find the binding energy of an `alpha `-particle from the following data .
Mass of the helium necleus m `(""._(2)He^(4))=4.001265u`
Mass of neutron `m_(n)= 1.008666 u`
(take , 1 u=931.4813 Me V )

Answer

`7.12855 MeV`

Explanation

Mass of two protons is `2 xx1.007277 =2.014554 a.m.u.`
Mass of two neutrons is `2 xx 1.008666 a.m.u. =2.017332 a.m.u.`
Total initial mass of two protons and two neutrons is `2.041554 +2.017332 =4.031866 a.m.u.`
Mass defect, `Delta m=4.031816 -4.001265 a.m.u.`
`=0.030621 a.m.u`
Now, binding energy of `alpha`-particles in `MeV` is given as
`Delta E_(n) =Delta m xx931.2 =28.5142 MeV`
Binding energy per nucleon `=28.5142//4=7.12855 MeV`.