Question

The ratio of the density of oxygen nucleus $\left({ }_8^{16} O \right)$ and helium nucleus $\left({ }_2^4 He \right)$ is

• A. $4: 1$
• B. $1: 1$
• C. $8: 1$
• D. $2: 1$

Answer 1

The Correct Option is B

1. Nuclear density is given by: $$\rho = \frac{\text{Mass}}{\text{Volume}} = \frac{A_u}{\frac{4}{3} \pi R^3}.$$
2. Substituting $R = R_0 A^{1/3}$, nuclear density becomes: $$\rho = \frac{A_u}{\frac{4}{3} \pi (R_0 A^{1/3})^3} = \frac{A_u}{\frac{4}{3} \pi R_0^3 A}.$$
3. Simplifying: $$\rho = \frac{3 A_u}{4 \pi R_0^3}.$$
4. Since nuclear density is independent of mass number $A$, the ratio is: $$\rho_{\text{O}} : \rho_{\text{He}} = 1 : 1.$$
Thus, the ratio is 1:1. Nuclear density is constant for all nuclei because the nuclear force binds nucleons in a fixed volume, irrespective of $A$.