Two radioactive materials have decay constant `5lambda&lambda`. If initially they have same no. of nuclei. Find time when ratio of nuclei become `((1)/(e))^(2)` :

Options

A. `(1)/(4lambda)`
B. `4 lambda`
C. `2 lambda`
D. `(1)/(2lambda)`

Answer

D

Explanation

Number of nuclei remained after time `t` can be written as
`N = N_(0)e^(-lambdat)`
where `N_(0)` is initial number of nucei of both the substances.
`N_(1) = N_(0)e^(-5 lambda t)` ..(i)
and `N_(2) = N_(0)e ^(-lambda t)` ..(ii)
Dividing Eq. (i) by Eq.(ii), we obtian
`(N_(1))/(N_(2)) = e^((-5 lambda+lambda)t) = e^(-4lambdat) = (1)/(e^(4lambdat))`
But, we have given
`(N_(1))/(N_(2)) = ((1)/(e))^(2) = (1)/(e^(2))`
Hence, `(1)/(e^(2)) = (1)/(e^(4lambda t))`
Comparing the powers, we get
`2 = 4 lambdat`
or `t = (2)/(4 lambda) = (1)/(2lambda)`