Atoms And Nuclei Question 345
Question: The nuclear fusion reaction $ 2H^{2}\to _{2}He^{4}+\text{Energy } $ , is proposed to be used for the production of industrial power. Assuming the efficiency of process for production of power is 20%, find the ass of the deuterium required approximately for a duration of 1 year. Given mass of $ _{1}H^{2} $ nucleus = 2.0141 a.m.u and mass of $ _{2}He^{4} $ nuclei = 4.0026 a.m.u and 1 a.m.u. = 31 MeV
Options:
A) 165kg
B) 138kg
C) 180kg
D) 60kg
Show Answer
Answer:
Correct Answer: B
Solution:
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Mass defect $ \Delta m=2\times 2.014-4.0026=0.0256,a.m.u. $
Energy released when two $ _{1}H^{2} $
nuclei fuse $ =0.0256\times 931=23.8MeV $
Total energy required to be produced by nuclear reaction in 1 year $ =2500\times 10^{6}\times 3.15\times 10^{7}=7.88\times 10^{16}J $
No. of nuclei of $ _{1}H^{2} $
required $ =\frac{7.88\times 10^{16}J}{23.8\times 1.6\times {{10}^{-13}}}\times 2=4.14\times 10^{28} $ Mass of Deuterium required $ =\frac{4.14\times 10^{28}}{6.02\times 10^{23}}\times 2\times {{10}^{-3}}kg=138kg $
