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 address this question, the relationship between nuclear mass density, mass number \( A \), and radius \( R \) must be understood.

Nuclear mass density \( \rho \) is defined as:

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

The mass of a nucleus is approximately proportional to its mass number \( A \), as \( A \) represents the combined count of protons and neutrons, which have nearly identical masses.

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

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

Empirical data indicate that the nuclear radius \( R \) is related to the mass number \( A \) by:

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

where \( R_0 \) is a constant. Consequently, 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 density formula yields:

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

Simplification results in:

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

This expression demonstrates that the density \( \rho \) is independent of \( A \), as the \( A \) terms cancel out.

Therefore, the conclusion is:

Independent of \( A \)