13.Nuclei
hard

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

A

$24$

B

$32$

C

$40$

D

$20$

(JEE MAIN-2024)

Solution

$R_1=\frac{R_2}{2}$

$R_0\left(A_1\right)^{1 / 3}=\frac{R_0}{2}\left(A_2\right)^{1 / 3}$

$A_1=\frac{1}{8} A_2$

$A_1=\frac{192}{8}=24$

Standard 12
Physics

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