I have a sample of U-238 of which my Geiger counter detects beta activity at 700 events per second. Based on the counter's efficiency of 98.6% for U-238, the activity would be about 710 becquerels, I think. Here I am going off of my understanding based on explanations from many different people.

From here I have figured I would just convert the beta energy in MeV to joules and then multiply it by the corrected activity (710Bq) to obtain the absolute absorbed energy per second. This is great and all and when considering the mass of the object absorbing the energy, we can get the number of Gy absorbed by an object.

What I want is the equivalent dose in sieverts per hour. I understand there are many weight factors for different types of tissues and all that. But, there is a lot of misinformation floating around on the Internet about the calculations.

Can someone point me in the right direction on how to go about calculating this?

  • $\begingroup$ You've left out the acceptance of the counter which certainly isn't better than 50% and is probably rather less than that. $\endgroup$ Jan 23, 2013 at 21:50
  • $\begingroup$ @dmckee When you say "acceptance" are you referring to the sensitivity or the overall efficiency? The main specification sheet give an efficiency of 25% but I think that is with respect to Cs-137. If you could clarify or inform me on what to look at, I may be able to come up with that number. Thanks :) $\endgroup$ Jan 23, 2013 at 22:40
  • $\begingroup$ Neither. Acceptance is almost always the geometric coverage of the detector. Even when you put the Geiger tuber near the source, some fraction of the decays miss the detector. The fraction that of raw events that trigger a count is a convolution of the acceptance, the quantum efficiency and the live time. At the rates you describe live time is close enough to 1 as makes no difference and for a Geiger tube the efficiency barely depends on geometry so overall rate is $r_\text{real} \approx r_\text{det} / f_e / f_A$ where the fractional acceptance can be found with a ruler. $\endgroup$ Jan 24, 2013 at 0:27
  • $\begingroup$ So in terms that make some good sense to me. We have efficiency, which is the number of events that pass through the detector which actually get reported. Then there's acceptance, which is the percentage of coverage the detector has on the point source that is radiating in all directions. If I am understanding this correctly then it definitely makes sense to say that the typical acceptance is far less than 25%. So this calculation is far more complicated than I had actually thought. Am I on the right track here? This seems more complicated than I had imagined now. $\endgroup$ Jan 24, 2013 at 2:23
  • $\begingroup$ Bang on. The acceptance factor is easier to calculate for a small solid angle (you can just measure the radius and take $\text{area of detector}/4 \pi R$), but of course, in that case the acceptance is smaller and you have to count longer to get enough statistics. $\endgroup$ Jan 24, 2013 at 2:29

1 Answer 1


For your dose calculation you will note that photons and electrons have a weighting factor of 1 and then go the the Particle Data Book chapter on "Passage of particles through matter" and read the sections on electrons and photons.

Now the table of the isotopes tells me that U-238 is mostly a alpha emitter, but that's a non-starter for both your Geiger tube and a person. There doesn't seem to be a single beta decay listed, and the double-beta branching ratio is or order $10^{-10}$. Ouch. So we look at the alpha-linked gamma and x-ray lines. Only two of those exceed 1% branching ratio and they are both below 20 KeV. Double ouch. None-the-less those are presumably most of what you are detecting. The betas would be much more energetic, but they'll be one for every hundred million of the gammas.

In the PDB you'll see that at those energies the photons dump essentially all there energy in any person close enough that the intervening air doesn't shield them.

  • $\begingroup$ I suppose the off chance is that these may not be beta particles that I am picking up but rather gamma rays and x rays. My detector (SE International Inspector Xtreme) is capable of picking up x rays. Anyway, the bottom line that I am getting here is that it's very difficult to calculate the actual dose received by a person. The reason I suspected (perhaps naively) that they were beta and not alpha or gamma is because a piece of wood or steel shields the emissions down to the normal background levels but about three feet of air does not. Gamma shouldn't be stopped that easily but beta should. $\endgroup$ Jan 24, 2013 at 7:05
  • $\begingroup$ But anyway, my question now with more insight on what needs to be considered is: What is the process for calculating the actual dosage (in sieverts) received by an individual exposed to this particular source? I now know about acceptance and that it's not beta particles I am detecting. I am attempting to determine if this particular sample is relatively safe to handle and gain knowledge so that I can calculate this risk for other known sources in the future. $\endgroup$ Jan 24, 2013 at 7:09
  • $\begingroup$ The handling risk for alpha emitters is always getting the stuff inside. Ingesting or inhaling even a small part of the source could making you very sick or even kill you. Take a look at the weighting factor for them and the energy of those lines; worse all that energy will be deposited internally. Beyond that I wouldn't carry the thing in my pocket or keep it near my person regularly, but as you've noticed almost any container will shield it. The mantra the DOE teaches at labs is ALARA (As Low As Reasonable Achievable). Good idea to have a dosimeter and keep records. $\endgroup$ Jan 24, 2013 at 14:47
  • $\begingroup$ I have it in a wooden box next to my other radioactive things. I did an 8 hour test just counting the number of events. 1 foot from the box in that period I received about 15,000 events and then my control which was placed in a normal living space received about 13,500 events. I would say that such a difference is expected. Anyway, later this evening I will attempt to accurately measure the dosing with all of the information you have provided. A fresh look at this information today has made it more clear. Thank you again :) $\endgroup$ Jan 24, 2013 at 17:41

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