Question

Energy released in fusion of 1 kg of deuterium nuclei.

• A. $8 \times 10^{13}J$
• B. $6 \times 10^{27}J$
• C. $2 \times 10^{7}KwH$
• D. $8 \times 10^{23}MeV$

Answer 1

Answer: (D)

$8 \times 10^{23}MeV$

Answer 2

Fusion reaction of deuterium is

$1H^{2} +_{1}H^{2} \rightarrow_{2}He^{3} +_{0}n^{1} + 3.27\mspace{6mu} MeV$

So $E = \frac{6.02 \times 10^{23} \times 10^{3} \times 3.27 \times 1.6 \times 10^{- 13}}{2 \times 2} = 7.8 \times 10^{13}J$

$= 8 \times 10^{13}J$.