A) 19.6 MeV
B) 2.4 MeV
C) 8.4 MeV
D) 17.3 MeV
Correct Answer: D
Solution :
Binding energy of \[L{{i}^{7}}=\](binding energy per nucleon) \[\times \](number of nucleons) \[=\text{ }5.60\times 7=39.20\text{ }MeV.\] Similarly, Binding energy of \[H{{e}^{4}}=7.06\times 4=28.24\,MeV.\] \[\therefore \]Binding energy of two \[H{{e}^{4}}\]nuclei \[=28.24\times 2\] = 56.48 MeV Hence, the energy of reaction is (56.48 - 39.20) = 17.28 MeV Hence, the correction option is [d].You need to login to perform this action.
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