Inside a large nucleus, a nucleon is attracted only by the nucleons that lie within the range of the nuclear force. A nucleon can have at most \(p\) such neighbours, so its binding energy is about \(pk\), where \(k\) is a constant. For large nuclei, the total binding energy of a nucleus of mass number \(A\) is roughly proportional to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(A\)
The nuclear force is short-ranged. A nucleon deep inside a large nucleus feels only the few neighbours close to it. Adding more nucleons to the nucleus does not change this number, so each nucleon keeps a binding energy of about \( pk \).
All \( A \) nucleons together give the total binding energy:
\( \begin{aligned} E_b &\approx A\times pk \\ E_b &\propto A \end{aligned} \)
From the above, \( E_b/A \approx pk \) stays nearly constant. Hence, the total binding energy of a large nucleus grows in proportion to \( A \).
If the force reached every nucleon of the nucleus, the number of bound pairs, and so \( E_b \), would grow as about \( A^{2} \). The short range of the nuclear force rules this out.
NCERT (2026-27):
Nuclear binding energy – “(ii) The constancy of the binding energy in the range 30 < A < 170 is a consequence of the fact that the nuclear force is short-ranged. … If a nucleon can have a maximum of p neighbours within the range of nuclear force, its binding energy would be proportional to p. … The binding energy per nucleon is a constant and is approximately equal to pk. …”