Two identical samples (same material and same amount) 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 1n ((A_P)/(A_Q))`
B. `T 1n ((A_Q)/(A_P))`
C. `T((A_P)/(A_Q))`
D. `T((A_Q)/(A_P))`

Answer

B

Explanation

`A_P=A_0e^(-lambdat_1)`
`A_Q=A_0e^(-lambdat_2)`
`lambda_(t_1)=1n(A_0//A_P)`
`:.` `t_1=1/lambda1n(A_0//A_P)=T1n(A_0//A_P)`
Similarly, `t_2=T1n(A_0//A_Q)`
`:.` `t_1-t_2=T1n((A_Q)/(A_P))`