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The percentage of quantity of a radioactive material that remains after 5 half-lives will be .A. `31%`B. `3.125%`C. `0.3%`D. `1%` |
Answer» Correct Answer - b `(N)/(N_(0))=(1)/(2^(5T//T))` `(N)/(N_(0)) =(1)/(2^(5))` `:. (N)/(N_(0)) xx 100=(100)/(32)=3.125`. |
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