Question

Find out the binding energy per nucleon of an $\alpha$-particle in MeV. It is given that the masses of $\alpha$-particle, proton and neutron are respectively $4.00150$ a.m.u., $1.00728$ a.m.u. and $1.00867$ a.m.u.

Answer 1

Binding Energy $=△m{c}^2$

$△m=$ mass of $\alpha$ particle $-$ mass of total number of proton and neutron

$=4.00150-2(1.00728+1.00867)=4.00150-4.0319=-0.0304u$

$E=△m{c}^2$$=0.0304\times 931MeV=28.30MeV$

So, Binding energy per nucleon $=\frac{E}{no.ofnucleon}=\frac{28.30}{4}=7.07MeV$