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

  • question_answer
    Number of neutrons in a parent nucleus \[X\], which gives \[_{7}{{N}^{14}}\] nucleus after two successive \[\beta \]  emissions would be [CBSE PMT 1998; MP PMT 2003]

    A)                 9             

    B)                 8

    C)                 7             

    D)                 6

    Correct Answer: A

    Solution :

              \[_{5}{{X}^{14}}{{\xrightarrow{-2\beta }}_{7}}{{N}^{14}}\] than no. of neutrons in \[_{5}{{X}^{14}}=14-5=9\].


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