A Radioactive element A says to B ''l am half of what you were when you are one fourth of what l was moreover l was `1.414` time than what you were '' If the half life of A is 8 days, what is the half life of 'B' is

Explanation

Let the initial number of nuclide of X be `x_(0 )`
Let the initial number of nuclide of Y be `y_(0)`
Let the number of nuclide of X after time t be `x_(t)`
Let the number of nuclide of Y after time t be `y_(t)`
`therefore x_(t) = ½y_(0) , yt = ¼x_(0) `
`x_(0) = 1.414 y_(0)`
Now, `x_(t)= x_(0) e^(-lambda_(1))t= ½ y_(0)" "..........(1)`
`y_(t)= y_(0) e^(-lambda_(2))t= ¼ x_(0)" "..........(2)`
Dividing equation. (1) by (2)
`(2y_(0))/x_(0)=(x_(0)e^((lamda_(2)-lamda_(1))t))/y_(0)rArr 2 = x_(0)^(2)/y_(0)^(2)e^((lamda_(2)-lamda_(1))t)`
`"since " x_(0) = sqrt(2) y_(0) rArr x_(0)^(2)=2y_(0)^(2)`
`therefore" " 2 = 2e^((lambda_(2)-lambda_(1))t)`
`therefore" " e^((lambda_(2)-lambda_(1))t)=1`
Since t is not zero, hence `lambda_(2) = lambda_(1)`. Therefore the half life of the other radioactive element is also 8 days.