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Two radioactive X(1) and X(2) have docay...

Two radioactive `X_(1)` and `X_(2)` have docay constants `10 lambda ` and `lambda` respecttively . If inittially they have the same number of noclei, then the radio of the number of nuclei of `X_(1)` to that of `X_(2)`will be `1//e` after a time .

A

`(1)/(10 lambda)`

B

`(1)/(11 lambda)`

C

`(11)/(10 lambda)`

D

`(1)/(9 lambda)`

Text Solution

Verified by Experts

The correct Answer is:
D
`N_(1) = N_(0) e^(-10 lambda_(1) ` and N_(2) = N_(0) e^(-10 lambda_(1)`
`:. (N_(1))/(N_(2)) = e^(-10 lambda_(1))/(e^(- lambda_(1))) = (1)/(e^(9 lambda_(1))`
`Given (N_(1))/(N_(2)) = (1)/(e ) :. (1)/(e^(lambda_(1)) = (1)/€`
`or , 9lambda_(1) = 1 , or t = ((1)/(9 lambda))`
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