Question

The mass defect in a particular reaction is \( 0.4 \, \text{g} \). The amount of energy liberated is \( n \times 10^7 \, \text{kWh} \), where \( n = \) _____. (speed of light \( = 3 \times 10^8 \, \text{m/s} \))

Answer 1

Correct Answer: 1

The energy liberated is given by:

\[ E = \Delta m c^2 \]

Substituting the values:

\[ E = 0.4 \times 10^{-3} \times (3 \times 10^8)^2 \]

\[ E = 3600 \times 10^7 \, \text{kWs} \]

Converting to kWh:

\[ E = \frac{3600 \times 10^7 \, \text{kWh}}{3600} = 1 \times 10^7 \, \text{kWh} \]