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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2009

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

    A)  \[0g\]             

    B)  \[0.125g\]

    C)  \[1g\]

    D)  \[0.5g\]

    Correct Answer: B

    Solution :

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


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