NEET AIPMT SOLVED PAPER MAINS 2011

  • question_answer
    Two radioactive nuclei P and Q, in a given sample decay into a stable nucleous R. At time t = 0, number of P species are 4No and that of Q are \[{{N}_{0}}\] Half-life of P (for conversion to K) is 1 min where as that of Q is 2 min. Initially there are no nuclei of R present in the sample. When number of nuclei of P and Q are equal, the number of nuclei of R present in the sample would be                                           

    A) \[\text{3}{{\text{N}}_{\text{0}}}\]                          

    B) \[\frac{\text{9}{{\text{N}}_{\text{0}}}}{\text{2}}\]          

    C) \[\frac{\text{5}{{\text{N}}_{\text{0}}}}{\text{2}}\]                          

    D) \[\text{2}{{\text{N}}_{\text{0}}}\]

    Correct Answer: B

    Solution :

    Initially \[P\to 4{{N}_{0}}\] \[Q\to {{N}_{0}}\]                 Half life\[{{T}_{p}}\to 1\,\min \]                 \[{{T}_{Q}}\to 2\,\min \] Let after time t number of nuclei of P and Q are equal ie,                \[\frac{4{{N}_{0}}}{{{2}^{t/1}}}=\frac{{{N}_{0}}}{{{2}^{t/2}}}\]                                 \[4={{2}^{t/2}}\]                                 \[{{2}^{2}}={{2}^{t/2}}\]                                 \[\frac{t}{2}=2\]                                 \[t=4\,\min \]    Disactive nucleus or Nuclei of R                                 \[=\left( 4{{N}_{0}}-\frac{4{{N}_{0}}}{{{2}^{4}}} \right)+\left( {{N}_{0}}-\frac{{{N}_{0}}}{{{2}^{2}}} \right)\]                                 \[=4{{N}_{0}}-\frac{{{N}_{0}}}{4}+{{N}_{0}}-\frac{{{N}_{0}}}{4}\]                                 \[=5{{N}_{0}}-\frac{{{N}_{0}}}{2}\] \[=\frac{9}{2}{{N}_{0}}\]


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