NEET AIPMT SOLVED PAPER SCREENING 2010

  • question_answer
    The activity of a radioactive sample is measured as \[{{N}_{0}}\] counts per minute at t = 0 and \[{{N}_{0}}/e\] counts per minute at t = 5 min. The time (in minute) at which the activity reduces to half its value is

    A) \[\text{lo}{{\text{g}}_{\text{e}}}\text{2/5}\]     

    B)        \[\frac{\text{5}}{\text{lo}{{\text{g}}_{\text{e}}}\text{2}}\]         

    C)        \[5{{\log }_{10}}2\]                         

    D) \[5{{\log }_{e}}2\]

    Correct Answer: D

    Solution :

    Fraction remains after n half lives \[\frac{N}{{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{n}}={{\left( \frac{1}{2} \right)}^{t/T}}\] Given    \[N=\frac{{{N}_{0}}}{e}\Rightarrow \frac{{{N}_{0}}}{e{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{5/T}}\] or            \[\frac{1}{e}={{\left( \frac{1}{2} \right)}^{5/T}}\] Taking log on both sides, we get                 \[\log 1-\log e=\frac{5}{T}\log \frac{1}{2}\]                                 \[-1=\frac{5}{T}(-\log 2)\] \[\Rightarrow \]                               \[T=5{{\log }_{e}}2\] Now, let t' be the time after which activity reduces to half            \[\left( \frac{1}{2} \right)={{\left( \frac{1}{2} \right)}^{t'/5{{\log }_{e}}2}}\] \[\Rightarrow \]                               \[t'=5{{\log }_{e}}2\]


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