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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2000

  • question_answer
    A nucleus X initially at rest, undergoes alpha decay according to the equation \[_{92}{{X}^{A}}{{\xrightarrow{{}}}_{z}}{{Y}^{228}}+\alpha \] Then, the values of A and Z are:

    A)  94, 230                

    B)         232, 90

    C)  190, 32                

    D)         230, 94

    E)  none of the above

    Correct Answer: B

    Solution :

    The decay equation is \[_{92}{{X}^{A}}{{\xrightarrow{{}}}_{z}}{{Y}^{228}}+\alpha \] \[\alpha -\]particle is nucleus of \[_{2}H{{e}^{4}}\] \[_{92}{{X}^{A}}{{\xrightarrow[{}]{{}}}_{Z}}{{Y}^{228}}{{+}_{2}}H{{e}^{4}}\] \[\therefore \]  \[A=228+4=232\] \[Z=92-2=90\]


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