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Manipal Medical Manipal Medical Solved Paper-2009

  • question_answer
    What is the respective number of\[\alpha \]and\[\beta \] particles emitted in the following radioactive decay? \[_{90}{{X}^{200}}{{\xrightarrow[{}]{{}}}_{80}}{{Y}^{168}}\]

    A)  6 and 8         

    B)  8 and 8

    C)  6 and 6         

    D)  8 and 6

    Correct Answer: D

    Solution :

    \[{{n}_{\alpha }}=\frac{A-A}{4}\] \[=\frac{200-168}{4}\] \[=8\] \[{{n}_{\beta }}=2{{n}_{\alpha }}-Z+Z\] \[=2\times 8-90+80=6\]


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