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WB JEE Medical WB JEE Medical Solved Paper-2011

  • question_answer
    In       the       nuclear       reaction \[_{7}^{14}N+X\xrightarrow{{}}_{6}^{14}C+_{1}^{1}H,\]the X will be

    A)  \[_{-1}^{0}e\]                                  

    B)  \[_{1}^{1}H\]

    C)  \[_{1}^{2}H\]                                   

    D)  \[_{0}^{1}n\]

    Correct Answer: D

    Solution :

                     In the reaction, we have to conserve the mass number and the atomic number. Thus, we have \[_{7}^{14}N+_{b}^{a}X\xrightarrow{{}}_{6}^{14}C+_{1}^{1}H\]                 So,          \[14+a=14+1\] Hence a = 1 and 7 + b = 6 + 1 Hence b = 0. So, we have \[_{b}^{a}X=_{0}^{1}X=_{0}^{1}n.\]


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