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AMU Medical AMU Solved Paper-2011

  • question_answer
    The radioactive decay constant of \[_{38}^{90}Sr\]is\[7.88\times {{10}^{-10}}{{s}^{-1}}.\]The activity of 15 mg of this isotope will be

    A)  1.5 Ci                                   

    B)  2.13 Ci

    C)  7.88 Ci                                 

    D)  8.76 Ci

    Correct Answer: B

    Solution :

                     Activity \[\lambda \,\,.\,\,N\] \[=7.88\times {{10}^{-10}}\times \frac{15\times {{10}^{-3}}}{90}\times 6.023\times {{10}^{23}}\] \[=7.91\times {{10}^{-10}}dps\] \[=\frac{7.91\times {{10}^{10}}}{3.7\times {{10}^{10}}}\] = 2.13 Ci


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