CET Karnataka Engineering CET - Karnataka Engineering 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)  32 min                                 

    B)  48 min

    C)  12 min                                 

    D)  24 min

    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                            \[{{\left( \frac{1}{4} \right)}^{4}}={{\left( \frac{1}{2} \right)}^{\frac{192}{{{t}_{1/2}}}}}\] or,                          \[4=\frac{192}{{{t}_{1/2}}}\] \[\therefore \]                  \[{{t}_{1/2}}=\frac{192}{4}=48\] minutes

adversite


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