CET Karnataka Medical CET - Karnataka Medical Solved Paper-2006

  • question_answer
    Activity of a radioactive sample decreases to (1/ 3)rd 1 of its original value in 3 days. Then, in 9 days its activity will become:

    A)  (1 / 27) of the original value

    B)  (1 / 9) of the original value

    C)  (1 / 18) of the original value

    D)  (1 / 3) of the original value

    Correct Answer: A

    Solution :

    \[R={{R}_{0}}{{e}^{-\lambda t}}\] \[\Rightarrow \]\[\left( \frac{1}{3} \right)={{e}^{-\lambda \times 3}}={{e}^{-3\lambda }}\] ?(i) Again,\[\frac{R}{{{R}_{0}}}={{e}^{-\lambda \times 9}}={{e}^{-9\lambda }}={{({{e}^{-3\lambda }})}^{3}}\] \[={{\left( \frac{1}{3} \right)}^{3}}\][from Eq. (i)] \[=\frac{1}{27}\] \[\Rightarrow \] \[R=\frac{{{R}_{0}}}{27}\] Hence, in 9 days activity will become \[\left( \frac{1}{27} \right)\]of the original value.


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