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. `3/4N_(0)`
B. `(N_(0))/9`
C. `N_(0)`
D. `(N_(0))/6`

Answer

B

Explanation

The radio actice disintegrations follow first order kinetics. The integrated rate law is given as `N=N_(0)e^(-lamdat)`
N=Activity number at time t.
`N_(0)=` original activity number
`(N_(0))/3=N_(0)e^(-9lamda)`
`implies1/3=e^(-9lamda)`
Now after further 9y
`N.=N_(0)e^(-18lamda)`
`N.=N_(0)(e^(-9lamda))^(2)`
`N.=N_(0)(1/3)^(2)=(N_(0))/9`