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

Explanation

We know that `K = (0.693)/(t_(1//2)) = (0.693)/(3.8) = 0.182 day^(-1)`
Let the initial amountd of radon be `N_(0)` and the amount left
after `t` days be `N`d which is equal to `N_(0)//10`.
Applying the equation,
`t = (2.303)/(K) log (N_(0))/(N)`
`t = (2.303)/(0.182) log (N_(0))/(N_(0)//10)`
`= (2.303)/(0.182) log 10 = 12.65` days
Alternatively,
`(t_(1//2))/(t_(x%)) = (0.3)/(log(a)/(a - x))`
`implies (3.8)/(t_(x%)) = (0.3)/(log((a)/((1)/(10))a)` or `(3.8)/(t_(x%)) = (0.3)/(log 10)`
`:. t_(x%) = (3.80)/(0.3) = 12.65` days