Estimating Radiation Exposure from Multiple Sources

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SUMMARY

The discussion focuses on estimating total extra radiation exposure from three sources: Ba-133, Cs-137, and Co-60. The user applies the equation \(\frac{dD}{dt}=\frac{dD_{0}}{dt}e^{-cx}\) to calculate exposure based on dose rates and absorption coefficients. The calculated doses for each source over a 10-minute exposure are 0.0462 mR for Ba-133, 0.0195 mR for Co-60, and 0.0178 mR for Cs-137, resulting in a total extra exposure of 0.0835 mR. The reasoning and calculations presented are deemed correct by other forum participants.

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Von Neumann
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Question:

I'm looking to estimate my total extra exposure (dose above background) by taking into account account roughly how long I was near each source, the strength of the source, and its approximate distance from me. I believe it's useful to use the following equation,

\frac{dD}{dt}=\frac{dD_{0}}{dt}e^{-cx}

where \frac{dD_{0}}{dt} is the dose rate at x=0 and c is the absorption coefficient.

The dose rates measured at x=0 for the 3 sources, as well as the calculated absorption coefficients are

Ba-133: 277 μR/h , c=0.303/in.
Cs-137: 107 μR/h , c=0.375/in.
Co-60: 117 μR/h , c=0.360/in.

*note that the background radiation has been subtracted from the dose rates shown.

So if I take the time t spend near each source as 10 minutes, and the distance from each source as x=0 in. (as I was holding the sources as we varied the distance in the first part of the experiment) I can approximate the exposure as


\frac{dD_{0}}{dt}\cdott

and then sum these for from each source. So therefore I get,

Ba-133: 277 μR/h * (10 min) * (1 h/ 60 min) = 0.0462 mR
Co-60: 107 μR/h * (10 min) * (1 h/ 60 min) = 0.0195 mR
Cs-137: 117 μR/h * (10 min) * (1 h/ 60 min) = .0178 mR

Thus as my total extra exposure I get,

0.0462 mR + 0.0195 mR + 0.0178 mR = 0.0835 mR

Is this reasoning correct? Thank you in advance for any input.
 
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Looks reasonable.
 

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