Question

What is meant by binding energy of nucleus? Determine the binding energy per nucleon of an $\alpha$ particle if the masses of proton, neutron and $\alpha$- particle are $1.00728amu$, $1.00867amu$ and $4.00150amu$ respectively.
(Given: $1amu=931MeV$)

Answer 1

Nuclear binding energy is the minimum energy that would be required to disassemble the nucleus of an atom into its component parts. These component parts are neutrons and protons, which are collectively called nucleons.

Binding energy for $\alpha$-particle ia given by $\Delta m.c^2$.

$\Delta m=4.00150-2.(1.00728+1.00867)$

$\Delta m=0.0304amu$

Total binding energy is $=0.0304\ast 931$ $=28.3$

So binding energy per nucleon is $\frac{28.3}{4}=7.0756$.