Question

What is meant by the binding energy of a nucleus? If mass defect for a nucleus is \( 10^{-6} \, \text{kg} \), then find its binding energy in electron volt.

Hint:

The binding energy of a nucleus is the energy required to break it into individual nucleons, and it is related to the mass defect by \( E_b = \Delta m c^2 \).

Answer 1

Step 1: Definition of Binding Energy.
The binding energy of a nucleus is the energy required to disassemble a nucleus into its constituent protons and neutrons. It is a measure of the stability of the nucleus. The binding energy is related to the mass defect, which is the difference in mass between the nucleus and its constituent particles.
Step 2: Formula for Binding Energy.
The binding energy \( E_b \) is given by the relation: \[ E_b = \Delta m c^2 \] where: - \( \Delta m \) is the mass defect, - \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)).
Step 3: Conversion of Mass Defect to Energy.
Given mass defect \( \Delta m = 10^{-6} \, \text{kg} \), we can calculate the binding energy in joules: \[ E_b = 10^{-6} \times (3 \times 10^8)^2 = 9 \times 10^{10} \, \text{J} \]
Step 4: Conversion to Electron Volts.
To convert joules to electron volts, use the conversion factor \( 1 \, \text{J} = 6.242 \times 10^{12} \, \text{eV} \): \[ E_b = 9 \times 10^{10} \, \text{J} \times 6.242 \times 10^{12} \, \text{eV/J} = 5.6178 \times 10^{23} \, \text{eV} \]