Uttarakhand PMT Uttarakhand PMT Solved Paper-2007

  • question_answer
    The decay constant of\[^{226}Ra\]is\[1.36\times {{10}^{-11}}{{s}^{-1}}\]. A sample of Ra- 226, having an activity of 1.5 \[mCi\]will contain

    A)  \[4.0\times {{10}^{18}}atoms\]

    B)  \[205\times {{10}^{15}}atoms\]

    C)  \[3.5\times {{10}^{7}}atoms\]

    D)  \[4.7\times {{10}^{10}}atoms\]

    Correct Answer: A

    Solution :

     \[\because \] \[1Ci=\frac{6.023\times {{10}^{23}}}{226}\]atoms \[\therefore \]\[1.5\times {{10}^{-3}}=\frac{1.5\times {{10}^{-3}}\times 6.023\times {{10}^{23}}}{226}\] \[=3.997\times {{10}^{18}}atoms\]


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