The activity of a radioative element decreases to one third of the original activity `I_(0)` in a period of nine years. After a further 1apse of nine years, its activity will be

Options

A. `A_(0)`
B. `(2)/(3)A_(0)`
C. `(A_(0))/(9)`
D. `(A_(0))/(6)`

Answer

c

Explanation

`(A_(0))/(3) =A_(0)((1)/(2))^((9)/(T_(1//2))`
`A'=(A_(0))/(3) ((1)/(2))^((9)/(T_(1//2))`
Dividing, we get
`(A' xx3)/(A_(0)) =(1)/(3)` or `A' =(A_(0))/(9)`.