The ratio of the surface area of the \({}^{64}_{30}\mathrm{Zn}\) nucleus to that of the \({}^{27}_{13}\mathrm{Al}\) nucleus is about
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(16:9\)
The radius of a nucleus depends on its mass number as \( R=R_0A^{1/3} \). A nucleus is taken as a sphere, so its surface area is \( 4\pi R^{2} \). The surface area is then proportional to \( A^{2/3} \).
\( S\propto R^{2}\propto A^{2/3} \)
\( \begin{aligned} \frac{S_{\mathrm{Zn}}}{S_{\mathrm{Al}}} &= \left(\frac{64}{27}\right)^{2/3} \\ &= \left(\frac{4}{3}\right)^{2} \\ &= \frac{16}{9} \end{aligned} \)
NCERT (2026-27):
Size of the Nucleus – “It has been found that a nucleus of mass number A has a radius”
\( R=R_0A^{1/3} \)\( \qquad (13.5) \)