Question

If the mass defect of a nucleus is \(2 \times 10^{-6}\) kg, then calculate its binding energy in (i) joule and (ii) electron-volt.

Hint:

Use \( E = mc^2 \) to calculate binding energy, and remember the conversion factor \( 1 \, \text{J} = 6.24 \times 10^{18} \text{ eV} \).

Answer 1

Step 1: The binding energy is calculated using Einstein’s equation: \[ E = mc^2 \] \[ = (2 \times 10^{-6}) (3 \times 10^8)^2 \] \[ = 1.8 \times 10^{11} \, \text{J} \] Step 2: Converting joules to electron-volts: \[ 1 \text{ J} = 6.24 \times 10^{18} \text{ eV} \] \[ E = 1.8 \times 10^{11} \times 6.24 \times 10^{18} \] \[ = 1.12 \times 10^{30} \, \text{eV} \] \[ \therefore \text{The correct answers are } 1.8 \times 10^{11} \text{ J and } 1.12 \times 10^{30} \text{ eV}. \] \[ \boxed{1.8 \times 10^{11} \text{ J}, \, 1.12 \times 10^{30} \text{ eV}} \]