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The ratio of volume occupied by an atom to the volume of the nucleus is
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Verified Answer
The correct answer is:
$10^{15}: 1$
As we knows, radius of an atom,
$\gamma_{\mathrm{A}} \approx 10^{-10} \mathrm{~m}$
radius of nucleus, $\gamma_{\mathrm{B}} \approx 10^{-15} \mathrm{~m}$
So, ratio of their volumes
$\begin{aligned}
& \frac{\mathrm{V}_{\mathrm{A}}}{\mathrm{V}_{\mathrm{N}}}=\frac{\frac{4}{3} \pi \mathrm{r}_{\mathrm{A}}^{3}}{\frac{4}{3} \pi \mathrm{r}_{\mathrm{N}}^{3}}=\left(\frac{\mathrm{r}_{\mathrm{A}}}{\mathrm{I}_{\mathrm{N}}}\right)^{3}=\left(\frac{10^{-10}}{10^{-15}}\right)^{3} \\
\therefore \quad & \mathrm{V}_{\mathrm{A}}: \mathrm{V}_{\mathrm{N}}=10^{15}: 1
\end{aligned}$
$\gamma_{\mathrm{A}} \approx 10^{-10} \mathrm{~m}$
radius of nucleus, $\gamma_{\mathrm{B}} \approx 10^{-15} \mathrm{~m}$
So, ratio of their volumes
$\begin{aligned}
& \frac{\mathrm{V}_{\mathrm{A}}}{\mathrm{V}_{\mathrm{N}}}=\frac{\frac{4}{3} \pi \mathrm{r}_{\mathrm{A}}^{3}}{\frac{4}{3} \pi \mathrm{r}_{\mathrm{N}}^{3}}=\left(\frac{\mathrm{r}_{\mathrm{A}}}{\mathrm{I}_{\mathrm{N}}}\right)^{3}=\left(\frac{10^{-10}}{10^{-15}}\right)^{3} \\
\therefore \quad & \mathrm{V}_{\mathrm{A}}: \mathrm{V}_{\mathrm{N}}=10^{15}: 1
\end{aligned}$
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