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How can I describe a distribution of decay times for an individual unstable nuclei 
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#1
Jul1510, 05:20 AM

P: 57

We all know that for an ensemble of particles we can use the exponential decay law. However, the problem I am working on requires to look at nuclei from an atomic perspective. Wikipedia says that this is a stochastic process on the atomic level. I do not have much knowledge of stochastic calculus so I would need help here....
Suppose we have a bunch of nuclei which are decaying. I wish to know a distribution function for the time in which a nuclei from the bunch would decay. How do I do this? 


#2
Jul1510, 05:36 AM

P: 477

Hi there,
Just a few words about your post. I dont' think that wikipedia says that the decay of nuclei has anything to do with the level of the atoms. A nuclear decay happens in the nucleus of an atom, which there you have it right, it has to do with the unstability of the nucleons. For events to happen, we know that a bunch of unstable nuclei will decay at a certain rate. What we don't know is when each and every one of them will decay: the stochastic process. Therefore, if you are looking for a way to know which one will decay when, well we all are, and specially the nuclear business would love to know. But what the decay time tells you about each specific nucleus is the probability of decay in the next time interval, that's it. Cheers 


#3
Jul1510, 06:04 AM

P: 57

By "viewing it as an atomic process" i simply mean, concentrating on a single decaying nuclei. I understand that it cannot be determined when a nuclei decays, but that might not mean some sort of stochastic modeling cannot be done...
and that is what I'm looking for. Suppose some nuclei decays at time t1, another at t2, then is there a to be a distribution for t ... ? 


#4
Jul1510, 06:11 AM

P: 477

How can I describe a distribution of decay times for an individual unstable nuclei
Hi there,
Stochastic means that the process is random. Therefore, when looking at just one nucleus, you can only use probabilistic methods to evaluate its decay. So the stochastic model that you are looking for would be base on the probability of decay in the next so long. And this is exactly what the decay constanct provides. Cheers 


#5
Jul1510, 11:35 AM

P: 4,663

There are actually three basic types of statistical time distributions; binomial, poisson, and gaussian. These are reviewed in
http://www.asp.ucar.edu/colloquium/1...pdf/chapt3.pdf This article is very brief, but does cover the essentials. Gaussian time distribution should only be used if the number of atoms N that are radioactive, and the number of atoms that decay D in time interval t are both large numbers. If N is very large (over say 10,000), but if D is small (under ~10), then poisson statistics should be used. This is closely related to time interval statistics. If both N and D are small (under 100 and 10 respectively), then binomial statistics should be used. The best thorough discussion of counting statistics I have seen is in the textbook "The Atomic Nucleus" by Evans, published about 1955. This edition is available via Amazon.com for ~$20.00. Bob S 


#6
Dec1410, 11:29 AM

P: 57

Hey, just for the record I was able to do it. Since the exponential decay is a truly random memoryless process, we have to start with a method which mimics this. I was able to "solve" this problem (of recreating the exponential distribution from the individual atom's "perspective") by using two different monte carlo approaches.
One was what we can call a "time discretization" approach. In this, you take discrete time steps, and associate each unstable atom in your ensemble with a parameter [tex] u [/tex]which has a uniform random distribution and compare it to [tex]\lambda [/tex] (decay constant) for each atom, then increment the number of times [tex] u>\lambda[/tex] to get your distribution. The second is a more mathematical and less programmatic one. Its called the "inverse transform method". I have the programs written down for both of these in ROOT (basically C++) so if anyone needs them just send me a PM. 


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