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If $M(A, Z)$, $M_p$ and $M_n$ denote the masses of the nucleus $^{A}_{Z}X$, proton and neutron respectively in units of u ($1\,u = 931.5$ MeV/$c^2$) and BE represents its binding energy in MeV, then -
A
$M(A, Z) = ZM_p + (A - Z)M_n - BE$
B
$M(A, Z) = ZM_p + (A - Z)M_n + BE/c^2$
C
$M(A, Z) = ZM_p + (A - Z)M_n - BE/c^2$
D
$M(A, Z) = ZM_p + (A - Z)M_n + BE$
Detailed Solution
A nucleus $^{A}_{Z}X$ has $Z$ protons and $(A - Z)$ neutrons.
Mass defect: $\Delta m = ZM_p + (A - Z)M_n - M(A, Z)$
Binding energy: $BE = \Delta m\,c^2$
$BE = [ZM_p + (A - Z)M_n - M(A, Z)]\,c^2$
$\Rightarrow \dfrac{BE}{c^2} = ZM_p + (A - Z)M_n - M(A, Z)$
$\Rightarrow M(A, Z) = ZM_p + (A - Z)M_n - \dfrac{BE}{c^2}$
The mass of the nucleus is less than the total mass of its nucleons by $BE/c^2$.
Mass defect: $\Delta m = ZM_p + (A - Z)M_n - M(A, Z)$
Binding energy: $BE = \Delta m\,c^2$
$BE = [ZM_p + (A - Z)M_n - M(A, Z)]\,c^2$
$\Rightarrow \dfrac{BE}{c^2} = ZM_p + (A - Z)M_n - M(A, Z)$
$\Rightarrow M(A, Z) = ZM_p + (A - Z)M_n - \dfrac{BE}{c^2}$
The mass of the nucleus is less than the total mass of its nucleons by $BE/c^2$.