Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

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:

M(A, Z) = Z M_p + (A-Z) M_n - BE/c^2

M(A, Z) = Z M_p + (A-Z) M_n + BE

M(A, Z) = Z M_p + (A-Z) M_n - BE

M(A, Z) = Z M_p + (A-Z) M_n + BE/c^2

Step-by-Step Solution

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. $\Delta m = [Z M_p + (A-Z) M_n] - M(A, Z)$
Binding Energy Relation: This mass defect is converted into binding energy ( $BE$ ) according to Einstein's mass-energy equivalence relation ( $E = mc^2$ ): $BE = \Delta m \times c^2$ $BE = \left( [Z M_p + (A-Z) M_n] - M(A, Z) \right) c^2$
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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