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:
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.