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Question

Question: Energy released in fusion of 1 kg of deuterium nuclei....

Energy released in fusion of 1 kg of deuterium nuclei.

A

8×1013J8 \times 10^{13}J

B

6×1027J6 \times 10^{27}J

C

2×107KwH2 \times 10^{7}KwH

D

8×1023MeV8 \times 10^{23}MeV

Answer

8×1023MeV8 \times 10^{23}MeV

Explanation

Solution

Fusion reaction of deuterium is

1H2+1H22He3+0n1+3.276muMeV1H^{2} +_{1}H^{2} \rightarrow_{2}He^{3} +_{0}n^{1} + 3.27\mspace{6mu} MeV

So E=6.02×1023×103×3.27×1.6×10132×2=7.8×1013JE = \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

=8×1013J= 8 \times 10^{13}J.