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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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