JEE Main & Advanced Chemistry Surface & Nuclear Chemistry / भूतल और नाभिकीय रसायन Question Bank Causes of radioactivity and Group displacement law

  • question_answer
    The number of neutrons in the parent nucleus which gives \[{{N}^{14}}\] on \[\beta \]-emission and the parent nucleus is [EAMCET 1985; MNR 1992; Kurukshetra CEE 1998; UPSEAT 2000, 01]

    A)                 \[8,\,{{C}^{14}}\]             

    B)                 \[6,\,{{C}^{12}}\]

    C)                 \[4,\,{{C}^{13}}\]             

    D)                 None of these

    Correct Answer: A

    Solution :

           \[_{6}{{C}^{14}}\to {{\,}_{7}}{{N}^{14}}+{{\,}_{+1}}{{e}^{0}}\]                                 No. of neutrons in \[{{C}^{14}}=14-6=8\].


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