Question

Show that the nuclear density is independent of mass number.

Answer 1

Step 1: Nuclear Mass Equation

The mass \( M \) of a nucleus is approximated by the mass number \( A \) multiplied by the mass of a nucleon. This can be expressed as:

\[ M \approx A \cdot m_{\text{nucleon}} \]

Where:

• \( A \) represents the mass number, which is the total count of protons and neutrons.
• \( m_{\text{nucleon}} \) is the mass of a single nucleon, approximately \( 1.67 \times 10^{-27} \, \text{kg} \).

Step 2: Nuclear Volume Equation

The volume \( V \) of a nucleus depends on its radius \( R \). The nuclear radius is described by the empirical formula:

\[ R = R_0 A^{1/3} \]

Here, \( R_0 \) is a constant, approximately \( 1.2 \, \text{fm} \), which is equivalent to \( 1.2 \times 10^{-15} \, \text{m} \).

For a spherical nucleus, the volume \( V \) is calculated as:

\[ V = \frac{4}{3} \pi R^3 \]

By substituting the expression for \( R \), the volume becomes:

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

Step 3: Nuclear Density Calculation

Nuclear density \( \rho \) is defined as mass per unit volume:

\[ \rho = \frac{M}{V} \]

Substituting the derived expressions for \( M \) and \( V \) yields:

\[ \rho = \frac{A \cdot m_{\text{nucleon}}}{\frac{4}{3} \pi R_0^3 A} \]

Upon simplification:

\[ \rho = \frac{3 m_{\text{nucleon}}}{4 \pi R_0^3} \]

Step 4: Result and Interpretation

It is observed that the mass number \( A \) cancels out in the final expression for \( \rho \), resulting in a constant value:

\[ \rho = \frac{3 m_{\text{nucleon}}}{4 \pi R_0^3} \]

Consequently, nuclear density is independent of the mass number. This indicates that the density of the nucleus is constant across all isotopes, irrespective of their size or mass number.