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neanderthalphysics

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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.

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)

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

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?

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?

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