Calculating Roentgens from an X-Ray source

[neanderthalphysics]

Hey,

I want to try to estimate the Roentgens/sec from an X-Ray tube. Here are my calculations, which I talk through so that I am not just throwing equations at you. I would be grateful if someone could comment if they are about right.

Stage 1: Estimation of emitted X-Ray strength
Assumed:
1. X-Ray tube input power: 1000W
2. X-Ray conversion efficiency: 1%
3. Total X-Ray energy: ~10W.

If the X-Ray source has an electron gun voltage of a few hundred keV, let's assume the Bremsstrahlung peak is at 100keV. Furthermore, for simplicity, let's use the 100keV value as the "characteristic" X-Ray wavelength. At this wavelength, our attenuation coefficient is 0.154 cm2/g (NIST tables for dry air)

Stage 2: Estimation of percent of X-Rays absorbed by 1kg of air
1 Roentgen is defined as 2.58e-4 C/kg

Dry air under these conditions has a density of 1.225 kg/m3. Therefore a 1kg cube of dry air has a length of 93.5cm on all sides.

The mass thickness of air = density of air x path length the radiation has to travel.
Therefore the mass thickness = 0.001225 x 93.5 = 0.1145 g/cc x cm

The ratio of the exiting X-Ray flux to the entering X-Ray flux, I/I0 = exp(-mass thickness x absorbance) = exp(-0.114 x 0.154) = 0.9825

Absorbed X-ray power = (1 - I/I0) x 10W = 0.175W

Stage 3: Converting the absorbed X-Rays into charges in the air
Average ionization energy of air = 34 eV = 5.45e-18 J

Assume that all the X-ray's energy is used to ionize the air at an energy of 34eV. I know that X-Rays are more energetic but because the proportion of ions to air molecules is very small (1x10-9) I am assuming that entropy prevails and all of the air molecules just absorb the first ionization energy.

Number of ion pairs created per second = 0.175W/5.45e-18 = 3.21e16 pairs
Fundamental charge = 1.6e-19 C
Charge produced per sec = 3.21e16 x 1.6e-19 x 2 (each pair of charges produced counts as 2x the fundamental charge) = 0.01 C/sec

Roentgens of source = 0.01/2.58e-4 = 40 Roentgens

What do you think about these calculations? Wide off the mark or about right? Could you improve on it?

 

[gleem]

X-ray exposure rate specified in Roentgens.sec can be only determined reasonable accurately by measurement.

If the X-Ray source has an electron gun voltage of a few hundred keV, let's assume the Bremsstrahlung peak is at 100keV. Furthermore, for simplicity, let's use the 100keV value as the "characteristic" X-Ray wavelength.

Bremstrahlung has no peak. The characteristic x-ray are dependent on the anode material typically Tungsten which have characteristic x-rays between about 60 and 70 keV.

You did not indicate and therefore take into consideration the waveform of the power supply single phase three phase or constant potential. You assumed constant potential. The 1% efficiency is a ball park figure.

You cannot use a single photon energy to estimate exposure since the range of energies of the bremstrahlung are continuously monotonically decreasing from 0 to the KVp of the unit. These x-rays are further absorbed by the tube window as well as added filtration typically used to eliminate the low energy x-rays This absorption is highly energy dependent.

Roentgens of source = 0.01/2.58e-4 = 40 Roentgens

Sources of X-rays are not specified in Roentgens but in Roentgen/time unit at a specified distance usually about 1 meter.

I did not see where you took the inverse square law into consideration. You cannot just convert x-ray power produced in the tube into Roentgen/sec emitted, only a certain fraction (about 2%) of the energy is emitted in the useful direction which you have not addressed.

What do you think about these calculations? Wide off the mark or about right? Could you improve on it?

Way off, you do not fully understand the situation. The nature of the situation is too complex to do realistic calculations in this manner. Such a calculation would be best handled by Monte Carlo techniques.

 

[neanderthalphysics]

Hi Gleem, thank you for your feedback.

X-ray exposure rate specified in Roentgens.sec can be only determined reasonable accurately by measurement.

OK thanks. But I still would like to do some back-of-the-envelope estimations that would hopefully be within one order of magnitude of the true answer.

Bremstrahlung has no peak. The characteristic x-ray are dependent on the anode material typically Tungsten which have characteristic x-rays between about 60 and 70 keV.

By peak I mean this peak in the graph. Excuse me if there is a specific term for it.

Where this peak is, and where the maximum photon energy is, is dependent on the setup of the electron gun, in particular the driving voltage.

You did not indicate and therefore take into consideration the waveform of the power supply single phase three phase or constant potential. You assumed constant potential. The 1% efficiency is a ball park figure.

A time-varying potential across the electrodes can be seen as a summation of constant potentials at a given time. I am trying to keep the basic by-hand calculations simple but nevertheless amenable to Monte Carlo simulations later should I need to do so.

You cannot use a single photon energy to estimate exposure since the range of energies of the bremstrahlung are continuously monotonically decreasing from 0 to the KVp of the unit. These x-rays are further absorbed by the tube window as well as added filtration typically used to eliminate the low energy x-rays This absorption is highly energy dependent.

Understood. But as it is somewhat tedious to do by hand a summation of multiple photon energies, I am just taking one representative value. Later on in a computer algorithm I can put this in a do loop and iterate to an arbitrary precision.

