The binding energy per nucleon of a nucleus (A = 240) is 7.8 MeV. If it breaks into two equal fragments, each of A = 120 and binding energy per nucleon 8.6 MeV, then total gain in binding energy is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 192 MeV
The total binding energy of a nucleus is its mass number multiplied by the binding energy per nucleon. The gain is the total binding energy of the two fragments minus that of the original nucleus.
\( E_i = 240 \times 7.8\,\mathrm{MeV} = 1872\,\mathrm{MeV} \)
\( E_f = 2 \times 120 \times 8.6\,\mathrm{MeV} = 2064\,\mathrm{MeV} \)
\( \Delta E = 2064\,\mathrm{MeV} - 1872\,\mathrm{MeV} = 192\,\mathrm{MeV} \)
This energy is released in the fission, because the fragments are more tightly bound.
Shortcut: The number of nucleons does not change (240 before and after), so the gain is simply \( 240 \times (8.6 - 7.8)\,\mathrm{MeV} = 192\,\mathrm{MeV} \).
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. 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 Ebn for A = 240 nucleus is about 7.6 MeV, Ebn for the two A = 120 fragment nuclei is about 8.5 MeV. \( \therefore \) 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.”