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BHU PMT BHU PMT (Screening) Solved Paper-2006

  • question_answer
    After losing a number of\[\alpha \]and \[\beta \]-particles \[_{92}{{U}^{238}}\]is changed to 32 Pb206. The total number of\[\alpha \]-particles lost in this process is:

    A)  10                                         

    B)  5

    C)   8                                           

    D)  32

    Correct Answer: C

    Solution :

                     Suppose the number of a-particles emitted\[=x\]and number of P-particles emitted = y then\[_{92}{{U}^{238}}{{\xrightarrow{{}}}_{82}}P{{b}^{206}}+{{x}_{2}}{{\alpha }^{4}}+{{y}_{-1}}{{\beta }^{0}}\] Equating the mass number on both sides, we get     \[238=206+4x+0y\]or    \[4x=32\]     or    \[x=\frac{32}{4}=8\] Hence,\[8\alpha -\]particles will be emitted.


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