the count rate from a radioactive sample falls from `4.0xx10^6`per second to `1.0xx10^6` per second in 20 hours.What will be the count rate 100 hours after the begnning ?

Answer

`3.91xx10^(3)s^(-1)`

Explanation

Here, `A_(0)=4xx10^(6)s^(-1) , A=1.0xx10^(6)s^(-1)`,
`t=20hr.`
As `N/(N_(0))=(1/2)^(n)=(1.0xx10^(6))/(4.0xx10^(6))=4 :. n=2`
As `n=t//T :. T=t/n=20/2=10hrs`
Again, `n'=(t')/T=100/10=10`
`:. A'=A_(0) (1/2)^(n')=4xx10^(6)(1/2)^(10)`
`=3.91xx10^(3)s^(-1)`