The mass number of nucleus having radius equal to half of the radius of nucleus with mass number 192 is :
Solution
<p>$$\begin{aligned}
& \mathrm{R}_1=\frac{\mathrm{R}_2}{2} \\
& \mathrm{R}_0\left(\mathrm{~A}_1\right)^{1 / 3}=\frac{\mathrm{R}_0}{2}\left(\mathrm{~A}_2\right)^{1 / 3} \\
& \mathrm{~A}_1=\frac{1}{8} \mathrm{~A}_2 \\
& \mathrm{~A}_1=\frac{192}{8}=24
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Bohr's Model of Hydrogen Atom
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