Consider the nuclear reaction $X^{200} \rightarrow A^{110}\ + B^{80}$. If the binding energy per nucleon for X, A and B are 7.4 MeV, 8.2 MeV and 8.1MeV respectively, then the energy released in the reaction is

Question

Dr. Abubacker · Accepted Answer

(a) For $X$, energy = 200 $	imes$ 7.4 =1480 MeV
For A, energy = 110 $	imes$ 8.2 =902 MeV
For B, energy = 80 $	imes$ 8.1 =648 MeV
Therefore energy released
$= (902\  + 648) - 1480$
$= \ 1550\ –\ 1480\  = \ 70$ MeV

Q. Consider the nuclear reaction $X^{200} \rightarrow A^{110}\ + B^{80}$. If the binding energy per nucleon for X, A and B are 7.4 MeV, 8.2 MeV and 8.1MeV respectively, then the energy released in the reaction is

Explanation

Reviewed by Dr. Abubacker

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