A) 13.9 MeV
B) 26.9 MeV
C) 23.6 MeV
D) 19.2 MeV
Correct Answer: C
Solution :
Total binding energy of helium atom \[({{\,}_{2}}H{{e}^{4}})\] \[=4\times 7=28\,\text{MeV}\] Total binding energy of deuteron \[{{\,}_{1}}{{H}^{2}}(Ip+1n)\] \[=2\times 1.1=2.2\,\text{MeV}\] \[\therefore \]Binding energy of 2 deuterons \[=2\times 2.2=4.4\,\text{MeV}\] so, the energy released in forming helium nucleus from two deuterons\[=(28-4.4)MeV\] \[=23.6\,\text{MeV}\] Hence, the correction option is (c).You need to login to perform this action.
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