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BCECE Engineering BCECE Engineering Solved Paper-2003

  • question_answer
    The amount of substance that gives \[3.7\times {{10}^{7}}\] dps, is:

    A) one becquerel   

    B)        one curie            

    C)        one millicurie  

    D)        one Rutherford

    Correct Answer: C

    Solution :

    Key Idea:  One curie is the amount of radio active element which gives \[3.7\times {{10}^{10}}\] disintegration per second, \[\therefore \]  1 millicurie \[=3.7\text{ }\times \text{ }{{10}^{7}}\text{ }dps.\]


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