BHU PMT BHU PMT (Mains) Solved Paper-2011

  • question_answer
    In the uranium radioactive series, the initial nucleus is\[_{92}{{U}^{238}}\] and that the final nucleus is \[_{82}P{{b}^{206}}\]When uranium nucleus decays to lead, the number of \[\text{ }\!\!\alpha\!\!\text{ -}\]particles and \[\beta -\]particles emitted are

    A)  \[8\alpha ,6\beta \]                                      

    B)  \[6\alpha ,7\beta \]

    C)  \[6\alpha ,8\beta \]                                      

    D) \[4\alpha ,3\beta \]

    Correct Answer: A

    Solution :

                     Let number of\[\alpha -\]particles emitted be\[x\]and number of\[\beta -\]particles emitted be y. Difference in mass number, \[4x=238-206=32\] \[\Rightarrow \]               \[x=8\] Difference in charge number, \[2x-1y=92-82=10\] \[\Rightarrow \]               \[16-y=10\] \[\Rightarrow \]               \[y=6\]


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