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Two identical samples (same material and same amout) `P and Q` of a radioactive substance having mean life `T` are observed to have activities `A_P` and `A_Q` respectively at the time of observation. If `P` is older than `Q`, then the difference in their age isA. `T "ln"((A_(P))/(A_(Q)))`B. `T "ln"((A_(Q))/(A_(P)))`C. `T ((A_(P))/(A_(Q)))`D. `T ((A_(Q))/(A_(P)))` |
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Answer» Correct Answer - B `A_(Q) overset(t=?)rarrA_(p)` (T=mean life) `A = A_(0)e^(-lambdat) , A_(p) = A_(Q)e^(-lambdat)` `rArr e^(lambdat) = (AQ)/(AP)` `lambdat = "ln"((A_(Q))/(A_(P)))t = T "ln" ((A_(Q))/(A_(P)))` `lambda = (1)/(T)` |
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