Question

In a given sample, two radioisotopes, A and B, are initially present in the ration of 1 : 4. The half lives of A and B are respectively 100 years and 50 years. Find the time after which the amounts of A and B become equal.

Answer 1

Let N_A be the concentration of A after t_A time and N_B be the concentration of B after t_B time.

So, N_A = N0e^−λ_At_A

N_B = 4N₀e^−λ_B^t_B (as N_0B = 4N_0A)

Now half-life of A is 100 years and B is 50 years.

`So lambda_A = (ln^2)/100 and lambda_B = (ln^2)/50`

Dividing we get

`(lambda_A)/(lambda_B)= 1/2 or lambda_B =2lambda_A`

Now let after t years N_A = N_B

So`(N_A)/(N_B) = e^(-lambdaA')/(4e^(lambdaB')`

`N_A =N_B`

`4e^(-lambdaB') =e^(-lambdaA') `

`4= e^-(lambda_A -lambda_B)`

`ln4 = -(+lambda_A - 2lambda_A)t          (because lambda_B = 2lambda_A)`

`ln4 = lambda_At`

`t = (ln4)/(ln2) xx 100         (because lambda_A = (ln2)/100)`

   = 200 years