The stability of heavy nuclei requires the ratio of number of neutrons to number of protons to be around
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 3 : 2
In a nucleus the nuclear force acts only between neighbouring nucleons, but the Coulomb repulsion acts between every pair of protons. As the number of protons grows, the repulsion grows faster than the attraction, so a heavy nucleus needs extra neutrons, which add attraction without adding repulsion.
For light stable nuclei N ≈ Z, but for heavy stable nuclei the ratio rises to about 1.5:
\( {}^{238}_{92}\mathrm{U}: \; N/Z = 146/92 \approx 1.59 \)
\( {}^{208}_{82}\mathrm{Pb}: \; N/Z = 126/82 \approx 1.54 \)
So N : Z is around 3 : 2.
NCERT (2026-27):
Points to Ponder – “… Stability requires the ratio of neutron to proton to be around 1:1 for light nuclei. This ratio increases to about 3:2 for heavy nuclei. (More neutrons are required to overcome the effect of repulsion among the protons.)”