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Two identical samples (same material and same amount) P and Q of a radioactive substance having mean life T are observed to have activities `R_(p) andR_(Q)` respectively, at the time of observation. If P is older than Q, Then the difference in their age is

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)))`

Text Solution

Verified by Experts

The correct Answer is:
B

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