## In This post you will get answer of Q.

Nuclear radius of $^{27}A 1$ is 3/5 times that of $^{A}X$. Value of the mass number A is

Solution:

## Radius of a nucleus is given by, $R = R_0 : A^{1/3}$ or $R propto A^{1/3}$

$therefore , frac{R_{A1}}{R_{X}} = left(frac{A_{A1}}{A_{X}}right)^{1 /3}$

Given that, $ frac{R_{A1}}{R_{X}} = frac{3}{5} , A_{A1} = 27 , A_{X} = ? $

So, $frac{3}{5} = left(frac{27}{A_{X}}right)^{1 3} or left(frac{3}{5}right)^{3} = frac{27}{A_{X}}$

$ therefore , A_{X} = 125 $