Question

The ratio of the surface area of the nuclei ${}_{52}T{e}^{125}$ to that of ${}_{13}A{l}^{27}$ is:

• A. $\frac{5}{3}$
• B. $\frac{125}{17}$
• C. $\frac{1}{4}$
• D. $\frac{25}{9}$
• E. $\frac{3}{5}$

Answer 1

Answer: (D)

$\frac{25}{9}$

By nuclear size
The radius of the nucleus
$R=R_o{A}^{1/3}$
where $R_o=1.1\times {10}^{-15}m$ is an empirical constant and $A$ is the mass number.
So, $R\propto A^{1/3}$
Thus $R_{Te}=A_{Te}^{1/3}$ ...(i)
Similarly, $R_{Al}=A_{Te}^{1/3}$ ...(ii)
$\frac{R_{Te}}{R_{Al}}={\left(\frac{125}{27}\right)}^{1/3}=\frac{5}{3}$
Surface area $A\propto R^2$
$\therefore \frac{A_{Te}}{A_{Al}}=(\frac{R_{Te}}{R_{Al}}{)}^2={\left(\frac{5}{3}\right)}^2=\frac{25}{9}$