If M(A, Z), M_p, and M_n denote the masses of the nucleus $_{Z}^{A}X$, proton, and neutron respectively in uni

Question

If M(A, Z), M_p, and M_n denote the masses of the nucleus $_{Z}^{A}X$, proton, and neutron respectively in units of u (1 u = 931.5 MeV/c^2) and BE represents its binding energy in MeV. Then:

Accepted Answer

1.  **Definition of Mass Defect ($\Delta m$):** The mass of a stable nucleus, $M(A, Z)$, is always less than the sum of the masses of its constituent nucleons (protons and neutrons) in their free state. This difference is called the mass defect.
    $$ \Delta m = [Z M_p + (A-Z) M_n] - M(A, Z) $$
2.  **Binding Energy Relation:** This mass defect is converted into binding energy ($BE$) according to Einstein's mass-energy equivalence relation ($E = mc^2$):
    $$ BE = \Delta m 	imes c^2 $$
    $$ BE = \left( [Z M_p + (A-Z) M_n] - M(A, Z) ight) c^2 $$
3.  **Rearrangement:** To find the mass of the nucleus $M(A, Z)$, rearrange the equation:
    $$ \frac{BE}{c^2} = Z M_p + (A-Z) M_n - M(A, Z) $$
    $$ M(A, Z) = Z M_p + (A-Z) M_n - \frac{BE}{c^2} $$

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