For nuclei of average mass, the binding energy per nucleon is about 8 MeV. The energy needed to remove the electron from a hydrogen atom in its ground state is 13.6 eV. The ratio of the binding energy per nucleon to this energy is about
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(5.9\times10^{5}\)
Both energies must be in the same unit before they are divided. Here \( 8\ \mathrm{MeV}=8\times10^{6}\ \mathrm{eV} \).
\( \begin{aligned} \text{Ratio} &= \frac{8\times10^{6}\ \mathrm{eV}}{13.6\ \mathrm{eV}} \\ &\approx 5.9\times10^{5} \end{aligned} \)
From the above, a nucleon is bound in a nucleus about \( 6\times10^{5} \) times more strongly than the electron in a hydrogen atom. So the force that holds a nucleus together must be of a different kind from the Coulomb force on electrons.
NCERT (2026-27):
Nuclear Force – “… In Section 13.4, we have seen that for average mass nuclei the binding energy per nucleon is approximately 8 MeV, which is much larger than the binding energy in atoms. …”