Nucleus A has mass number \( 220 \) and its binding energy per nucleon is \( 5.6 \) MeV. It splits into two fragments B and C of mass numbers \( 105 \) and \( 115 \), respectively. If two fragments have the same binding energy of \( 6.4 \) MeV per nucleon, then the energy released in each fission will be:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 176\,\mathrm{MeV} \)
The energy released in fission is the increase in total binding energy: binding energy of the products minus that of the original nucleus.
\( BE_{\mathrm{A}} = 220\times5.6\,\mathrm{MeV} = 1232\,\mathrm{MeV} \)
\( BE_{\mathrm{B}} + BE_{\mathrm{C}} = (105 + 115)\times6.4\,\mathrm{MeV} = 1408\,\mathrm{MeV} \)
\( Q = 1408\,\mathrm{MeV} - 1232\,\mathrm{MeV} = 176\,\mathrm{MeV} \)
Shortcut: The number of nucleons is conserved (220 before and after), so \( Q = 220\times(6.4 - 5.6)\,\mathrm{MeV} = 220\times0.8\,\mathrm{MeV} = 176\,\mathrm{MeV} \).