Calculate the time reuired for `40%` of a radioactive sample to disintegrate. Half life of the sample is `1.6xx10^(10)` years.

Explanation

`T_(1)/(2)=1.6xx10^(10)` years.
Let t be the time in which `40%` of the sample disintegrates
`therefore` after t years `60%` of the sample will be left over
i.e., `(60)/(100)N_(0)=N_(0)e^(-lambda t)`
`therefore lambda t =2.303(log_(10)100-log_(10)60)=2.303(0.2219)`
`therefore t=(2.303xx0.2219)/(lambda)=(2.303xx0.2219xx1.4xx10^(10))/(0.693)`
`t=1.032xx10^(10)` years.