Question

Mass numbers of two nuclei are in the ratio of \(4 : 3\). Their nuclear densities will be in the ratio of

• A. \(4 : 3\)
• B. \(\bigg(\frac{3}{4}\bigg)^{\frac{1}{3}}\)
• C. \(1:1\)
• D. \(\bigg(\frac{4}{3}\bigg)^{\frac{1}{3}}\)

Answer 1

The Correct Option is C

To solve this problem, we need to understand the concept of nuclear density. Nuclear density is defined as the mass of the nucleus divided by its volume. The formula for nuclear density \( \rho \) is given by:

\(\rho = \frac{M}{V}\),

where \( M \) is the mass of the nucleus, and \( V \) is its volume.

The volume of a nucleus is proportional to its mass number \( A \), i.e., \( V \propto A \). The volume of a nucleus can be expressed as:

\(V = \frac{4}{3}\pi R^3\),

where \( R \) is the radius of the nucleus. The radius \( R \) is related to the mass number \( A \) by the relation:

\(R = R_0 A^{1/3}\),

where \( R_0 \) is a constant.

Thus, the volume \( V \) becomes:

\(V = \frac{4}{3}\pi (R_0 A^{1/3})^3 = \frac{4}{3}\pi R_0^3 A\).

Substituting this into the formula for density, we get:

\(\rho = \frac{M}{\frac{4}{3}\pi R_0^3 A}\).

Since mass number \( A \) is proportional to mass \( M \), the actual density becomes:

\(\rho = \text{constant}\).

This shows that nuclear density is independent of the mass number (and thus the size) of the nucleus. Therefore, the nuclear densities of two nuclei, regardless of their mass numbers, will always be the same.

Thus, the ratio of the nuclear densities of the two nuclei is:

\(1:1\).

Hence, the correct answer is \(1:1\).