Poisson distribution (radioactive decay)

  • Thread starter steamyoshi
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  • #1

Homework Statement


I am a freshman in physics, just done a lab about radioactive decay.
I've measured the # of beta particles per second 400 times and got the frequency of each number K using Excel.
I'm supposed to take the data and fit it to a puason distribution in MATlab.
The data points themselves seem to be on a nice curve.
The problem I'm having is that the equation for the probabilty doesn't seem to curve at all.

Homework Equations


Puason distribution in radioactive decay: the chance that K beta particles will be detected in 1 second is
P(k) = e^-λ * λ^K/K!


The Attempt at a Solution


I've made an excel chart which shows λ^K/K! for many different K values, and then tweaked the lambda value.
For all the values of λ I've tried, λ^K/K! always increases for increasing values of K, meaning the graph never curves back down. What am I doing wrong?
 
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Answers and Replies

  • #2
Simon Bridge
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Puason distribution in radioactive decay: the chance that K beta particles will be detected in 1 second is
P(k) = e^-λ * λ^K/K!
... that would be poisson distribution, like the fish.
Careful to keep the same variables through your expressions.

$$P(k)=\frac{\lambda^k e^{-k}}{k!}$$

I've made an excel chart which shows λ^K/K! for many different K values, and then tweaked the lambda value.
For all the values of λ I've tried, λ^K/K! always increases for increasing values of K, meaning the graph never curves back down. What am I doing wrong?
##\lambda^k/k!## does increase for positive k, if ##\lambda > 1##but the poisson distribution function has a negative exponential in it which makes it converge for large k.
 
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  • #3
I'm sorry, I don't understand why e is ^-k, all the formulae I've found have e^-λ, which is a constant
thanks for the reply
 
  • #4
Simon Bridge
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That's because I'm an idiot... I should know better than to answer questions at 2am.
concentrating:

When I plot P vs k, I get a decreasing exponential for small values of lambda, and an approximate gaussian for large values of lambda.

You probably have a mistake in your code.
Check - sounds like a misplaced minus sign.

Time for bed.
 
  • #5
Thanks, it WAS a code problem.
 
  • #6
Simon Bridge
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No worries.
And I got to demonstrate not to be afraid of making dumb mistakes too :)
 

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