The ratio of the volume of a \({}^{125}_{53}\mathrm{I}\) nucleus to the volume of a \({}^{216}_{84}\mathrm{Po}\) nucleus is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 125 : 216
The radius of a nucleus of mass number \( A \) is \( R=R_0A^{1/3} \). The volume of a sphere is \( \frac{4}{3}\pi R^{3} \). So the volume of a nucleus grows in direct proportion to \( A \).
\( \begin{aligned} V &= \frac{4}{3}\pi R_0^{3}A \\ \frac{V_{\mathrm{I}}}{V_{\mathrm{Po}}} &= \frac{A_{\mathrm{I}}}{A_{\mathrm{Po}}} \\ &= \frac{125}{216} \end{aligned} \)
Therefore, the volumes are in the ratio 125 : 216, the same as the ratio of the mass numbers. The radii are in the ratio 5 : 6.
NCERT (2026-27):
Size of the Nucleus – “It has been found that a nucleus of mass number A has a radius”
\( \displaystyle R=R_0A^{1/3} \qquad (13.5) \)
“where R0 = 1.2 × 10–15 m (=1.2 fm; 1 fm = 10–15 m). This means the volume of the nucleus, which is proportional to R3 is proportional to A. …”