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. `TIn((R_(p))/(R_q))`
B. `TIn((R_(q))/(R_p))`
C. `T((R_(p))/(R_q))`
D. `T((R_(q))/(R_p))`

Answer

B

Explanation

Activity, `R=R_(0)e^(-lamdat)`
`because" "(R_(p))/(R_(q))=(R_(0)e^(-lamdat_(1)))/(R_(0)e^(-lamdat_(2)))`
m `because" "e^(lamda(t_(1)-t_(2)))=(R_(q))/(R_(p))" "(becauselamda=(1)/(T))`
`therefore"The difference image",(t_(1)-t_(2))=T1n((R_(q))/(R_(p)))`