The activity of a radioactive sample decreases to `1//3` of the original activity, `A_(0)` in a period of 9 years. After 9 years more, its activity `A_(0)//x`. Find the value of `x`.

Answer

`9`

Explanation

`t = (2.303)/(lambda) log (A_(0))/(A_(1)0`
`9 = (2.303)/(lambda) log 3 implies lambda = 0.122`
`18 = (2.303)/(0.122) log (A_(0))/(A_(2))`
`(A_(2))/(A_(0)) = (1)/(9), :. A_(2) = (A_(0))/(9) ("given" (A_(0))/(x))`
`:. X = 9`