Q87 / 90 · JEE Main 2024 (31 Jan Shift 2) · Physics

The mass number of nucleus having radius equal to half of the radius of nucleus with mass number $192$ is:

Solution

Given: $R_{1} = \frac{R_{2}}{2}$

As we know, size of the nucleus is given by $R = R_{0} {(A)}^{1 / 3}$

Therefore, we can write $R_{0} {(A_{1})}^{1 / 3} = \frac{R_{0}}{2} {(A_{2})}^{1 / 3}$

$\Rightarrow A_{1} = \frac{1}{8} A_{2}$

$\Rightarrow A_{1} = \frac{192}{8} = 24$

Hence, the correct answer is option: 'A'.