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Manipal Engineering Manipal Engineering Solved Paper-2009

  • question_answer
    2 g of\[a\cdot \]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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