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 is

Options

A. `Tln((A_(p))/(A_(Q)))`
B. `Tln((A_(Q))/(A_(P)))`
C. `1/Tln((A_(P))/(A_(Q)))`
D. `T((A_(P))/(A_(Q)))`

Answer

B

Explanation

`A_(P)=A_(Q) e^(-gamma t)=A_(0)e^(1/Tt) " " :. t=Tln(A_(Q))/(A_(P))`