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The half life of a radioactive substance is `20` minutes . The approximate time interval `(t_(2) - t_(1))` between the time `t_(2)` when `(2)/(3)` of it had decayed and time `t_(1)` when `(1)/(3)` of it had decay isA. `7 min`B. `14 min`C. `20 min`D. `28 min` |
Answer» Correct Answer - C `(2)/(3)N_(0)=N_(0)e^(-lambdat_(1))` `(1)/(3)N_(0)=N_(0)e^(-lambdat_(2))` `2=e^(lambda(t_(2)-t_(1)))` `lambda(t_(2)-t_(1))= l n 2` `(t_(2)-t_(1))= ( l n 2)/(lambda)=20 "min"` |
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