Filtration, etc of the outgoing photons is something which can again be added later to an algorithm. For now I want to capture the essence of the calculations.

Sources of X-rays are not specified in Roentgens but in Roentgen/time unit at a specified distance usually about 1 meter.

What about say dosimeter pens where the workers clearly can not maintain over the course of a day, X-ray sources 1 meter away from them?

I did not see where you took the inverse square law into consideration. You cannot just convert x-ray power produced in the tube into Roentgen/sec emitted, only a certain fraction (about 2%) of the energy is emitted in the useful direction which you have not addressed.

Good point re: inverse square law. Three points:
1. The inverse square law can be included in a computer algorithm later.
2. Furthermore I note that dosimeters do not account for the inverse square law either. It can not even be calibrated out, because the circumstances of exposure are so variable.
3. Is the emission from the tungsten target isotropic? I don't think it is, in which case the inverse square law, even if it does apply, would have some anisotropy to the intensity as well.

Way off, you do not fully understand the situation. The nature of the situation is too complex to do realistic calculations in this manner. Such a calculation would be best handled by Monte Carlo techniques.

I will be happy if the calculations are OK for a quick back-of-the envelope calculation. Issues like the continuous distribution of X-Ray energies, inverse square law, etc. can be handled by Monte Carlo iteration on a computer later.

P.S. Mods would this post be more appropriate in the Nuclear Physics subforum?

 

[anorlunda]

P.S. Mods would this post be more appropriate in the Nuclear Physics subforum?

No. Nuclear engineering is appropriate.

 

[gleem]

You want only to be within an order of magnitude? What is the point of such a calculation?

But I'll tell you what, I'll give you a number for the output from a typical medical x-ray device operating at 100 kVp that you can aim for. The exposure rate at 1 meter is about 12 milli-R/mAs. The mAs is milliamperes second. Milliamperes referring to the electron beam current and the seconds refers to the time that the current is flowing i.e. exposure time.

 

[neanderthalphysics]

You want only to be within an order of magnitude? What is the point of such a calculation?

But I'll tell you what, I'll give you a number for the output from a typical medical x-ray device operating at 100 kVp that you can aim for. The exposure rate at 1 meter is about 12 milli-R/mAs. The mAs is milliamperes second. Milliamperes referring to the electron beam current and the seconds refers to the time that the current is flowing i.e. exposure time.

Thanks.

OOM calculations that capture the essence of the physics are useful. Like that Bremsstrahlung graph I put up...say we want to calculate the area. On a computer we would write a do loop to break it down into hundreds or more trapezoids until we reach an arbitrary precision. On paper to demonstrate the point, we roughly break it up into a two trapeziums and get an OOM calculation. We can deal with precision later.

Thanks for the info about the medical X-Ray machine. If I may:

1. Do you know the electrical power draw of the entire machine? (i.e. before the high voltage transformers, electron gun, etc.)

2. The 100 kVp is the potential across the electron gun, or to the Bremsstrahlung "peak"? If the former, do you know what the "peak" X-Ray emitted energies are?

3. What are the dimensions of the dosimeter that give the 12 mR/mAs reading?

4. By any chance do you have the intensity - photon energy plots, with corresponding exposure rates, at a given setting on your machine for me to check the results? Any non-confidential data would be helpful, even the user manuals.

I found a plot of the emission of X-Rays from a tungsten target with angle...
Ref:
Electron range evaluation and X-ray conversion optimization in tungsten transmission-type targets with the aid of wide electron beam Monte Carlo simulations
Oct 2014
DOI: 10.13140/2.1.3991.7125
Soffienko, A., Jarvis, C., Voll, A.

 

[gleem]

1. Do you know the electrical power draw of the entire machine? (i.e. before the high voltage transformers, electron gun, etc.)

The power needed to produce the x-ray is limited by the tube ability to dissipate the heat. A tube of a typical constant potential generator can accept about 80 kW of heat for a short time of about 0.1 sec. If you what a long time of say 10 sec you need to reduce the power to about 20 kW.

2. The 100 kVp is the potential across the electron gun, or to the Bremsstrahlung "peak"? If the former, do you know what the "peak" X-Ray emitted energies are?

100 is the potential difference between the cathode and anode. Typically the "peak" or the maximum number of x rays occur at about 1/3 of the kVp.

3. What are the dimensions of the dosimeter that give the 12 mR/mAs reading?

It can be any size as long as the exposure rate is acceptably constant over the volume typically for x- rays it is 10's of cc's assuming a typical air ionization chamber. The smaller the dosimeter the smaller the charge collected. Of course the chamber must be properly calibrated. The chamber reading is meant to give the average exposure (or rate) over its volume.

4. By any chance do you have the intensity - photon energy plots, with corresponding exposure rates, at a given setting on your machine for me to check the results? Any non-confidential data would be helpful, even the user manuals.

see https://pubs.rsna.org/doi/pdf/10.1148/radiographics.13.6.8290728 The plots give the relative number of x-rays at a given energy. The 14 mR/mas is the sum of the exposure from all of the x-rays in the spectrum.

 

[neanderthalphysics]

OK thanks for your help Gleem. I need to think about it and see if I can come up with a method to approximate the dose from the source.