JEE Physics 2023 Question with Solution

Given: Mass of one nucleon is $m = 1.6 \times 10^{-27} \, \text{kg}$ and nuclear radius is $R = 1.5 \times 10^{-15} A^{1/3} \, \text{m}$.

Find: The value of $n$ in the ratio of nuclear density to water density, written as $n \times 10^{13}$.

The density of a nucleus is

$$\rho = \frac{\text{mass of nucleus}}{\text{volume of nucleus}}$$

1. Mass of the nucleus:

$$\text{Mass} = A m = A \times 1.6 \times 10^{-27} \, \text{kg}$$

2. Volume of the nucleus:

$$V = \frac{4}{3} \pi R^3$$

where

$$R = 1.5 \times 10^{-15} A^{1/3} \text{ m}$$

So,

$$V = \frac{4}{3} \pi \left(1.5 \times 10^{-15} A^{1/3}\right)^3$$ $$= \frac{4}{3} \pi \left(1.5\right)^3 \times 10^{-45} A \; \text{m}^3$$

3. Nuclear density:

$$\rho = \frac{A \times 1.6 \times 10^{-27}}{\frac{4}{3} \pi \left(1.5\right)^3 \times 10^{-45} A}$$

The factor $A$ cancels, so density is independent of $A$.

$$\rho \approx \frac{1.6 \times 10^{-27}}{14.14 \times 10^{-45}} \approx 0.113 \times 10^{18} \, \text{kg/m}^3$$

4. Water density:

$$\rho_w = 10^3 \, \text{kg/m}^3$$

5. Ratio:

$$\frac{\rho}{\rho_w} = \frac{0.113 \times 10^{18}}{10^3} = 0.113 \times 10^{15} = 11.31 \times 10^{13}$$

Therefore, the approximate value of $n$ is $11$.

Given: Nuclear radius varies as $R \propto A^{1/3}$.

Find: Why the ratio comes out as a constant.

Since the mass of the nucleus is proportional to the number of nucleons,

$$\text{mass} \propto A$$

and since

$$R \propto A^{1/3}$$

its volume is

$$V \propto R^3 \propto A$$

Hence,

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

This is why nuclear density is nearly the same for all nuclei. Substituting the given numerical values gives

$$\frac{\rho}{\rho_w} \approx 11.31 \times 10^{13}$$

So the required approximate value is $11$.