JIPMER Jipmer Medical Solved Paper-2009

  • 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 \[{{\left( \frac{1}{e} \right)}^{2}}\]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            \[t=\frac{2}{4\lambda }=\frac{1}{2\lambda }\]   


You need to login to perform this action.
You will be redirected in 3 sec spinner