If $m, m_{n}$ and $m_{p}$ are the masses of $_{Z} X^{A}$ nucleus, neutron and proton respectively:

m = (A - Z) m_{n} + {Z m}_{p}

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m < (A - Z) m_{n} + {Z m}_{p}

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m > (A - Z) m_{n} + {Z m}_{p}

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m = (A - Z) m_{n} + {Z m}_{n}

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Solution

The correct option is B $m < (A - Z) m_{n} + {Z m}_{p}$
Mass of the nucleus is less than the sum of the masses of the constituents particles.
Given that:
$m_{n}$ and $m_{p}$ are masses of neutron and proton respectively.
$Z$ and $A$ are atomic number and mass number

Sum of masses $= Z m_{p} + (A - Z) m_{n}$
If $m$ is the real mass of the nucleus, the mass defect is:
$Δ m = Z m_{p} + (A - Z) m_{n} - m$
$\Rightarrow Δ m + m = Z m_{p} + (A - Z) m_{n}$
$\Rightarrow m < Z m_{p} + (A - Z) m_{n}$

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