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Binding Energy

If MA, Z, M...

Question

If M(A, Z), $M_{p}$ and $M_{n}$ denote the masses of the nucleus, proton and neutron respectively in units of $u (1 u = 931.5 M e V / c^{2})$ and BE represents its binding energy in MeV, then.

M (A, Z) = Z M_{p} + (A - Z) M_{n} - B E

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M (A, Z) = Z M_{p} + (A - Z) M_{n} + B E / c^{2}

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M (A, Z) = Z M_{p} + (A - Z) M_{n} - B E / c^{2}

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M (A, Z) = A M_{p} + (A - Z) M_{n} + B E

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Solution

The correct option is C $M (A, Z) = Z M_{p} + (A - Z) M_{n} - B E / c^{2}$
$Z M_{p} + (A - Z) M_{n} = M (A, Z)$
Mass effect= $\frac{B . E .}{c^{2}}$
$M (A, Z) = Z M_{p} + (A - Z) M_{n} - B E / c^{2}$

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