Arrange the following nuclei in decreasing order of their binding energies per nucleon.
(A) \( {}^{6}\mathrm{Li} \)
(B) \( {}^{238}\mathrm{U} \)
(C) \( {}^{56}\mathrm{Fe} \)
(D) \( {}^{4}\mathrm{He} \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (C), (B), (D), (A)
The binding energy per nucleon curve rises steeply for light nuclei, peaks at about \( 8.8\ \mathrm{MeV} \) near mass number 56 (iron), and falls slowly for heavy nuclei.
\( {}^{56}\mathrm{Fe}:\ \approx 8.8\ \mathrm{MeV} \)
\( {}^{238}\mathrm{U}:\ \approx 7.6\ \mathrm{MeV} \)
\( {}^{4}\mathrm{He}:\ \approx 7.1\ \mathrm{MeV} \)
\( {}^{6}\mathrm{Li}:\ \approx 5.3\ \mathrm{MeV} \)
So the decreasing order is (C), (B), (D), (A).
Helium-4 is a sharp peak on the curve: it is far more tightly bound than its neighbours such as lithium-6, and that is why it lies above \( {}^{6}\mathrm{Li} \) even though both are light.
NCERT (2026-27):
Nuclear Binding Energy – “… We notice the following main features of the plot: (i) the binding energy per nucleon, Ebn, is practically constant, i.e. practically independent of the atomic number for nuclei of middle mass number (30 < A < 170). The curve has a maximum of about 8.75 MeV for A = 56 and has a value of 7.6 MeV for A = 238. (ii) Ebn is lower for both light nuclei (A<30) and heavy nuclei (A>170).”
