The radius of ${}_{32}X$ nuclide is measured to be twice the radius of ${}_4{}^{9}Be$. The number of neutrons in X is

radius of the nucleus is directly proportional  $A^{\frac{1}{3}}$ where A is the mass no of the element.

$\frac{r_1}{r_2}={\left(\frac{A_1}{A_2}\right)}^{\frac{1}{3}}$    then ${\left(\frac{r_1}{r_2}\right)}^3=\frac{A_1}{A_2}$

$$\begin{gathered} as r_1=2r_2 8=\frac{A_1}{A_2}=\frac{A_1}{9} A_1=72 \\ 72= n_p+n_n=30+n_n \\ no of neuron n_n is 42 \end{gathered}$$