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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2009

  • question_answer
    2 g of a radioactive sample having half-life of 15 days was synthesized on 1st Jan 2009. The amount of the sample left behind on 1st  March, 2009 (including both the days) is

    A)  0 g                                       

    B)  0.125 g

    C)  1 g                                        

    D)  0.5 g

    Correct Answer: B

    Solution :

    \[N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{n}}\] Given,   \[{{N}_{0}}=2\,\,g\]                 \[{{t}_{1/2}}=15\] days                 \[T=60\] days \[\Rightarrow \]               \[n=\frac{60}{15}=4\] \[\therefore \]  \[N=2{{\left( \frac{1}{2} \right)}^{4}}\] or            \[N=0.125\,g\]


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