Half life of a radio-active substance is `20` minutes. The time between `20 %` and `80 %` decay will be

Options

A. 20 min
B. 30 min
C. 40 min
D. 25 min

Answer

C

Explanation

According to radioactive decay, `N = N_(0) e^(-lambda t)`
where, `N_(0)` = Number of radioactive nuclei present in the sample at t = 0, N = Number of radioactive nuclei left undecayed after time t, `lambda` = decay constant.
For 20% decay, `(80 N_(0))/(100) = N_(0) e^(-lambda t_(1))" "...(i)`
For 80% decay, `(20 N_(0))/(100) = N_(0) e^(-lambda t_(2))" "...(ii)`
Dividing equation (i) by (ii), we get
`4 = e^(-lambda (t_(1) - t_(2))) rArr 4 = e^(lambda (t_(2) - t_(1)))`
Taking natural logarithms of both sides of above equation, we get
1n `4 = lambda (t_(2) - t_(1)) or 2 1n 2 = (1n 2)/(T_(1//2)) (t_(2) - t_(1))`
`rArr" "t_(2) - t_(1) = 2 xx T_(1//2) = 2 xx 20` min = 40 min.