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. $
