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The percentage of quantity of a radioactive material that remains after 5 half-lives will be .A. `1%`B. `3%`C. `5%`D. `20%` |
Answer» Correct Answer - B `N = N_(0) ((1)/(2))^(n) = N_(0)((1)/(2))^(5) = (N_(0))/(32)` `(N)/(N_(0)) xx 100 = (100)/(32) ~~ 3%` |
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