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The unit of length convenien on the nuclear scale is a fermi : $1 f=10^{-15} \mathrm{~m}$. Nuclear sizes obey roughly the following empirical relation, $r=r_o \mathrm{~A}^{\frac{1}{3}}$, where $r$ is the radius of the nucleus, $A$ its mass number, and $r_0$ is a constant equal to $1.2 \mathrm{f}$. Show that the rule implies that nuclear mass density is nearly constant for different nuclei. Estimate the mass density of sodium nucleus. Compare it with average mass density of sodium atom.
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Let $\mathrm{m}=$ average mass of proton or neucleon
$\therefore \quad$ Nuclear mass, $\mathrm{M}=\mathrm{mA}$
and radius of nucleus, $r=r_0 \mathrm{~A}^{1 / 3}$
Nuclear density, $\rho=\frac{\text { mass of nucleus }}{\text { volume of nucleus }}$
$$
=\frac{\mathrm{M}}{\frac{4}{3} \pi r^3}=\frac{\mathrm{mA}}{\frac{4}{3} \pi\left(r_0 \mathrm{~A}^{1 / 3}\right)^3}=\frac{3 \mathrm{~m}}{4 \pi r_0^3}
$$
as $r_0$ and $\mathrm{m}$ are constant, hence nuclear density would be constant for all nuclei.
Now, $\mathrm{m}=1.60 \times 10^{-27} \mathrm{~kg}$
and $r_0=1.2 \mathrm{f}=1.2 \times 10^{-15} \mathrm{~m}$
$$
\begin{aligned}
\therefore \quad \rho=\frac{3 \mathrm{~m}}{4 \pi \mathrm{r}_0{ }^3} &=\frac{3 \times 1.66 \times 10^{-27}}{4 \times 3.14 \times\left(1.2 \times 10^{-15}\right)^3} \\
&=2.29 \times 10^{17} \mathrm{~kg} \mathrm{~m}^{-3} .
\end{aligned}
$$
From last question, density of sodium atom $=0.58 \times 10^3 \mathrm{~kg} / \mathrm{m}^3$
$$
\begin{gathered}
\therefore \quad \frac{\text { Nuclear mass density }}{\text { Atomic mass density }}=\frac{2.3 \times 10^{17}}{0.58 \times 10^3} \\
=3.96 \times 10^{14}
\end{gathered}
$$
$\therefore$ Nuclear density is around $10^{15}$ times the atomic density of matter.
$\therefore \quad$ Nuclear mass, $\mathrm{M}=\mathrm{mA}$
and radius of nucleus, $r=r_0 \mathrm{~A}^{1 / 3}$
Nuclear density, $\rho=\frac{\text { mass of nucleus }}{\text { volume of nucleus }}$
$$
=\frac{\mathrm{M}}{\frac{4}{3} \pi r^3}=\frac{\mathrm{mA}}{\frac{4}{3} \pi\left(r_0 \mathrm{~A}^{1 / 3}\right)^3}=\frac{3 \mathrm{~m}}{4 \pi r_0^3}
$$
as $r_0$ and $\mathrm{m}$ are constant, hence nuclear density would be constant for all nuclei.
Now, $\mathrm{m}=1.60 \times 10^{-27} \mathrm{~kg}$
and $r_0=1.2 \mathrm{f}=1.2 \times 10^{-15} \mathrm{~m}$
$$
\begin{aligned}
\therefore \quad \rho=\frac{3 \mathrm{~m}}{4 \pi \mathrm{r}_0{ }^3} &=\frac{3 \times 1.66 \times 10^{-27}}{4 \times 3.14 \times\left(1.2 \times 10^{-15}\right)^3} \\
&=2.29 \times 10^{17} \mathrm{~kg} \mathrm{~m}^{-3} .
\end{aligned}
$$
From last question, density of sodium atom $=0.58 \times 10^3 \mathrm{~kg} / \mathrm{m}^3$
$$
\begin{gathered}
\therefore \quad \frac{\text { Nuclear mass density }}{\text { Atomic mass density }}=\frac{2.3 \times 10^{17}}{0.58 \times 10^3} \\
=3.96 \times 10^{14}
\end{gathered}
$$
$\therefore$ Nuclear density is around $10^{15}$ times the atomic density of matter.
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