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. `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)))`

Answer

B

Explanation

`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)`