Question

The mass of ${}_{3}L{i}^7$ nucleus is $0.042$ a.m.u. less than the sum of masses of its nucleons. Find the binding energy per nucleon.

Answer 1

We have Mass number = 7 and Atomic number= 3

If mass, $m=1u$ and coulomb force $c=3$X${10}^{8}m{s}^{-1}$, then

$E=931MeV$ that is, $1u=931MeV$

Binding Energy = $0.042$X$931=39.10MeV$

Therefore, BInding Energy per Nucleon $=\frac{BindingEnergy}{MassNumber}=\frac{39.10}{7}=5.58=5.6MeV$