The mass defect $\Delta m$ of a nucleus $\,^{A}_{Z}\mathrm{X}$ with mass $M$ is defined as
A$\Delta m = M - (Z m_p + (A - Z) m_n)$
B$\Delta m = Z m_p - (A - Z) m_n$
C$\Delta m = M - A\,u$
D$\Delta m = (Z m_p + (A - Z) m_n) - M$
Answer & Solution
Correct answer: D. $\Delta m = (Z m_p + (A - Z) m_n) - M$
1. The mass defect captures the fact that an assembled nucleus weighs LESS than the sum of its free constituents — the "missing" mass shows up as the nuclear binding energy.
2. NCERT §13.4 defines: $\Delta m = (\text{sum of nucleon masses}) - M_{\text{nucleus}}$.
3. Sum of free nucleon masses: $Z$ protons of mass $m_p$ plus $(A-Z)$ neutrons of mass $m_n$, giving $Z m_p + (A-Z) m_n$.
4. Subtract the actual nucleus mass $M$: $\Delta m = Z m_p + (A-Z) m_n - M$.
5. By construction, $\Delta m > 0$ for any stable nucleus. Option A has the sign reversed. Options C, D don't represent the right combination of free constituents.
_Source: NCERT Class 12 Physics Part 2, Ch 13, §13.4 (Mass-Energy and Nuclear Binding Energy, definition of mass defect), p. 5._
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