If a nucleus with mass number A = 240 with binding energy per nucleon…

The correct answer is Option (1) → 216 MeV

The energy released in fission equals the increase in total binding energy: the fragments are more tightly bound than the parent nucleus. Total binding energy = mass number × binding energy per nucleon.

    \( E_{\mathrm{b,\ parent}} = 240 \times 7.6\,\mathrm{MeV} = 1824\,\mathrm{MeV} \)

    \( E_{\mathrm{b,\ fragments}} = 2 \times 120 \times 8.5\,\mathrm{MeV} = 2040\,\mathrm{MeV} \)

    \( Q = 2040\,\mathrm{MeV} - 1824\,\mathrm{MeV} = 216\,\mathrm{MeV} \)

Shortcut: The number of nucleons does not change (240 = 2 × 120), so multiply it by the gain in binding energy per nucleon: \( 240 \times (8.5 - 7.6)\,\mathrm{MeV} = 240 \times 0.9\,\mathrm{MeV} = 216\,\mathrm{MeV} \).

Note: A similar question was also asked in CUET (UG) 2025 on 03-06-2025 (Slot 1).

Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.

NCERT (2026-27):
Fission – “… Let us take a nucleus with A = 240 breaking into two fragments each of A = 120. Then E_bn for A = 240 nucleus is about 7.6 MeV, E_bn for the two A = 120 fragment nuclei is about 8.5 MeV. ∴ Gain in binding energy for nucleon is about 0.9 MeV. Hence the total gain in binding energy is 240×0.9 or 216 MeV.”