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useful information for x-ray shielding!

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LAN_403

Ruskie

Wed Feb 06 2013, 12:34AM

Registered Member #9290 Joined: Sat Jan 05 2013, 03:42PM
Location:
Posts: 60

I've been looking around here in the archives for weeks for this information and, other than finding bits and pieces, I didn't find anything as complete as what I'm about to share. A friend of mine shared this very useful information regarding calculations for proper thickness of x-ray shielding:

If you look at "type of data", you can find linear attenuation
coefficient for various x-ray energies and then use the attenuation
formula to see if shielding is sufficient.

Attenuation= exp (-ux), where u is linear attenuation coefficient and x
is thickness of material in cm.

Example: If am running an 80 KV @ 4 mA 1mm Be x-ray tube with 150mm equi-distant shielding of 3 mm thick lead on all sides, would I be safe standing directly in front of the Pb sealed aperture for an hour?

First, we'll find u from the website referenced above. When I have the data shown in the form of "HTML tables", the closest KV is 81.6564 with a u of 2.538 mm-1 (I am using mm instead of cm since shielding is in mm- this would have been a u of 25.38 cm-1 originally).

Now if we plug into the original equation:

Attenuation = exp (-2.538 * 3) = 4.93 x 10-4 or 0.000493

Now, get a rough idea of x-ray output from tube here:

Assuming these calculations are for standing directly in the output path, a tube with a 1mm Be window, 80 KV @ 4 mA output, 3 mm Pb shielding at 15 cm distance from tube, you would NOT be safe 1434 R/hr. Just quadrupling distance to 60 cm would drop exposure to 89.5 R/hr.

This has certainly cleared up my uncertainy issues regarding shielding thickness.

I looked for this specific information for quite some time, but was unable to find it. I hope others will be able to use it too. I thank a friend of mine for this info, who doesn't belong to the forum, and will remain anonymous.

Bill

Vlad

Wed Feb 06 2013, 02:08PM

Registered Member #9711 Joined: Sat Jan 19 2013, 03:27PM
Location:
Posts: 65

Thanks for sharing this, particularly the linear att coeff site! One question though: the final attenuation number, would I multiply my tube output by that number (in other words, if my tube is outputting 45 R/min, would I multiply that by the 4.93x10-4)?

The second site for the Radpro calculator is interesting as well. It's interesting to see how much more lead is generally needed to stop 60 KV x-rays over ones at 40 KV and less. Seems like there is a large jump in penetration within that range.

I tended to think of Pb, even say 1mm of it, as being the all stop for just about any x-ray intensity, but Radpro shows that is definitely not the case.

Vlad

Ruskie

Wed Feb 06 2013, 09:09PM

Registered Member #9290 Joined: Sat Jan 05 2013, 03:42PM
Location:
Posts: 60

Vlad wrote ...

Thanks for sharing this, particularly the linear att coeff site! One question though: the final attenuation number, would I multiply my tube output by that number (in other words, if my tube is outputting 45 R/min, would I multiply that by the 4.93x10-4)?
Vlad

Now that I'm not 100% sure of, but it would seem likely. If you multiplied both, you'd get 22.2 mR/min. That does seems a reasonable figure based on calculations I've done so far.

Wolfram

Thu Feb 07 2013, 12:17AM

Registered Member #33 Joined: Sat Feb 04 2006, 01:31PM
Location: Norway
Posts: 971

This method works for shielding calculations, but it overestimates the transmission by orders of magnitude by assuming that all of the x-rays are emitted at the peak energy, where in reality most of the x-rays are emitted at lower energies where the attenuation is significantly higher.

Simulations using the Tucker/Barnes model show that the attenuation is around 1.6 * 10^7, making the dose around 70ÂµSv/hr at the surface of the shielding. This is a bit above background, but with shorter runtimes and a bit of distance I wouldn't worry too much. If you're running the tube for hours, 15cm away from yourself and pointed directly towards you with some lead in between, you have larger problems than a bit of radiation. Also keep in mind that the dose figures are not whole-body figures, as it's impossible to have every point of your body 15cm away from the focal spot.

klugesmith

Thu Feb 07 2013, 01:07AM

Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716

Good chart there.
It shows why it's not trivial to apply the linear attenuation coefficients to Coolidge tube output. The x-ray source spectrum is broad and bottom-heavy; kVp on the tube is just the upper bound of the photon energy in keV.
The spectrum is even more bottom-heavy in AC operation (traditional dental heads, for example), where the anode voltage is less than the kVp value most of the time.

Another complication is hinted at by references that give two different linear or mass attenuation values. IIRC, one is just about energy absorbed, and the other includes energy scattered off-beam. If you dig into the kinds of scattering, some result in photons with reduced energy.

I think the modeling in RadProCalculator tool (perhaps including Tucker/Barnes?) takes those factors into account.
Where things get too complicated or not completely understood, I bet the program is conservative about safety, i.e. better to overestimate than to underestimate the flux after shielding.
An interesting experiment would be to see if the second and third millimeters of lead attenuate by the same factor as the first mm. I would think not, since the spectrum is progressively hardened. What did Friedrich Nietzsche say about what doesn't kill us?

Wolfram

Thu Feb 07 2013, 02:04AM

Registered Member #33 Joined: Sat Feb 04 2006, 01:31PM
Location: Norway
Posts: 971

RadPro uses only the peak energy, not the energy spectrum, so it's going to give a conservative and unrealistic estimate as well.

They also claim that x-ray tube output is hard to predict, this is incorrect as well. Bremsstrahlung production is a simple and well understood process, and as long as the major variables (anode angle, anode material, attenuation from materials in the beam path etc.) are taken into consideration, good estimates can be made.

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