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_(p)=((dN)/(dt))_(p)= - lambdaN_(i)e^(-lambda(t+Deltat))` and
`A_(q)=((dN)/(dt))= -lambda N_(i)e^(-lambda(t))`
Dividing both and solving, we get `._(Delta t)`