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

Options

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

Answer

B

Explanation

(b) Here `T_(1//2) 20` minutes , we know `(N)/(N_0) = ((1)/(2))^(t//T_(1//2))`
For `20 %` decay `(N)/(N_0) = (80)/(100) = ((1)/(2))^(t_1//20)` …(i)
For `80 %` decay `(N)/(N_0) = (20)/(100) = ((1)/(2))^(t_1//20)` …(ii)
Dividing (ii) and (i)
`(1)/(4) = ((1)/(2))^(((t_2 - t_1))/(20))` , on solving we get `t_2 - t_1 = 40 min`.