# Nuclear decay activity - value for money?

1. Nov 20, 2012

### Silversonic

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

Radioactive nuclei A are produced at a rate R per second in a nuclear reactor. They decay with probability λ per second.

(Qu 1-2 involving deriving the rate of change and number particles at any given time, I've done this).

3) Show the activity tends to R, a constant, as t is large. (I've done this!)

4) If the accelerator is costly to run, how long would you irradiate to get the best value for money. i.e. maximum activity per buck (I'm stuck/confused by this).

2. Relevant equations

$N(t) = \frac {R}{λ}(1-e^{-λt})$

N'(t) = R - λN

Activity = λN = $R(1-e^{-λt})$

3. The attempt at a solution

I'm confused by what 4) is actually asking. What would actually be the purpose of this whole arrangement then? Radioactive particles are produced by an accelerator which is costly to run. That's fair enough. But what are the radioactive particles wanted for? Do we want to use the nuclei before they decay? Or do we want to use the decay results of the nuclei?

The actual reason behind this confuses me, I don't understand exactly the aim behind the arrangement and thus how we would get "value for money". It depends what we want this to be used for.

When t gets large, N(t) stays relatively constant and no more "new" nuclei are added to the sum total. This is because the activity reaches the rate at which nuclei enter the system. I have the answer and it tells me that after time t = 1/λ activity reaches 63% of equilibrium, t = 2/λ it reaches 85%, then 95% and so on. So "you don't win much by irradiating longer than t = 2-3/λ)". What is there to win?

If we want to increas the number of nuclei, then yes, it wouldn't make much sense to keep it on too long because the number of nuclei won't increase. But if we turn the accelerator off then R goes to zero and the number of nuclei drops to zero due to normal decay. Which would surely be counter-productive if what we "want" is a large number of nuclei? Or do we want the activity to rise because we want the decay properties for something? Even if the activity doesn't change much for large t, turning the accelerator off would mean the activity eventually drops to zero anyway which would also surely be counter productive.

This may/may not be introductory. It was basically a revision question for radioactive decay laws which is pretty much based on A level ("highschool") syllabuses.

2. Nov 21, 2012

### haruspex

I would interpret the question thus:
The reactor will be run for some time T then switched off. The cost of that is cT, some constant c. The reaction products will then be sold for a price, pλN, proportional to the activity remaining at that time. Find the T that maximises profit.