If `20 gm` of a radioactive substance due to radioactive decay reduces to `10 gm` in `4` minutes, then in what time `80 gm` of the same substance will reduce to `10 gm` ?

Options

A. 8 minutes
B. 12 minutes
C. 16 minutes
D. 20 minutes

Answer

B

Explanation

(b) `20 gm` substance reduces to `10 gm` (i.e., becomes half in `4 min`. So `T_(1//2) = 4 min`. Again `M = M_0 ((1)/(2))^(t//T_1//2)`
`rArr 10 = 80((1)/(2))^(t//4) rArr (1)/(8) = ((1)/(2))^3 = ((1)/(2))^(t//4)`
`rArr t = 12 min`.