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Binding Energy

If M A, Z, ...

Question

If M (A, Z), $M_{p}, a n d, M_{n}$
denote the masses of
the nucleus
$_{Z}^{A} X$ proton and neutron
respectively in units of u $(1 u = 931.5 M e V / C^{2})$
and BE represents its bonding 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) = Z M_{p} + (A - Z) M_{n} + B E

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Solution

The correct option is C $M (A, Z) = Z M_{p} + (A - Z) M_{n} - B E / C^{2}$
Mass defect
$Δ m = Z M_{p} + (A - Z) M_{n} - M (A, Z)$
Binding energy $= Δ m C^{2}$
$B E = [Z M_{p} + (A - Z) M_{n} - M (A, Z)] . C^{2}$
$M (A, Z) = Z M_{p} + (A - Z) M_{n} - \frac{B E}{C^{2}}$

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Similar questions

m_{n}

m_{p}

_{Z} X^{A}

M (A; Z), M_{p}

M_{n}

_{Z}^{A} X

(1 u = 931.5 M e V / c^{2})

If m, $m_{n}$ and $m_{p}$ are the masses of $_{Z} X^{A}$ nucleus, neutron and proton respectively, then which of the following is correct?

M_{p}

M_{n}

u (1 u = 931.5 M e V / c^{2})

_{Z}^{A} X

M (A, Z)

M_{p}

M_{n}

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