The half life of radon is 3.8 days . After how many days will only one twentieth of radon sample be left over ?

Explanation

Given, half-life of sample, `T_(1//2) = 3.8` days
Given, `N= N_(0)/20`
Decay constant, `lambda = 0.693/T_(1//2)= 0.693/3.8 "day"^(-1)`
Let t be the disintegration time. Then
`N = N_(0)e^(-lambdat)`
`rArr N_(0)/20 = N_(0)e^(-lambdat)`
`rArr N_(0)/20 = N_(0)e^(-lambdat)`
`rArr e^(lambdat) =20`
Taking natural log on both sides, we get
`lambdat log_(e)e = log_(e)20`
`rArr lambdat = 2.303 xx log_(10)20`
`rArr t = (2.303 xx log_(10)20)/lambda`
`=(2.303 xx 1.3010 xx 3.8)/0.693` days
`=16.43` days