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Manipal Engineering Manipal Engineering Solved Paper-2012

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

    A) \[94,\,\,230\]                    

    B) \[232,\,\,90\]

    C) \[190,\,\,32\]    

    D)        \[230,\,\,94\]

    Correct Answer: B

    Solution :

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


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