The count rate of a radioactive sample falls from `4.0xx10^6 s^(-1)` to `1.0xx10^6 s^(-1)` in 20 hours. What will be the count rate after 100 hours from beginning ?

Options

A. `3.91xx10^3 s^(-1)`
B. `3.91xx10^2 s^(-1)`
C. `3.91xx10^4 s^(-1)`
D. `3.91xx10^6 s^(-1)`

Answer

A

Explanation

Here `R_0=4.0xx10^6 s^(-1) , R=1.0xx10^6 s^(-1)`
t=20 hours, t'=100 hours
As,`R/R_0(1/2)^n` or `(1.0xx10^6)/(4.0xx10^6)=(1/2)^n`
or `(1/2)^n = (1/2)^2 therefore n=2`
`T_(1//2)=t/n=20/2`=10 hours
Now, `n'="t'"/T_(1//2)=100/10`=10
`therefore R'=R_0(1/2)^(n') =4.0xx10^6 (1/2)^10 = 3.91xx10^3 s^(-1)`