For a nucleus of mass number $A$ and radius $R$, the mass density of nucleus can be represented as
- $\mathrm{A}^3$
- $\mathrm{A}^{\frac{1}{3}}$
- $\mathrm{A}^{\frac{2}{3}}$
- Independent of A
Solution
$\rho=$ Mass of nucleus / Volume of nucleus
For a spherical nucleus with radius R, the volume is:
$\text {Volume }= \left(\frac{4}{3}\right) \pi R^3$
The mass of the nucleus is related to the mass number A :
$\text { Mass }-\mathrm{A} \times \mathrm{m}_u$
(where \(\mathrm{m}_\mathrm{u}\) is the atomic mass unit, approximately $1.66 \times 10^{-27} \mathrm{~kg}$)
Therefore:
$\rho=\left(\mathrm{A} \times \mathrm{m}_\mathrm{u}\right) /\left[ \left(\frac{4}{3}\right) \pi \mathrm{R}^3\right]$
Now, there's an important relationship between the mass number A and nuclear radius R. Empirically, it has been found that:
$\mathrm{R}=\mathrm{Ro} \times \mathrm{~A}^{\left(\frac{1}{3}\right)}$
Where Ro is a constant approximately equal to $1.2 \times 10^{-15} \mathrm{~m}~(1.2$ fermi).
Substituting this into the density equation:
$\begin{aligned}
& \rho=\left(\mathrm{A} \times \mathrm{m}_\mathrm{u}\right) /\left[ \left(\frac{4}{3}\right) \pi\left(\mathrm{Ro} \times \mathrm{A}^{\left(\frac{1}{3}\right)}\right)^3\right] \\
& \rho=(\mathrm{A} \times \mathrm{m}_u) /\left[ \left(\frac{4}{3}\right) \pi \mathrm{Ro}^3 \times \mathrm{A}\right] \\
& \rho=\mathrm{m}_\mathrm{u} /\left[ \left(\frac{4}{3}\right) \pi \mathrm{Ro}^3\right]
\end{aligned}$
This shows that the nuclear density is approximately constant for all nuclei, regardless of their mass number. This is a fundamental property of nuclear matter known as nuclear saturation density.
The numerical value is:
$\rho=2.3 \times 10^{17} \mathrm{~kg} / \mathrm{m}^3$
Asked in: JEE Main 2025 (08 Apr Shift 2)