Atoms And Nuclei Question 343

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:

  • 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 $


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