CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2011

  • question_answer
    A radioactive sample at any instant has its disintegration rate 5000 disintegrations per minute. After 5 min, the rate becomes 1250 disintegration per minute. Then, its decay constant (per minute) is

    A)  \[0.8\,{{\log }_{e}}2\]

    B)  \[0.4\,{{\log }_{e}}2\]   

    C)        \[0.2\,{{\log }_{e}}2\]                    

    D)        \[0.1\,{{\log }_{e}}2\]

    E) \[0.6\,{{\log }_{e}}2\]

    Correct Answer: B

    Solution :

    Given R = 1250,\[{{R}_{0}}=5000\]and t = 5 min Hence   \[R={{R}_{0}}{{e}^{-\lambda t}}\] \[1250=5000{{e}^{-\lambda \times 5}}\] \[\lambda =0.4\,{{\log }_{e}}2\]


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