Nuclear Physics And Radioactivity Question 212
Energy released in fusion of 1 kg of deuterium [RPET 2000]
Options:
A) $ 8\times 10^{13},J $
B) $ 6\times 10^{27},J $
C) $ 2\times 10^{7},KwH $
D) $ 8\times 10^{23},MeV $
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Answer:
Correct Answer: D
Solution:
Fusion reaction of deuterium is $ _{1}H^{2}+ _{1}H^{2}\to _{2}He^{3}+ _{0}n^{1}+3.27\ MeV $ So $ E=\frac{6.02\times 10^{23}\times 3.27\times 1.6\times {{10}^{-13}}}{2\times 2} $ $ =7.8\times 10^{13}J $ $ =8\times 10^{13}J $ .
BETA
