A) 17.3 MeV
B) 8.4 MeV
C) 2.4 MeV
D) 19.6 MeV
Correct Answer: A
Solution :
Binding energy of \[L{{i}^{7}}=39.20\,MeV\] Binding energy of \[H{{e}^{4}}=28.24\,MeV\] Therefore binding energy of \[2H{{e}^{4}}=56.48\,MeV\] Now energy of reaction is \[=56.48-39.20=17.28\text{ }MeV\] \[=17.3\text{ }MeV\]You need to login to perform this action.
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