Q109 / 150 · MHT CET 2026 (16 April Shift 2) · Physics
The ratio of the density of oxygen nucleus $({}^{16}_{8}\text{O})$ and helium nucleus $({}^{4}_{2}\text{He})$ is
- Correct answerA$1 : 1$
- B$2 : 1$
- C$4 : 1$
- D$8 : 1$
Solution
The radius of a nucleus with mass number $A$ is given by $R = R_0 A^{1/3}$.
The volume of the nucleus is $V = \dfrac{4}{3} \pi R^3 = \dfrac{4}{3} \pi R_0^3 A$.
The mass of the nucleus is $M \approx A m$, where $m$ is the average mass of a nucleon.
The density of the nucleus is $\rho = \dfrac{M}{V} = \dfrac{A m}{\dfrac{4}{3} \pi R_0^3 A} = \dfrac{m}{\dfrac{4}{3} \pi R_0^3}$.
Since the density is independent of the mass number $A$, it is constant for all nuclei.
Therefore, the ratio of the density of the oxygen nucleus to the helium nucleus is $1 : 1$.
Answer: $1 : 1$
The volume of the nucleus is $V = \dfrac{4}{3} \pi R^3 = \dfrac{4}{3} \pi R_0^3 A$.
The mass of the nucleus is $M \approx A m$, where $m$ is the average mass of a nucleon.
The density of the nucleus is $\rho = \dfrac{M}{V} = \dfrac{A m}{\dfrac{4}{3} \pi R_0^3 A} = \dfrac{m}{\dfrac{4}{3} \pi R_0^3}$.
Since the density is independent of the mass number $A$, it is constant for all nuclei.
Therefore, the ratio of the density of the oxygen nucleus to the helium nucleus is $1 : 1$.
Answer: $1 : 1$
Hence, the correct answer is option: 'A'.