BCECE Engineering BCECE Engineering Solved Paper-2007

  • question_answer
    A radioactive substance has half-life of 60 min. During 3 h, the fraction of the substance that has to be decayed, will be

    A)  87.5%                  

    B)  52.5%                  

    C)         25.5%                  

    D)         8.5%

    Correct Answer: A

    Solution :

    From Rutherford and soddy law, the rate of decay of a radioactive substance is proportional to number of atoms left at that instant, using this we can arrive at \[N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{n}}\] where\[{{N}_{0}}\]is original number of atoms, n is number of half-lives \[n=\frac{t}{{{T}_{1/2}}}=\frac{180}{60}=3\]                 \[\therefore \]  \[\frac{N}{{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{n}}={{\left( \frac{1}{3} \right)}^{3}}=\frac{1}{8}\]                 \[\therefore \]  \[N=\frac{{{N}_{0}}}{8}=0.125\,{{N}_{0}}=12.5%\,N\] Amount decayed \[=100-12.5=87.5%.\]


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