JEE Physics 2023 Question with Solution

Given: oxygen nucleus $^{16}\!O$ and helium nucleus $^{4}\!He$.

Find: the ratio of their nuclear densities.

Nuclear density is given by

$$\rho = \frac{\text{Mass}}{\text{Volume}} = \frac{Au}{\frac{4}{3}\pi R^3}$$

Using $R = R_0 A^{1/3}$,

$$\rho = \frac{Au}{\frac{4}{3}\pi (R_0 A^{1/3})^3} = \frac{Au}{\frac{4}{3}\pi R_0^3 A}$$

So,

$$\rho = \frac{3u}{4\pi R_0^3}$$

Thus, nuclear density is independent of mass number $A$.

Therefore,

$$\rho_{O} : \rho_{He} = 1:1$$

So the correct option from the given choices is C. The solution says B, but the worked steps clearly give $1:1$, which matches option C.

For any nucleus, mass is proportional to the mass number $A$, while radius follows $R = R_0 A^{1/3}$. Therefore volume is proportional to

$$R^3 \propto (A^{1/3})^3 = A$$

Hence,

$$\rho \propto \frac{A}{A} = \text{constant}$$

So all nuclei have approximately the same density, and the ratio for oxygen and helium is $1:1$.