equivalent to

Find the binding energy per nucleon for $^{120}_{50}Sn$. Given: mass of proton $m_{p} = 1.00783 \, U$,mass of neutron $m_{n} = 1.00867 \, U$,and mass of tin nucleus $m_{Sn} = 119.902199 \, U$. (Take $1 \, U = 931 \, MeV$) (in $, MeV$)

$m_{p}$ and $m_{n}$ are the masses of a proton and a neutron, respectively. For an element of mass $M$ having $Z$ protons and $N$ neutrons, which of the following is true?

Binding energy per nucleon of ${ }_1^2 H$ and ${ }_2^4 He$ are $1.1 \ MeV$ and $7.0 \ MeV$ respectively. Energy released in the process ${ }_1^2 H + { }_1^2 H \rightarrow { }_2^4 He$ is: (in $MeV$)

The energy equivalent to $1\,mg$ of matter in $MeV$ is

$M_p$ denotes the mass of a proton and $M_n$ that of a neutron. $A$ given nucleus,of binding energy $B$,contains $Z$ protons and $N$ neutrons. The mass $M(N, Z)$ of the nucleus is given by ($c$ is the velocity of light):