A mixture consists of two radioactive materials `A_1` and `A_2` with half-lives of `20 s` and `10 s` respectively. Initially the mixture has `40 g` of `A_1` and `160 g` of `a_2`. The amount the two in the mixture will become equal after

Options

A. `60s`
B. `80s`
C. `20s`
D. `40s`

Answer

D

Explanation

Here, `T_(1)=20s, T_(2)=10s, N_(01)=40g`,
`N_(02)=160g`.
Let the amount of the two radioactive materials become equal after t sec.
form `N/(N_(0))=(1/2)^(n)=(1/2)^(t//T)`, we get
`N_(1)=N_(01)(1/2)^(t//20)=40(1/2)^(t//20)`
`N_(2)=N_(02)(1/2)^(t//10)=160(1/2)^(t//10)`
As `N_(1)=N_(2) :. 40(1/2)^(t//20)=160(1/2)^(t//10)`
or `(1/2)^(t//20)=4(1/2)^(t//20)=(1/2)^(t/10-2)`
`:. t/20=t/10-2 or t=40s`