Calculate the binding energy of an `alpha`-particle. Given that mass of proton = 1.0073 u, mass of neutron = 1.0087 u, and mass of `alpha`- particle = 4.0015 u.

Explanation

For `._(2)He^(4), A = 4, Z = 2, m_(P)=1.0073u`
`m_(n)=1.0087u, m_(n)=4.0015 u`
(i) `Delta M`
`= [Zm_(P)+(A-Z)m_(n)-M]`
`= [2(1.0073)+(402)(1.0087)-4.00260]`
`= [2xx1.0073+2xx1.0087-4.00260]`
`= (2.0146+2.0174)-4.0015`
`Delta M = [4.032-4.0015] = 0.0305 u`
(ii) `BE=Delta xx c^(2)=Delta M xx 931.5 MeV`
`= 0.0305xx931.5 MeV`
`= 0.0305xx931.5`
`therefore BE=28.41 MeV`