JEE Main 2025 — Physics Modern Physics

The mass density ( $\rho$ρ) of a nucleus is defined as the ratio of its mass to its volume:

$\rho=$ρ= Mass of nucleus / Volume of nucleus

For a spherical nucleus with radius R , the volume is:

$$\text { Volume }=(4 / 3) \pi R^3$$ Volume =(4/3)πR3

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}$1.66×10−27 kg)

Therefore:

$$\rho=\left(\mathrm{A} \times \mathrm{m} \_\mathrm{u}\right) /\left[(4 / 3) \pi \mathrm{R}^3\right]$$ρ=(A×m_u)/[(4/3)πR3]

Now, there's an important relationship between the mass number A and nuclear radius R. Empirically, it has been found that:

$$\mathrm{R}=\mathrm{Ro} * \mathrm{~A}^{\wedge}(1 / 3)$$R=Ro∗ A∧(1/3)

Where Ro is a constant approximately equal to $1.2 \times 10^{-15} \mathrm{~m}(1.2$1.2×10−15 m(1.2 fermi).

Substituting this into the density equation:

$\begin{aligned}
& \rho=\left(\mathrm{A} \times \mathrm{m} _\mathrm{u}\right) /\left[(4 / 3) \pi\left(\mathrm{Ro} \times \mathrm{A}^{\wedge}(1 / 3)\right)^3\right] \
& \rho=(\mathrm{A} \times \mathrm{m} \text { _u }) /\left[(4 / 3) \pi \mathrm{Ro}^3 \times \mathrm{A}\right] \
& \rho=\mathrm{m} _\mathrm{u} /\left[(4 / 3) \pi \mathrm{Ro}^3\right]
\end{aligned}$

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:

$$\rho=2.3 \times 10^{17} \mathrm{~kg} / \mathrm{m}^3$$ρ=2.3×1017 kg/m3