Question

For a nucleus of mass number $ A $ and radius $ R $, the mass density of the nucleus can be represented as:

• A. \( \frac{2}{3} A \)
• B. \( \frac{1}{3} A \)
• C. \( A^3 \)
• D. Independent of \( A \)

Hint:

For a nucleus, the mass density is independent of the mass number \( A \). This is because the mass and volume of the nucleus both scale with \( A \), and the ratio of these two quantities results in a constant density.

Answer 1

The Correct Option is D

The mass density \( \rho \) of a nucleus is defined as the mass per unit volume.
For a nucleus with mass number \( A \) and radius \( R \), we can express the mass and volume as follows: - The mass of the nucleus is proportional to the mass number \( A \), i.e., the mass \( M \) is proportional to \( A \), - The volume \( V \) of the nucleus is proportional to \( R^3 \), where \( R \) is the radius of the nucleus.
The volume is given by the formula for the volume of a sphere: \[ V = \frac{4}{3} \pi R^3 \] Since the mass number \( A \) is proportional to the volume, the mass density is given by: \[ \rho = \frac{\text{mass}}{\text{volume}} = \frac{A}{\frac{4}{3} \pi R^3} \] We know from the liquid drop model of the nucleus that the radius \( R \) is proportional to \( A^{1/3} \).
Thus, we have: \[ R \propto A^{1/3} \] Substituting this into the equation for density: \[ \rho \propto \frac{A}{R^3} = \frac{A}{A} = 1 \] Therefore, the mass density \( \rho \) is independent of \( A \). Thus, the correct answer is Option (4), which states that the mass density is independent of \( A \).