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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
    An isotope \[_{Y}{{A}^{X}}\] undergoes a series of \[m\]  alpha and \[n\] beta disintegration to form a stable isotope \[_{Y-10}{{B}^{X-32}}\].  The values of \[m\] and \[n\] are respectively              [MP PET 1995]

    A)                 6 and 8 

    B)                 8 and 10

    C)                 5 and 8 

    D)                 8 and 6

    Correct Answer: D

    Solution :

           \[_{\,Y}{{A}^{X}}\to {{\,}_{Y-10}}{{B}^{X-32}}+m{{\,}_{2}}H{{e}^{4}}+n{{\,}_{+1}}{{e}^{0}}\]                    Value of m = \[\frac{X-(X)-32}{4}=8\]                                 Value of n = Y ? Y ? 10 ? 2 × 8 = 6.


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