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

Options

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

Answer

B

Explanation

Here, half-life of radioactive substance, T = 20 min
We have, `N=N_(0)((1)/(2))^(t1//T)`
n So, for `20%` decay, `(N)/(N_(0))=80/100=((1)/(2))^(t1//20)`
For `80%` decay, `(N)/(N_(0))=20/100=((1)/(2))^(t1//20)`
On dividing Eq. (ii) by Eq. (i) we get
`((1)/(2))^(((t_(2)-t_(1))/(20)))=1/4`
or `((1)/(2))^(((t_(2)-t_(1))/(20)))=((1)/(2))^(2)`
`therefore` The time between `20% and 80%` decay `t_(2)-t_(1)`= 40 min