CET Karnataka Medical CET - Karnataka Medical Solved Paper-2004

  • question_answer
    A radioactive isotope decays at such a rate that after 192 minutes only 1/16 of the original amount remains:

    A) \[32min\]

    B) \[48min\]

    C) \[12min\]        

    D) \[24min\]

    Correct Answer: B

    Solution :

    Given, \[t=192\]minutes \[\frac{N}{{{N}_{0}}}=\frac{1}{16}\] \[{{t}_{1/2}}=?\] We know that, \[\frac{N}{{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{t/{{t}_{1/2}}}}\] \[\therefore \] \[\frac{1}{16}={{\left( \frac{1}{2} \right)}^{\frac{192}{{{t}_{1/2}}}}}\] or \[{{\left( \frac{1}{2} \right)}^{4}}={{\left( \frac{1}{2} \right)}^{\frac{192}{{{t}_{1/2}}}}}\] or \[4=\frac{192}{{{t}_{1/2}}}\] \[\therefore \] \[{{t}_{1/2}}=\frac{192}{4}=48\] minutes


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