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. `t_0`
B. `(2//3)t_0`
C. `(t_0//9)`
D. `(t_0//6)`

Answer

C

Explanation

( c) Activity `A` of a radioactive element is defined as its disintegration rate i.e.,
`A = -(dN//dt)`
But `N - N_0 e^(- I t)`
`- (dN)/(dt) lamda N_0^(-lamda t) = lamda N`
`:. A = lamda N`
Hence `(A)/(A_0) =(lamda N)/(lamda N) = (N)/(N_0)`
Fraction value period `t_f` is the time after which the number of mother nuclei left is a fraction `f` times that at start. Then after a lapse of time `t(=n_f t_f)` the number of surviving mother nuclei is given by :
`(N)/(N_0) = (f)^(n_f)`or `(A)/(A_0) = (f)^(n_f)`
Here `A_0 = I_0` and `t_f = 9 years`
`t = 9 + 9 = 18 `years
`n_f = (t)/(t_f) = (18)/(9) = 2`
Now `f = (1//3)`
`:. (A)/(t_0) = ((1)/(3))^2 = (1)/(9)` or `A = (t_0)/(9)`.