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

To find the ratio of the densities of the oxygen nucleus \(({_8^{16} O})\) and the helium nucleus \(({_2^4 He})\), we can use the concept of nuclear density. The density of a nucleus is given by:

\(\text{Density} = \frac{\text{Mass of nucleus}}{\text{Volume of nucleus}}\)

The volume of a nucleus is proportional to its mass number \((A)\) (as volume is proportional to \(r^3\) and \(r \propto A^{1/3}\)):

\(\text{Volume} \propto A\)

Thus, the density of a nucleus becomes:

\(\text{Density} \propto \frac{A}{A^{1}} = \text{constant}\)

This shows that nuclear density is approximately the same for all nuclei, as it only depends on a constant factor that is the same for all nuclei.

Therefore, the densities of both the oxygen nucleus and the helium nucleus are approximately equal.

Hence, the ratio of their densities is:

\(1:1\)

Thus, the correct answer is:

• $1:1$