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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. `14 min`B. `20 min`C. `28 min`D. `7 min` |
Answer» Correct Answer - B At `t_(1) (2)/(3)=(1)/(2^(t_(1)//20))` At `t_(2)=(1)/(3)=(1)/(2^(t_(2)//20))` `t_(2)-t_(1)=20 "mins"` |
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