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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.
Solution
Let NA be the concentration of A after tA time and NB be the concentration of B after tB time.
So, NA = N0e−λAtA
NB = 4N0e−λBtB (as N0B = 4N0A)
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 NA = NB
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
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