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

  • question_answer
    When the nucleus of a radium-226, which is at rest, decays, an \[\alpha \] particle and the nucleus of radon are created. The released energy during the decay is 4.87 MeV, which appears as the kinetic energy of the two resulted particles. Calculate the kinetic energy of a particle and radon nucleus.

    A)  4.78 MeV and 0.09 MeV

    B)  4.67 MeV and 0.2 MeV

    C)  4.84 MeV and 0.03 MeV

    D)  4.81 MeV and 0.06 MeV

    Correct Answer: A

    Solution :

    [a] \[{{k}_{\alpha }}=\frac{M}{M+4}Q=\frac{222}{226}\times 4.87=4.87\,\,MeV\]


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