Question

The Coulomb energy component in the binding energy of a nucleus is 18.432 MeV. If the radius of the uniform and spherical charge distribution in the nucleus is 3 fm, the corresponding atomic number (rounded off to the nearest integer) is \(\underline{\hspace{2cm}}\).
\text{(Given: } \frac{e^2}{4 \pi \epsilon_0} = 1.44 \, \text{MeV fm}).}

Hint:

To calculate the atomic number from the Coulomb energy, use the formula \( E_{\text{Coulomb}} = \frac{3Z^2 e^2}{5 R} \) and solve for \( Z \).

Answer 1

Correct Answer: 8

The Coulomb energy in the binding energy of the nucleus is given by: \[ E_{\text{Coulomb}} = \frac{3Z^2 e^2}{5 R}, \] where:
- \( E_{\text{Coulomb}} = 18.432 \, \text{MeV} \),
- \( Z \) is the atomic number,
- \( R = 3 \, \text{fm} \) is the radius of the nucleus.
Substitute the known values: \[ 18.432 = \frac{3Z^2 \times 1.44}{5 \times 3}. \] Simplifying the equation: \[ 18.432 = \frac{3Z^2 \times 1.44}{15}. \] Solving for \( Z^2 \): \[ Z^2 = \frac{18.432 \times 15}{3 \times 1.44} \approx 8. \] Thus, \( Z = 8 \). The atomic number is 8.