3
$\begingroup$

I'm attempting to determine the half-lives of two silver isotopes (Ag-108 and Ag-110). A lab manual for a similar experiment may be found here http://web.vu.lt/ff/a.poskus/files/2013/06/NP_No08.pdf

I have gathered my data, and I have counts/5 seconds on the y-axis and time elapsed on the x-axis. I have already subtracted background radiation. However, in certain cases, this has given me negative values for counts/time. I wish to fit an exponential curve to my data, but am unable to do so. I have read that

"After subtracting the background, some points in the region of large times may become negative, because nuclear decay is a random process, and because the background has been evaluated with an error. This is normal, and such points must not be excluded from analysis."

How can I take into account the errors in order to be able to plot the exponential? My guess is that the principle is something akin to weighted least squares, in which certain measurements carry more weight because of lesser uncertainty. However, my lab manual suggests that the standard deviations for this lab are calculated under the assumption of Poisson distribution. I don't know how one can take the square root of negative values. Any guidance would be greatly appreciated. Thank you!

$\endgroup$
1
$\begingroup$

Poisson errors apply to total counts, rather than to background-subtracted counts. So if your constant background is 20 counts per interval, and you have an interval that only had 16 counts in it, the associated uncertainty is $\sqrt{16}=4$, and that interval enters your fit as –4±4 counts.

$\endgroup$
  • $\begingroup$ Thank you rob. Now I'm wondering how to take error into account when plotting a semi-log graph of my 2 decay rates, but I'll work on it. Thank you again! $\endgroup$ – user37257 Apr 27 '14 at 16:31
  • $\begingroup$ You plot the measured value on the graph with a cross showing the +/- error . See omatrix.com/spmanual/serrorbars.htm $\endgroup$ – anna v Jun 4 '14 at 3:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.