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After an interval of one day , `1//16th` initial amount of a radioactive material remains in a sample. Then, its half-life is .A. `6h`B. `12 h`C. `1.5 h`D. `3h` |
Answer» Correct Answer - a `N=(N_(0))/(2^(t//T))` `(N_(0))/(16) =(N_(0))/(2^(t//T))` `2^(t//T) =16=2^(4)` or `T=(t)=(4)` or `T=(t)/(T)=(24)/(4) h=6h` . |
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