A nucleus with mass number 240 and binding energy per nucleon 7.6 MeV splits into two fragments each of mass number 120 with binding energy per nucleon 8.5 MeV. The amount of energy released in this process will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 216 MeV
In fission the energy released equals the increase in the total binding energy, because the products are more tightly bound than the parent nucleus. Total binding energy = mass number \( \times \) binding energy per nucleon.
\( \displaystyle \text{BE (parent)} = 240 \times 7.6\,\mathrm{MeV} = 1824\,\mathrm{MeV} \)
\( \displaystyle \text{BE (fragments)} = 2 \times 120 \times 8.5\,\mathrm{MeV} = 2040\,\mathrm{MeV} \)
\( \displaystyle Q = 2040 - 1824 = 216\,\mathrm{MeV} \)
Note: A similar question was also asked in CUET (UG) 2026 on 26-05-2026 (Slot 2).
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 – “… The energy released (the Q value ) in the fission reaction of nuclei like uranium is of the order of 200 MeV per fissioning nucleus. This is estimated as follows: 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. ∴ 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.”