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JEE Main & Advanced Physics Nuclear Physics And Radioactivity Question Bank Self Evaluation Test - Nuclei

  • question_answer
    A proton is bombarded on a stationary Lithium nucleus. As a result of collision two \[\alpha \]-particles are produced. The direction of motion of the \[\alpha \]- particles with the initial direction of motion makes an angle\[{{\cos }^{-1}}\frac{1}{4}\]. If B.E./Nucleon for \[L{{i}^{7}}\] and \[\,H{{e}^{4}}\] are 5.60 MeV and 7.06 MeV respectively, then:

    A)  Kinetic energy of striking proton is 17.28 MeV

    B)  Kinetic energy of striking proton is 8.64 MeV

    C)  Kinetic energy of striking proton is 4.32 MeV

    D)  Kinetic energy of striking proton is 2.16 MeV

    Correct Answer: A

    Solution :

    [a] Q value of the reaction \[Q=(2\times 4\times 7.06-7\times 5.6)=17.28\] MeV....(i) \[{{K}_{P}}+Q=2{{K}_{\alpha }}\]                              .......(ii) \[\sqrt{2{{m}_{P}}{{K}_{p}}}=2\sqrt{2{{m}_{\alpha }}{{K}_{\alpha }}}\cos \alpha \] \[{{K}_{P}}={{K}_{\alpha }}\]                                     ?...(iii) So   \[{{K}_{P}}=17.28\text{ }MeV.\]


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