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CMC Medical CMC-Medical VELLORE Solved Paper-2011

  • question_answer
    The following equation represents induced transmutation\[_{4}B{{e}^{9}}{{+}_{2}}H{{e}^{4}}{{\xrightarrow[{}]{{}}}_{6}}{{C}^{12}}+X\]In this equation, X represents

    A)  one negative \[\beta \text{-}\]particle

    B)  \[\alpha \]-particle

    C)  a positron

    D)  a neutron

    E)  \[\gamma -\]particle

    Correct Answer: D

    Solution :

                    \[_{4}B{{e}^{9}}+{{\,}_{2}}H{{e}^{4}}\xrightarrow{{}}{{\,}_{0}}{{C}^{12}}+X\] From conservation of mass number, mass number of\[X=9+4-12=1\] Similarly, atomic number of \[X\] \[=4+2-6=0\] So, \[X\] is \[_{0}{{X}^{1}},\]i.e., neutron\[({{\,}_{0}}{{n}^{-1}})\].


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