A) \[26.9\text{ }MeV\]
B) \[25.8\text{ }MeV\]
C) \[23.6\text{ }MeV\]
D) \[12.9\text{ }MeV\]
Correct Answer: C
Solution :
: Binding energy of deuteron \[{{(}_{1}}{{H}^{2}})\] \[=2\times 1.1MeV=2.2MeV\] Binding energy of helium atom \[{{(}_{2}}H{{e}^{4}})\] \[=4\times 7MeV=28MeV\] \[_{1}{{H}^{2}}{{+}_{1}}{{H}^{2}}{{\to }_{2}}H{{e}^{4}}+Energy\,released\] Energy released = BE of helium \[-2\times \] BE of deuteron \[=28MeV-2\times 2.2MeV\] \[=28MeV-4.4Me=23.6MeV\]
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