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

To solve this question, we need to understand how the mass density of a nucleus relates to its mass number \( A \) and its radius \( R \).

The mass density \( \rho \) of a nucleus is given by the formula:

\(\rho = \frac{\text{mass of nucleus}}{\text{volume of nucleus}}\)

We know that the mass of a nucleus is approximately proportional to its mass number \( A \), because the mass number represents the sum of protons and neutrons, which have nearly equal mass.

The volume \( V \) of a spherical nucleus is calculated using the formula:

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

For a nucleus, it is empirically found that the radius \( R \) is related to the mass number \( A \) by the equation:

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

where \( R_0 \) is a constant. Thus, the volume \( V \) can be expressed as:

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

Substituting this into the formula for density, we get:

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

Upon simplification, it becomes:

\(\rho = \frac{1}{\frac{4}{3} \pi R_0^3}\)

From this expression, we can see that the density \( \rho \) is independent of \( A \) since the \( A \) terms cancel each other out.

Therefore, the correct answer is:

Independent of \( A \)

Answer 2

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 \( A \).