Atoms And Nuclei Question 336
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 deuterons
$ =4\times 1.15=4.60,MeV $
The total binding energy of an alpha particle is $ 28.4,MeV $
The energy released in the process $ =28.4-4.60=23.8,MeV $ .
BETA
