A) \[13.9\text{ }MeV\]
B) \[26.9\text{ }MeV\]
C) \[23.6\text{ }MeV\]
D) \[19.2\text{ }MeV\]
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\[{{(}_{1}}{{H}^{2}})\] \[=1.1MeV\] \[\therefore \]Total binding energy of one deuteron nucleus \[=2\times 1.1=2.2MeV\] The binding energy per nucleon of helium\[{{(}_{2}}H{{e}^{4}})\] \[=7MeV\] \[\therefore \] 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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