# Desperate: Residual activity of daughter in decay

1. Oct 5, 2009

### tramar

1. The problem statement, all variables and given/known data
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.

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

3. 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.

2. Oct 6, 2009

### 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.