EAMCET Medical EAMCET Medical Solved Paper-1999

  • question_answer
    Radioactivity of an isotope falls to 1/8 of its original value in 48 days. Its half life period is (in days):

    A)  48                                         

    B)  16

    C)  12                                         

    D)  8

    Correct Answer: B

    Solution :

                     \[N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{n}}\] \[\frac{N}{N}={{\left( \frac{1}{2} \right)}^{n}}\] \[\frac{1}{8}={{\left( \frac{1}{2} \right)}^{n}}\] \[{{\left( \frac{1}{2} \right)}^{3}}={{\left( \frac{1}{2} \right)}^{n}}\] \[n=3\] \[n=\frac{Total\,time}{Half\,-life}\]                 \[Half=life=\frac{Total\,time}{n}\]                                 \[=\frac{48}{3}=16\]day


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