Atoms And Nuclei Question 354

Question: Consider the following reaction $ _{1}H^{2} + _{1}H^{2} \to _{2}H^{4}+Q. $ If $ m, _{1}H^{2}=2.014lamu; $ $ m, _{2}H^{4}=4.0024lamu. $ The energy Q released (in MeV) in this fusion reaction is

Options:

A) 12

B) 6

C) 24

D) 48

Show Answer

Answer:

Correct Answer: C

Solution:

  • $ _{1}H^{2}+ _{1}H^{2}\to _{2}He^{4}+Q $

$ \Rightarrow \Delta m=m _{2}He^{4}-2m _{1}H^{2} $

$ \Rightarrow \Delta m=4.0024-2(2.0141) $
$ \Rightarrow \Delta m=-0.0258,amu $ Since, $ Q=c^{2}\Delta m $
$ \Rightarrow ,Q=(0.0258)(931.5)MeV\Rightarrow Q=24,MeV. $



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