The ratio of the density of oxygen nucleus ($_8^{16}O$) and helium nucleus ($_2^{4}\mathrm{He}$) is
Solution
Nuclear density is independent of mass number<br/><br/>
As nuclear density $=\frac{\mathrm{Au}}{\frac{4}{3} \pi \mathrm{R}^3}$<br/><br/>
Also, $\mathrm{R}=\mathrm{R}_0 \mathrm{~A}^{\frac{1}{3}}$<br/><br/>
And $R^3=R_0^3 A$<br/><br/>
$\Rightarrow$ Nuclear density $=\frac{\mathrm{Au}}{\frac{4}{3} \pi \mathrm{R}_0^3 \mathrm{~A}}$<br/><br/>
Nuclear density $=\frac{3 \mathrm{u}}{4 \pi \mathrm{R}_0^3}$<br/><br/>
$\Rightarrow$ Nuclear density is independent of $\mathrm{A}$
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Bohr's Model of Hydrogen Atom
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