A) 13.9 MeV
B) 26.9 MeV
C) 23.6 MeV
D) 19.2 MeV
Correct Answer: C
Solution :
As given \[_{1}{{H}^{2}}{{+}_{1}}{{H}^{2}}{{\xrightarrow[{}]{{}}}_{2}}H{{e}^{4}}+energy\] The binding energy per nucleon of a deuteron \[{{(}_{1}}{{H}^{2}})\] \[=1.1\,MeV\] \[\therefore \]Total binding energy of one deuteron nucleus \[=2\times 1.1=2.2MeV\] The binding energy per nucleon of helium\[{{(}_{2}}H{{e}^{4}})\] \[=7\text{ }MeV\] \[\therefore \] Total binding energy \[=4\times 7=28\text{ }MeV\] Hence, energy released in the above process \[=28-2\times 2.2\] \[=28-4.4=23.6\text{ }MeV\]
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