A nucleus `._Z^A X` has mass represented by `M(A, Z)`. If `M_p` and `M_p` denote the mass of proton and neutron respectively and `BE` the binding energy (in MeV), then :

Options

A. `BE = [M(A, Z) - ZM_p - (A -Z) M_n] c^2`
B. `BE = [ZM_p + (A - Z) M_n - M(A, Z)] c^2`
C. `BE = [ZM_p + AM_n - M(A, Z)] c^2`
D. `BE = M (A, Z) - ZM_p - (A - Z) M_n`

Answer

B

Explanation

(b) In the case of formation of a nucleus the evolution of energy equal to the binding energy of the nucleus take place due to disappearance of a fraction of the total mass. If the quantity of mass dissappearing is `Delta M`. Then the binding energy is
`BE = Delta Mc^2`
From the above discussion, it is clear that the mass of the nucleus must be less than the sum of the masses of the constituent neutrons and protons, we can then write
`Delta M = ZM_p + NM_n - M(A, Z)`
where `M(A , Z)` is the mass of the atom of mass number `A` and atomic number `Z`. Hence, the binding energy of the nucleus is
`BE = [ZM_p + NM_n - M(A , Z)]c^2`
`BE = [ZM_p + (A - Z) M_n - M(A, Z)]c^2`
Where `N = A - Z` = number of neutrons.