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WB JEE Medical WB JEE Medical Solved Paper-2011

  • question_answer
    A radioactive nucleus of mass number A, initially at rest, emits an \[\text{ }\!\!\alpha\!\!\text{ -}\]particle with a speed v. The recoil speed of the daughter nucleus will be                                  

    A)  \[\frac{2v}{A-4}\]                           

    B)  \[\frac{2v}{A+4}\]

    C)   \[\frac{4v}{A-4}\]                          

    D)  \[\frac{4v}{A+4}\]

    Correct Answer: C

    Solution :

                     The initial momentum of the nucleus would be equal to \[4{{u}_{H}}v,\] where \[{{\text{u}}_{\text{H}}}\] is the unit mass of hydrogen. Since the mass of hydrogen is 1, so the initial momentum can be given by 4v. The final momentum of the nucleus would be equal to (A - 4)V, where V is the  final velocity. Thus, we get from the conservation of momentum, (A - 4)V = 4v, so      \[V=\frac{4v}{(A-4)}.\]


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