Desperate: Residual activity of daughter in decay

[tramar]

Homework Statement

Consider harvesting a radioactive daughter product from a radioactive parent at regular intervals of time, \theta. If n samples of accumulated daughter product are removed at time \theta, 2\theta, ..., n\theta, show that the sum of their residual activities at time n\theta is just that which would have accumulated in the parent material had no separations been made.

Homework Equations

A_{1}(t)=A_{10}e^{-\lambda_{1}t}
A_{2}(t)=\frac{\lambda_{1A}}{\lambda_1}\frac{\lambda_{2}}{(\lambda_{2}-\lambda_{1})}A_{1}(t)(1-e^{-(\lambda_{2}-\lambda_{1})(t-t_{1})})

The Attempt at a Solution

I've tried to take the equation for A_{2}(t) given above and perform a sum from \theta all the way to n\theta for all of the residual activities. I tried to reduce that expression down so that it gave me A_{1}(t) but it seems impossible for the following reasons:

1) You end up left with the factor \frac{\lambda_{1A}}{\lambda_1}\frac{\lambda_{2}}{(\lambda_{2}-\lambda_{1})} on the left side and it is not on the right side.
2) The equation for A_{1}(t) has no \lambda_{2}, so I don't see how it can possibly equal any residual activity of the daughter.

This leads me to believe that I've interpreted the problem statement incorrectly...
If anyone can give me some insight, it would be greatly appreciated.

 

[willem2]

The quantity of the daughter at time t is A_1(0) \frac {\lambda_1 } { \lambda_2 - \lambda_1} (e^{- \lambda_1 t} - e^{- \lambda_2 t})

I don't see how you get that expression for A_2(t). What is the amount of the parent left at the time of removing batch k?. What is the amount of daughter in batch k at the time of its removal? The amount of daughter left in batch k after time n \theta

It seems all a rather pointless exercise, since it doesn't matter at all for radioactive decay if you separate the atoms in batches.