CMC Medical CMC-Medical Ludhiana Solved Paper-2015

  • question_answer
    The activity of a radioactive sample drops to\[\frac{1}{128}\] th of its original value in 2h. The decay constant will be

    A)  \[1.484\times {{10}^{3}}{{s}^{-1}}\]       

    B)  \[6.7\times {{10}^{-4}}{{s}^{-1}}\]

    C)  \[6.7\times {{10}^{3}}{{s}^{-1}}\]                            

    D)  None of these

    Correct Answer: B

    Solution :

                    \[{{N}_{0}}=1,\,N=\frac{1}{128},\]\[t=2h=2\times 60\times 60\,s\] \[\lambda =\frac{2.303}{t}\log \frac{{{N}_{O}}}{N}\] \[\lambda =\frac{2.303}{2\times 60\times 60}\log \frac{1}{1/128}\] \[\lambda =6.7\times {{10}^{-4}}{{s}^{-1}}\]


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