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Binding Energy

If M A ; Z,...

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 $(1 u = 931.5 M e V / c^{2})$ 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 $M (A, Z) = Z M_{p} + (A - Z) M_{n} - B E / c^{2}$
Binding energy $B E = (m a s s d e f e c t) c^{2}$
$B E = [Z M_{p} + (A - Z) M_{n} - M (A, Z)] c^{2}$
$B E / c^{2} = Z M_{p} + (A - Z) M_{n} - M (A, Z)$
$M (A, Z) = Z M_{p} + (A - Z) M_{n} - B E / c^{2}$

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M_{p}

M_{n}

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

m_{n}

m_{p}

_{Z} X^{A}

M_{p}, a n d, M_{n}

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(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?

_{Z}^{A} X

m (A, Z)

m_{p}

m_{n}

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