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

Options

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

Answer

D

Explanation

(d) number of nuclei remained after time t can be written as
`N=N_(o)e^(-lamdat)`
where ,`N_(0)` is initial number of nuclei of both the substances
`N_(1)=N_(0)e^(-5lamdat) `
`and N_(2)= N_(0)e^(-lamdat) `
On dividing Eq (i) by eq (ii) we get
`(N_(1))/(N_(2))=e^((-5lamda +lamda )it =e^(-4lamda t)=(1)/(e^(4lamda t )`
But , we have given
`(N_(1))/(N_(2))=((1)/(e))^(2)=(1)/(e^(2))`
comparing the power , we get
`2=4lamda t implies t=(2)/(4lamda )=(1)/(2lamda)`