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A mixture consists of two radioactive ma...

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

A

60s

B

80s

C

20s

D

40s

Text Solution

Verified by Experts

The correct Answer is:
D
`(N_(1))/(N_(01)) = ((1)/(2))^((t)/(20)), (N_(2))/(N_(02)) = ((1)/(2))^(t//10)`
`N_(1)= N_(2)`
`rArr (40)/((2)^(t//20))= (160)/((2)^(t//10))`
`2^(t//20) = 2^(((t)/(10)-2))`
`rArr (t)/(20) = (t)/(10)-2 rArr t = 40`
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