Nuclear Physics And Radioactivity Question 406
The binding energy of deuteron $ _{1}^{2}H $ is 1.15 MeV per nucleon and an alpha particle $ _{2}^{4}He $ has a binding energy of 7.1 MeV per nucleon. Then in the reaction $ _{1}^{2}H+ _{1}^{2}H\to _{2}^{4}He+Q $ the energy released Q is:
Options:
A) 5.95 MeV
B) 26.1 MeV
C) 23.8 MeV
D) 289.4 MeV
Show Answer
Answer:
Correct Answer: C
Solution:
Given, $ _{1}H^{2}+ _{1}H^{2}\to _{2}He^{4}+Q $ The total binding energy of the deutrons $ =2\times 1.15=2.30,MeV $ The total binding energy of alpha particle $ =1\times 7.1=7.1,MeV $ The energy released in the process $ =7.1-2.30=4.8,MeV $ .
BETA
