Question

In a nuclear fusion reaction mass defect is 0.2%. What will be amount of energy released in the fusion of 1 kg mass?

Hint:

Use \( E = \Delta m\, c^2 \) with \( \Delta m = 0.2\% \) of 1 kg, i.e. \( 2 \times 10^{-3} \) kg.

Answer 1

Step 1 (Set up the fraction directly): Since only 0.2% of the mass is lost, the released energy is 0.2% of the total rest energy of 1 kg. So \(E = 0.002 \times (m c^2)\) with \(m = 1\) kg.

Step 2 (Rest energy of 1 kg): First find the full rest energy,
\[E_0 = m c^2 = 1 \times (3\times10^{8})^2 = 9 \times 10^{16}\ \text{J}.\]

Step 3 (Take 0.2% of it):
\[E = 0.002 \times 9 \times 10^{16} = 1.8 \times 10^{14}\ \text{J}.\]

Step 4 (Result): The same value is obtained, confirming the answer.
\[\boxed{E = 1.8 \times 10^{14}\ \text{J}}\]