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Assuming the nuclei to be spherical in shape, how does the surface area of a nucleus of mass number $A_{1}$ compare with that of a nucleus of mass number $A_{2}$

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${(A_{1} \times A_{2})}^{2}$

${(A_{1} \times A_{2})}^{\frac{2}{3}}$

${(\frac{A_{1}}{A_{2}})}^{\frac{2}{3}}$

${(\frac{A_{1}}{A_{2}})}^{2}$

answer is B.

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Detailed Solution

The surface area $S$ of a sphere is given by

$S = 4 {πR}^{2}$

Since, $R = R_{0} A^{1 / 3}$

$\Rightarrow S = 4 {πR}^{2}$ $\propto$ $A^{2 / 3}$

$\Rightarrow \frac{S_{1}}{S_{2}} = {(\frac{A_{1}}{A_{2}})}^{\frac{2}{3}}$

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