Question

A solution contains a mixture of two isotopes A (half-life=10days) and B (half-life=5days). Total activity of the mixture is `10^10` disintegration per second at time `t=0`. The activity reduces to 20% in 20 days. Find (a) the initial activities of A and B, (b) the ratio of initial number of their nuclei.

Answer 1

Correct Answer - A::B::C::D
(a) Let `R_(A_0)` and `R_(B_0)` be the initial activities of A and B. Then,
`R_(A_0)+R_(B_0)=10^10dps` …(i)
Activity of A after time `t=20` days (two half-lives of A) is
`R_A=(1/2)^2R_(A_0)=0.25R_(A_0)`
Similarly, activity of B after `t=20` days (four half-lives of B) is
`R_B=(1/2)^4R_(B_0)=0.0625R_(B_0)`
Now, it is given that `R_A+R_B=20%` of `10^10`
or `0.25R_(A_0)+0.0625R_(B_0)=0.2xx10^10dps` ...(ii)
Solving Eqs. (i) and (ii), we get
`R_(A_0)=0.73xx10^10dps`
and `R_(B_0)=0.27xx10^10dps`
(b) `(R_(A_0))/(R_(B_0))=(lambda_AN_(A_0))/(lambda_BN_(B_0))=((t_(1//2))_B)/((t_(1//2))_A)*(N_(A_0))/(N_(B_0))`
`:.` `(N_(A_0))/(N_(B_0))=((R_(A_0))/(R_(B_0)))((t_(1//2))_A)/(t_(1//2))_B=((0.73)/(0.27))((10)/(5))`
`=5.4`