NEET Sample Paper NEET Sample Test Paper-14

  • question_answer If the binding energy per nucleon in \[\text{L}{{\text{i}}^{\text{7}}}\]and\[\text{H}{{\text{e}}^{\text{4}}}\] nuclei is respectively 5.60 MeV and 7.06 MeV, then the energy of the reaction \[L{{i}^{7}}+p\to {{2}_{2}}H{{e}^{4}}\]

    A)  19.6 MeV                   

    B)  2.4 MeV

    C)  8.4 MeV                     

    D)  17.3 MeV

    Correct Answer: D

    Solution :

    Binding energy of \[L{{i}^{7}}=\](binding energy per nucleon) \[\times \](number of nucleons) \[=\text{ }5.60\times 7=39.20\text{ }MeV.\] Similarly, Binding energy of \[H{{e}^{4}}=7.06\times 4=28.24\,MeV.\] \[\therefore \]Binding energy of two \[H{{e}^{4}}\]nuclei \[=28.24\times 2\] = 56.48 MeV Hence, the energy of reaction is (56.48 - 39.20) = 17.28 MeV Hence, the correction option is [d].

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