RAJASTHAN PMT Rajasthan - PMT Solved Paper-2010

  • question_answer
    57.      A nucleus\[z{{X}^{A}}\] emits 9 \[\alpha -\]-particles and 5 \[\beta \]-particles. The ratio of total protons and neutrons in the final nucleus is

    A)  \[\frac{Z-13}{(A-Z-23)}\]                             

    B)  \[\frac{Z-18}{(A-36)}\]

    C)  \[\frac{Z-13}{(A-36)}\] 

    D)         \[\frac{Z-13}{(A-Z-13)}\]

    Correct Answer: A

    Solution :

    \[_{z}{{X}^{A}}{{\xrightarrow{9\alpha }}_{Z-18}}{{X}^{A-36}}{{\xrightarrow{5\beta }}_{Z-13}}{{X}^{A-36}}\] Number of protons\[=(Z-13)\] Number of neutrons                 \[=(A-36)-(Z-13)=(A-Z-23)\] \[\therefore \]  \[\frac{P}{N}=\frac{(Z-13)}{(A-Z-23)}\]


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