Question

The binding energy of ${}_{17}{Cl}^{35}$ nucleus is $298$ MeV. Find its atomic mass. The mass of hydrogen atom $\left({}_1{H}^1\right)$ is $1.008143$ a.m.u. Given $1$ a.m.u. $=931$ MeV.

Answer 1

The ${}_{17}{Cl}^{35}$ atom has $17$ proton and $18$ neutrons in its nucleus.
Mass of $17{}_1{H}^1$ atom has $17$ protons and $18$ neutrons in its nucleus
Mass of $17$ proton$=17\times 1.008143$ amu$=17.138431$ amu
Mass of $18$ neutrons$=18\times 1.008986=18.161748$ a.m.u
Total$=35.300179$ a.m.u
Mass defect $\Delta m=298/931=0.320085$ a.m.u
The atomic mass of ${}_{17}{Cl}^{35}$ would be the sum of equivalent of the binding energy of the nucleus. Hence atomic mass of
${}_{17}{Cl}^{35}=35.300179-0.320085=34.980094$ a.m.u