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Half-lives of two radioactive substances...

Half-lives of two radioactive substances `A` and `B` are respectively `20` min and `40` min. Initially, the samples of `A` and `B` have equal number of nuclei. After `80` min the raatio of remaining number of `A` and `B` nuclei is

A

`4 : 1`

B

`1 : 2`

C

`8 : 1`

D

`16 : 1`

Text Solution

Verified by Experts

The correct Answer is:
A
`T_(1//2(A))= 40min, T_(1//2(B))= 20 min`
t = 80 min
`n_(A) = t/T_(1//2(A))=80/40 = 2`
`n_(B) = t/T_(1//2(B))=80/20 = 4`
`N_(A)/N_(B)=(N_(0)//2^(2))/(N_(0)//2^(4))=16/4 = 4 : 1`
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