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Two radioactive substance A and B have d...

Two radioactive substance A and B have decay constants `5lamdaand lamda,` respectively. At t=0 they have the same number of nuclei. The ratio of number of nuclei of A to those of B will be `((1)/(e))^(2)` efter a time interval.

A

` 4 lamda `

B

` 2 lamda `

C

` ( 1 ) /( 2lamda ) `

D

` ( 1 ) /( 4lamda ) `

Text Solution

Verified by Experts

The correct Answer is:
C
At t=0, `N=N_(0)` for both the substances A and B
`therefore N_A=N_(0)e^(-lambda_(gl)) and N_(B)=N_(a)e^(-lambda_(gl))`
`(N_(A))/(N_(B))=(e^(-lambda_(A^(t))))/(e^(-lambda_(g t)))=e^((lambda_(H)-lambda_(A))t)=e^((lambda-5lambda)t`
`"As" (N_(A))/(N_(B))=((1)/(e))^(2)["According to question"]`
`or t=(2)/(4lambda)=(1)/(2lambda)`
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