A sample of radioactive material has an activity of `9xx10^(12)` Becquerel. The material has a half life of 80s. How long will it take for the activity to falls to `2xx10^(10)` Becquerel ?

Answer

`173.6s`

Explanation

Activity `(dN)/(dt) propN :. N/(N_(0))=(2xx10^(12))/(9xx10^(12))=2/9`
form `N/(N_(0))=(1/2)^(n)`
`:. n(log 1 -log 2)=log.(N)/(N_(0))=log.(2)/(9)`
`=(log 2-log9)`
`n(0-0.3010)=(0.3010-0.9542)`
`n=0.6532/0.3010=2.17`
form `t=nT, t=2.17xx80=173.6s`