# How do you calculate the equivalent absorbed radiation dose from activity, type of emission, and the energy of the emission?

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?

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You've left out the acceptance of the counter which certainly isn't better than 50% and is probably rather less than that. –  dmckee Jan 23 '13 at 21:50
@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 :) –  Michael J. Gray Jan 23 '13 at 22:40
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. –  dmckee Jan 24 '13 at 0:27
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. –  Michael J. Gray Jan 24 '13 at 2:23
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. –  dmckee Jan 24 '13 at 2:29

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.