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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2003

  • question_answer
    The number of \[\alpha \] and \[\beta \]particles emitted during the transformation of \[_{90}T{{h}^{232}}\]to \[_{82}P{{b}^{208}}\]are respectively:

    A)  \[4,2\]             

    B)  \[2,2\]

    C)  \[8,6\]              

    D)  \[6,4\]

    Correct Answer: D

    Solution :

    No. of a-particles \[=\frac{1}{4}\times \] (Difference in mass number) \[=\frac{1}{4}\times (232-208)=\frac{1}{4}\times 24=6\] No.  of  \[\beta \]-particles  \[=2\times \alpha \]-particles (Difference in atomic number) \[=2\times 6-(90-82)=12-8=4\]


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