JCECE Medical JCECE Medical 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,                \[\text{n =}\frac{\text{elapsed}\,\text{time}}{\text{half}-\text{life}}\] \[\Rightarrow \] \[2=\frac{16}{{{T}_{1/2}}}\] \[\Rightarrow \] \[{{T}_{1/2}}=8\,days\]


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