A) 30.2 MeV
B) 32.4 MeV
C) 23.6 MeV
D) 25.8 MeV
Correct Answer: C
Solution :
[c] \[{}_{1}{{H}^{2}}+{}_{1}{{H}^{2}}\to {}_{2}H{{e}^{4}}\] Total binding energy of two deuterium nuclei \[=1.1\times 4=4.4\,\,MeV\] Binding energy of a \[({}_{2}H{{e}^{4}})\] nuclei \[=4\times 7=28\,\,MeV\] Energy released in this process \[=28-4.4=23.6\,\,MeV\]You need to login to perform this action.
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