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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 a and b-particles. \[_{92}{{U}^{238}}\] is changed to \[_{82}P{{b}^{206}}\]. The total number of a-particles lost in this process is     [UPSEAT 1999, 2000]

    A)                 10          

    B)                 5

    C)                 8             

    D)                 32

    Correct Answer: C

    Solution :

           Suppose the no. of a-particles emitted = x and the no. of b-particles emitted = y, then                         \[_{92}{{U}^{238}}{{\to }_{82}}P{{b}^{206}}+x{{\,}_{+2}}{{\alpha }^{4}}+y{{\,}_{-1}}{{\beta }^{0}}\]                    Equating the mass number on both sides, we get                    238 = 206 + 4x + 0y or 4x = 32 or \[x=\frac{32}{4}=8\]                                 Hence 8 a-particles will be emitted.


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