After a certain lapse of time, fraction of radioactive polonium undecayed is found to be `12.5%` of the initial quantity. What is the duration of this time lapsed if the half life of polonium is 138 days ?

Explanation

Given, half-life of polonium, `t_(1//2) = 138` days Amount of undecayed Po after 138 days = 50% of initial amount
Amount of undecayed Po after `138 xx 2 = 276` days - 25% of initial amount.
Amount of undecayed. Then number of atoms.
`N = 25 of N_(0) = N_(0)/4`
Let n be the number of half-lives passed. Then
`N = N_(0) (1/2)^(n)`
`rArr N_(0)/4 = N_(0)(1/2)^(n)`
`rArr (1/2)^(2) =(1/2)^(n)` or n=2
Given, half-life, `t_(1//2) = 3.8` days
`therefore t=t_(1//2) xx n = 3.8xx 2 = 7.6` days.