NEET Sample Paper NEET Sample Test Paper-16

  • question_answer The binding energies per nucleon of deuteron \[({{\,}_{1}}{{H}^{2}})\] and helium atom \[({{\,}_{2}}H{{e}^{4}})\] are 1.1 MeV and 7 MeV If two deuteron atoms react to form a single helium atom, then the energy released is:

    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).

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