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BCECE Engineering BCECE Engineering Solved Paper-2007

  • question_answer
    \[_{\text{7}}{{\text{N}}^{\text{14}}}\]is bombarded with \[_{2}H{{e}^{\text{4}}}.\] The resulting nucleus is \[_{8}{{O}^{17}}\] with the emission of

    A)  neutrino            

    B)         antineutrino     

    C)         proton                

    D)         neutron

    Correct Answer: C

    Solution :

    Key Idea: Mass number and atomic number in nuclear reactions are conserved.        The nuclear reaction is as follows.                 \[{{\,}_{7}}{{N}^{14}}+{{\,}_{2}}H{{e}^{4}}\xrightarrow{{}}{{\,}_{8}}{{O}^{17}}+{{\,}_{q}}{{X}^{p}}\] Conservation of mass number gives                                 \[p=14+4-17=1\] Conservation of atomic number gives                                 \[q=7+2-8=1\] Hence, particle is a proton \[{{\,}_{1}}{{H}^{1}}.\]           


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