Compute the binding energy of `""_(2)^(4)"He"` nucleus using the following data: Atomic mass of Helium atom, `M_(A) (He) = 4.00260 u` and that of hydrogen atom, `m_(H) = 1.00785 u`.

Explanation

Binding energy BE = `[Zm_(H) + Nm_(n) - M_(A)]c^(2)`
For helium nucleus, `Z = 2, N = A -Z = 4 - 2 = 2`
Mass defect
`Delta m = [(2 xx 1.00785u )] + (2 xx 1.008665u ) - 4.00260 u] Delta m = 0.03038 u`
`B.E = 0.03038u xx c^(2)`
`B.E = 0.03038 xx 931 MeV = 28 MeV`
`[therefore 1 uc^(2) = 931 MeV]`
The binding energy of the `""_(2)^(4)"He"` nucleus is 28 MeV.