Question

Show that the nuclear density is independent of mass number.

Answer 1

Step 1: Expression for mass of the nucleus

The mass \( M \) of a nucleus is approximately equal to the mass number \( A \) times the mass of a nucleon (proton or neutron), i.e.,

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

Where:

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

Step 2: Expression for volume of the nucleus

The volume \( V \) of a nucleus is related to its radius \( R \). The radius of the nucleus is given by the empirical formula:

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

Where \( R_0 \) is a constant approximately equal to \( 1.2 \, \text{fm} = 1.2 \times 10^{-15} \, \text{m} \).

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

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

Substituting the expression for \( R \), we get:

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

Step 3: Expression for nuclear density

The nuclear density \( \rho \) is defined as the mass per unit volume:

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

Substituting the expressions for \( M \) and \( V \), we get:

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

Simplifying:

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

Step 4: Conclusion

Notice that in the final expression for \( \rho \), the mass number \( A \) cancels out, and we are left with a constant value:

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

Therefore, the nuclear density is independent of the mass number. This shows that the density of the nucleus remains constant for all isotopes, regardless of their size or mass number.