[Poisson Stats] Error on half-life for radioactive decay

In summary, the speaker has conducted a radioactive decay experiment and obtained two half-lives. They are now trying to determine the error associated with the half-life and have been advised to use Poisson statistics. They have researched and found that the standard deviation for a Poisson distribution is the square root of the mean. However, they are unsure if they should take the square root of the individual half-lives to get the error. They have provided some websites they have used for reference.
  • #1
henrym_
1
0
Hi there, not sure whether this is in the right section but:

I've made two runs of a radioactive decay experiment where I've got a log(N) vs. time plots. From this I've got the decay constants and hence the half-life. I've averaged these two half-lives ( = 160 secs) and now I'm trying to work out the error associated with the half-life.

Usually I'd take the error on the fits of the two graphs and then use the error propagation equations to get the error ( = 2 secs) but I've been advised to use Poisson statistics as the error I have is likely to be an underestimate (which I agree with).

I've done some research and I've read that the standard deviation (i.e. the error) for a Poisson distribution is the square root of the mean. Does that mean I square root the two individual half-lives to get the error? (Giving me an error of approx. 16 secs)

If not, can anybody lend me help? How do I work out the final error on the half-life using Poisson statistics?

Some of the sites I've used:
https://www.colorado.edu/physics/phys2170/phys2170_fa06/downloads/poisson.pdf
https://www.phys.ufl.edu/courses/phy4803L/group_I/gamma_spec/poisson.pdf
https://ned.ipac.caltech.edu/level5/Leo/Stats2_2.html
https://www.colorado.edu/physics/phys2150/phys2150_sp14/phys2150_lec6.pdf

Thanks :)
 
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  • #2
Taking the square root of the half-life is not the right thing to do. You should consider in your plot of log(N) vs T that each value of N has an error of size sqrt(N), so the relative error is sqrt(N) / N = 1/sqrt(N). Then, when you do the least-squares fit of your plot of log(N) vs T you need to include the errors in the values of N. There are many descriptions on line of how to estimate the error in your slope when the points themselves have errors. At least, this is how I would approach it.
 

1. What is the Poisson distribution used for in radioactive decay?

The Poisson distribution is used to model the probability of a certain number of radioactive particles decaying within a given time interval. This is because radioactive decay follows a random process, making it suitable for modeling using a Poisson distribution.

2. How is the half-life of a radioactive substance calculated using Poisson statistics?

The half-life of a radioactive substance can be calculated using the Poisson distribution by finding the mean number of decays within a given time interval and using the formula t1/2 = ln(2)/λ, where λ is the decay constant. This is because the Poisson distribution can be used to model the probability of a certain number of decays occurring within a time interval.

3. How does the error on the half-life for radioactive decay change with increasing sample size?

The error on the half-life for radioactive decay decreases with increasing sample size. This is because the larger the sample size, the more accurate the estimate of the mean number of decays will be, resulting in a smaller error on the half-life calculation.

4. Can the Poisson distribution be used to predict the exact number of decays in a given time interval?

No, the Poisson distribution can only provide a probability distribution of the number of decays within a given time interval. It cannot predict the exact number of decays, as radioactive decay is a random process.

5. What are the assumptions made when using Poisson statistics for radioactive decay?

The assumptions made when using Poisson statistics for radioactive decay include a constant decay rate, a large number of particles decaying, and independence between decays. Additionally, the time interval must be small enough that only one decay can occur at a time. Violation of these assumptions can lead to inaccurate results.

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