Atoms And Nuclei Question 159

Question: The binding energy of deuteron $ _{1}^{2}H $ is 1.112 MeV per nucleon and an $ \alpha - $ particle $ _{2}^{4}He $ has a binding energy of 7.047 MeV per nucleon. Then in the fusion reaction $ _{1}^{2}H+ _{1}^{2}H\to _{2}^{4}He+Q $ , the energy Q released is [MP PMT 1992; Roorkee 1994; IIT 1996; AIIMS 1997]

Options:

A) 1 MeV

B) 11.9 MeV

C) 23.8 MeV

D) 931 MeV

Show Answer

Answer:

Correct Answer: C

Solution:

Mass of $ _{1}H^{2}=2.01478\ a.m.u. $

Mass of $ _{2}He^{4}=4.00388\ a.m.u. $

Mass of two deuterium $ =2\times 2.01478=4.02956 $

Energy equivalent to $ 2_{1}H^{2} $

$ =4.02956\times 1.112\ MeV=4.48\ MeV $

Energy equivalent to $ _{2}H^{4} $

$ =4.00388\times 7.047\ MeV=28.21\ MeV $

Energy released $ =28.21-4.48=23.73\ MeV $ = 24 MeV



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