If M(A, Z),M_p,and M_n denote the masses of t | vedclass

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(C) The mass defect $\Delta m$ of a nucleus is given by the difference between the sum of the masses of its constituent nucleons and the actual mass o

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

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(C) The mass defect $\Delta m$ of a nucleus is given by the difference between the sum of the masses of its constituent nucleons and the actual mass of the nucleus: $\Delta m = [ZM_p + (A - Z)M_n] - M(A, Z)$.The binding energy $BE$ is related to the mass defect by the equation $BE = \Delta m \cdot c^2$,which implies $\Delta m = BE/c^2$.Substituting this into the mass defect equation: $BE/c^2 = [ZM_p + (A - Z)M_n] - M(A, Z)$.Rearranging the terms to solve for the nuclear mass $M(A, Z)$,we get: $M(A, Z) = [ZM_p + (A - Z)M_n] - BE/c^2$.

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

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