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JEE Main & Advanced Chemistry Surface & Nuclear Chemistry / भूतल और नाभिकीय रसायन Question Bank Causes of radioactivity and Group displacement law

  • question_answer
    After losing a number of \[\alpha \]and \[\beta \]-particles, \[_{92}{{U}^{238}}\] changed to \[_{82}P{{b}^{206}}\]. The total number of particles lost in this process is    [MNR 1985]

    A)                 14          

    B)                 5

    C)                 8             

    D)                 32

    Correct Answer: A

    Solution :

           \[_{92}{{U}^{238}}\to {{\,}_{82}}P{{b}^{206}}+x{{\,}_{+2}}{{\alpha }^{4}}+y{{\,}_{-1}}{{\beta }^{0}}\]                    no. of a-particles = \[\frac{238-206}{4}=8\]                    no. of b-particles = \[92-82-2\times 8=6\]                                 Total no. of particles = \[8+6=14\].


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