Atoms And Nuclei Question 234

Question: Energy released in fusion of 1 kg of deuterium nuclei [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 $

Show Answer

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 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 $ .



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