Timeline for How do you calculate the equivalent absorbed radiation dose from activity, type of emission, and the energy of the emission?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Jan 24, 2013 at 17:41 | vote | accept | Michael J. Gray | ||
| Jan 24, 2013 at 3:10 | history | edited | dmckee --- ex-moderator kitten |
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| Jan 24, 2013 at 3:04 | answer | added | dmckee --- ex-moderator kitten | timeline score: 0 | |
| Jan 24, 2013 at 2:53 | comment | added | Michael J. Gray | let us continue this discussion in chat | |
| Jan 24, 2013 at 2:53 | comment | added | Michael J. Gray | So, if for example the object is evenly radiating and causes 1000 events per minute and the surface area of the object is 100cm while the detector's surface area 10cm, we have an acceptance of 10%, an efficiency of 98.6%, an adjusted emission of 1000/0.886 becquerels? If that's correct, then great. I would imagine that counting for statistics would be fine if done over a period of 2 hours or something, but again I could be wrong. However, this still leaves more of the problem. I'm imagining that a person isn't exposed to 100% of the radiated energy from an object, even without shielding. | |
| Jan 24, 2013 at 2:29 | comment | added | dmckee --- ex-moderator kitten | 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. | |
| Jan 24, 2013 at 2:23 | comment | added | Michael J. Gray | 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. | |
| Jan 24, 2013 at 0:27 | comment | added | dmckee --- ex-moderator kitten | 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. | |
| Jan 23, 2013 at 22:40 | comment | added | Michael J. Gray | @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 :) | |
| Jan 23, 2013 at 21:58 | review | First posts | |||
| Jan 24, 2013 at 0:29 | |||||
| Jan 23, 2013 at 21:50 | comment | added | dmckee --- ex-moderator kitten | You've left out the acceptance of the counter which certainly isn't better than 50% and is probably rather less than that. | |
| Jan 23, 2013 at 21:40 | history | asked | Michael J. Gray | CC BY-SA 3.0 |