CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2006

  • question_answer
    In artificial radioactivity,\[1.414\times {{10}^{6}}\]nuclei are disintegrated into\[{{10}^{6}}\]nuclei in 10 min. The half-life in minutes must be:

    A)  5       

    B)                                         20

    C)  15                         

    D)         30

    E)  25

    Correct Answer: A

    Solution :

    From Rutherford-Soddy law \[N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{n}}\] \[n=\frac{t}{T}\] \[\therefore \]  \[{{10}^{6}}=1.414\times {{10}^{6}}{{\left( \frac{1}{2} \right)}^{t/T}}\] \[\Rightarrow \]               \[\frac{1000}{1414}={{\left( \frac{1}{2} \right)}^{t/T}}\] \[\Rightarrow \]         \[{{\left( \frac{1}{2} \right)}^{2}}={{\left( \frac{10}{12} \right)}^{2}}\]            (Approximately) \[\Rightarrow \]               \[n=2\] \[\Rightarrow \]               \[n=\frac{t}{T}=2\] \[\Rightarrow \]               \[T=\frac{10}{2}=5\,\min \]


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