The radius of the \( {}^{4}_{2}\mathrm{He} \) nucleus is 3 fermi. The radius of the \( {}^{108}_{47}\mathrm{Ag} \) nucleus will be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 9 fermi
The radius of a nucleus of mass number A is \( R=R_0A^{1/3} \), so the radius is proportional to the cube root of the mass number. The ratio of the two radii depends only on the mass numbers 108 and 4.
\( \begin{aligned} \frac{R_{\mathrm{Ag}}}{R_{\mathrm{He}}} &= \left(\frac{108}{4}\right)^{1/3} \\ &= 27^{1/3} \\ &= 3 \end{aligned} \)
\( \begin{aligned} R_{\mathrm{Ag}} &= 3\times3\,\mathrm{fermi} \\ &= 9\,\mathrm{fermi} \end{aligned} \)