The half-life period of `U^(234)` is `2.5 xx 10^(5)` years. In how much is the quantity of the isotope reduce to 25% of the original amount?

Explanation

Initial amount of this isotope `N_(0) = 100`
Final amount of the isotop `N = 25`
We know that `N = ((1)/(2))^(n) N_(0)`
So `25 = ((1)/(2))^(n) xx 100`
or `(25)/(100) = ((1)/(2))^(n)`
or `(25)/(100) = ((1)/(2))^(n)`
or `((1)/(2))^(2) = ((1)/(2))^(n)`
or `n = 2`
`= T = n xx t_(1//2) = 2 xx 2.5 xx 10^(5) = 5 xx 10^(5)` year
Alternatively
We can use `(t_(1//2))/(t_(x%)) = (0.3)/("log"(a)/(a - x))`
`= (t_(1//2))/(t_(25%)) = (0.3)/("log"(100)/(25))`
`(2.5 xx 10^(5))/(t_(25%)) = (0.3)/("log"4)`
`t_(25%)` or time required to reduced amount to 25%
`= (2.5 xx 10^(5))/(0.3) xx "log" 4`
`= 2.5 xx (10^(5))/(0.3) xx 0.602`
`= 5.0 xx 10^(5)` years