2
$\begingroup$

I'm working on a project, and I understand that half life of a source is typically calculated with time intervals (eg 5 sec intervals for 20 minutes) and then using that data to find the decay constant using exponential decay and then calculating the half life using ln2/decay constant, but would it be possible to calculate the half life of a source using simply one long measurement? For example if you measured a source for 20 minutes and summed up the counts of the photopeak after 20 minutes, is there anyway to get the half life from just those 2 values?

I initially thought that it seems feasible that it would be possible since it's just something like Total counts after x minutes so there has to be some relationship but if I think about it I'm actually not sure since I don't have an initial counts to use, so I can't use the exponential decay formula (at least not in the way I would expect?) and I don't have any other tools that I can think of to get the half life aside from that.

$\endgroup$
  • $\begingroup$ Yes, it is possible, see my answer. $\endgroup$ – Gert Mar 30 at 19:22
  • $\begingroup$ @Gert I saw your answer and I wanted to wait for more people to just get some certainty, but I see you deleted it, I understnad why it's not completely applicable but I'd still like to see it again if possible because I did like the derivation and I'd like to mention it in my report as an assumption that I could make! thanks! $\endgroup$ – Choco-Elliot Wyvern Mar 31 at 11:22
  • $\begingroup$ I've undeleted it. Ta. $\endgroup$ – Gert Mar 31 at 12:19
  • $\begingroup$ You can delete it again now if you want to! I got it thank you! $\endgroup$ – Choco-Elliot Wyvern Mar 31 at 14:44
0
$\begingroup$

You need at least two data points to fix the two free parameters that characterize an exponential function. So the answer is no, you would need at least one more data point to say something about the properties of that decay.

$\endgroup$

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.