Nuclear Physics And Radioactivity Question 413
Question: If the binding energy per nucleon in $ _{3}^{7}Li $ and $ _{2}^{4}He $ nuclei are 5.60 MeV and 7.06 MeV respectively, then in the reaction $ p+ _{3}^{7}Li \rightarrow 2 _{2}^{4}He $ energy of proton must be
Options:
A) 28.24 MeV
B) 17.28 MeV
C) 1.46MeV
D) 39.2MeV
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Let E be the energy of proton, then $ E+7\times 5.6=2\times [4\times 7.06] $
$ \Rightarrow E=56.48-39.2=17.28,MeV $
BETA
