The half-life of the radioactive radon is 3.8 days. The time, at the end of which `1//20th` of the radon sample will remain undecayed, is (given `log_10 e=0.4343`)

Explanation

We know that `lambda =0.693//T_(1//2)`
Here, `T_(1//2) =3.8 day`
`:. Lambda =(0.693)/(3.8)=0.182 per day`
If initially `(at t=0)` the number of atoms present be `N_(0)`, then the number of atoms N left after a time given by
`N =N_(0) e^(-lambda t)`
`(N)/(N_(0))= e^(-lambda t)` or `(1)/(20) =e^(- lambda t)`
Taking log, `lambda t=log_(e) 20 =2.3026 log_(10) 20`
`:. t=(2.3026 log_(10)20)/(lambda) = (2.3026 log_(10)20)/(0.182) =16.45 days`.