Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2008

  • question_answer
    Two radioactive substances A and B have decay constants \[5\lambda \] and \[\lambda \] respectively. At t = 0 they have the same number of nuclei. The ratio of   number of nuclei of A to those of B will be \[\left( \frac{1}{e} \right)\]  after a time interval

    A) \[\frac{1}{4\lambda }\]                                 

    B) \[4\lambda \]

    C) \[2\lambda \]                                   

    D) \[\frac{1}{2\lambda }\]

    Correct Answer: D

    Solution :

    Number of nuclei remained after time t can be written as \[N={{N}_{0}}{{e}^{-\lambda t}}\] where \[{{N}_{0}}\] is initial number of nuclei of both the substances. \[{{N}_{1}}={{N}_{0}}{{e}^{-5\lambda t}}\]             ...(i) and          \[{{N}_{2}}={{N}_{0}}{{e}^{-\lambda t}}\]             ...(ii) Dividing Eq. (i) by Eq. (ii), we obtain                 \[\frac{{{N}_{1}}}{{{N}_{2}}}={{e}^{(-5\lambda +\lambda )t}}={{e}^{-4\lambda t}}=\frac{1}{{{e}^{4\lambda t}}}\] But, we have given                 \[\frac{{{N}_{1}}}{{{N}_{2}}}={{\left( \frac{1}{e} \right)}^{2}}=\frac{1}{{{e}^{2}}}\] Hence,  \[\frac{1}{{{e}^{2}}}=\frac{1}{{{e}^{4\lambda t}}}\] Comparing the powers, we get                 \[2=4\lambda t\] or            \[t=\frac{2}{4\lambda }=\frac{1}{2\lambda }\]


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