Question

Find out the binding energy per nucleon of an $\alpha$-particle in MeV. It is given that the masses of $\alpha$-particles, 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 $E=△m{c}^2$

$△m=massof\alpha particle-(massofproton+massofneutron)△m=4.00150-2(1.00728+1.00867)=4.00150-4.0319=-0.0304amu$

$E=△m{c}^2$

$E=-0.0304\times 931E=28.30MeV$