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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. `Tln((A_(p))/(A_(Q)))`B. `Tln((A_(Q))/(A_(P)))`C. `1/Tln((A_(P))/(A_(Q)))`D. `T((A_(P))/(A_(Q)))` |
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Answer» Correct Answer - B `A_(P)=A_(Q) e^(-gamma t)=A_(0)e^(1/Tt) " " :. t=Tln(A_(Q))/(A_(P))` |
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