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#1 Posted : 12 October 2008 12:05:00(UTC)
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Posted By Brian M. Jones Is there a relatively staright forward way to calculate the safe distance from a radio-graphic source. There is a proposal to bring a gamma-ray source to site that is rated at 9.58 CI and the NDT company are quoting, from their procedure, a safety radius of 35 metres around the test piece. I have no reason to doubt the procedure as they are a reputable company. The barrier will be set so that the dose rate, at the barrier, will not exceed 7.5 µSv/Hr. What I am trying to establish if there is a theoretical calculation that one can carry out to determine the safe distance from a source of a known activity to achieve such a dose rate. I am working over seas and there is very little in the way of regulation and certainly no guidance on such issues. Thank you for you time.
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#2 Posted : 12 October 2008 13:31:00(UTC)
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Posted By Alan Nicholls Brian By no means an expert in this field but if you, Look at the Atomic Energy Licencing Board Site.Exposure = Intensity(Dose rate.I)x Time(T) (Calculations using Inverse Square Law). If the company is registered, and comply with current regulations and practices it should be ok. You might ask how they calculated the safe distance, and what they do to ensure it is not exceeded. Measurements should be taken at the boundary. Just as a pointer. Good luck Alan
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#3 Posted : 12 October 2008 14:36:00(UTC)
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Posted By William I think it was always if you double the distance then you reduced the intensity by half, there are also other factors such is their any sheilding which would reduce the intensity.
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#4 Posted : 13 October 2008 07:23:00(UTC)
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Posted By Jane Blunt You have not given enough information. First you find out what the source is, because the output from the source is dependent on the radionuclide. For example, the output from a 400 GBq source of Iridium 192 is 50 mS/h at 1m, whereas the output from a 400GBq Cobalt 60 source is 138 mSv/h. Next you need to convert from Ci to Bq; 1 Ci is 3.7 x 10^10 Bq. Finally you need to calculate the reduction due to distance. It is an inverse square law, so at 35m, the dose is 1/35 x 1/35 times what it is at 1m. Jane
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#5 Posted : 13 October 2008 13:33:00(UTC)
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Posted By Robert K Lewis Jane Now I remember why I never wanted to cover this area:-) Bob
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