JEE Main 2025 — Nuclear Physics Question with Solution

The mass density ( $\rho$) of a nucleus is defined as the ratio of its mass to its volume:
$\rho =$ Mass of nucleus / Volume of nucleus
For a spherical nucleus with radius R , the volume is:
$\text{ Volume }=(4/3)\pi R^3$
The mass of the nucleus is related to the mass number A :
\text { Mass }-\mathrm{A} \times \mathrm{m} \text { _u }
(where m_u is the atomic mass unit, approximately $1.66\times 10^{-27}\mathrm{kg}$)
Therefore:
$\rho =\left(\mathrm{A}\times \mathrm{m}\_\mathrm{u}\right)/\left[(4/3)\pi {\mathrm{R}}^3\right]$
Now, there's an important relationship between the mass number A and nuclear radius R. Empirically, it has been found that:
$\mathrm{R}=\mathrm{Ro}\ast {\mathrm{A}}^{\wedge }(1/3)$
Where Ro is a constant approximately equal to $1.2\times 10^{-15}\mathrm{m}(1.2$ fermi).
Substituting this into the density equation:
$\begin{aligned}
& \rho=\left(\mathrm{A} \times \mathrm{m} \_\mathrm{u}\right) /\left[(4 / 3) \pi\left(\mathrm{Ro} \times \mathrm{A}^{\wedge}(1 / 3)\right)^3\right] \\ & \rho=(\mathrm{A} \times \mathrm{m} \text { _u }) /\left[(4 / 3) \pi \mathrm{Ro}^3 \times \mathrm{A}\right] \\ & \rho=\mathrm{m} \_\mathrm{u} /\left[(4 / 3) \pi \mathrm{Ro}^3\right]
\end{aligned}$<br/>Thisshowsthatthenucleardensityisapproximatelyconstantforallnuclei,regardlessoftheirmassnumber.Thisisafundamentalpropertyofnuclearmatterknownasnuclearsaturationdensity.<br/>Thenumericalvalueis:<br/>$\rho=2.3 \times 10^{17} \mathrm{~kg} / \mathrm{m}^3$