JCECE Engineering JCECE Engineering Solved Paper-2002

  • question_answer
    Certain radioactive substance reduces to \[25%\] of its value in \[16\] days. Its half-life is:

    A)  32 days                               

    B)  8 days

    C)   64 days                              

    D)   28 days

    Correct Answer: B

    Solution :

    From Rutherford-Soddy law                 \[N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{n}}\] where \[n\] is number of half-lives. Given    \[\frac{N}{{{N}_{0}}}=\frac{25}{100}=\frac{1}{4}\] \[\therefore \]  \[\frac{1}{4}={{\left( \frac{1}{2} \right)}^{n}}\] \[\Rightarrow \]               \[\frac{1}{{{2}^{2}}}=\frac{1}{{{2}^{n}}}\] \[\Rightarrow \]               \[n=2\] Also,                 \[n=\frac{elapsed\,\,times}{half-life}\] \[\Rightarrow \]               \[2=\frac{16}{{{T}_{1/2}}}\] \[\Rightarrow \]               \[{{T}_{1/2}}=8\,\,days\]


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