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The half-life of a radioactive substance is 3h and its activity is `1mu Ci`. Then the activity after 9h will be (in `mu Ci`)-A. `1/9`B. `1/27`C. `1/3`D. `1/8` |
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Answer» Correct Answer - D `A=A_(0)e^(-lambdat)` `lambda=(ln2)/(t_(1//2)) t=9h` then `A=(A_(0))/8` |
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