I am using a photo diode with known quantum efficiency to measure the intensity of my x-ray source. The diode's signal is measured in volts on a scope with known impedance.
Can anyone help with, how to calculate the counts per second for the photons of the x-ray source.
Many thanks,
Rad
Hi!
See the website or attached file.
Again, find out the energy requirement to create an electron- hole pair in your photo diode material.
one can use that data to calculate the same with your efficiency correction
http://www.aps.anl.gov/Xray_Science_Division/Detectors/Detector_Pool/Detector_Information/Calibrated_PIN_Diode/Calibrated_PINdiode_QuickGuide_V2.pdf
Must have an X-ray detector with energy resolution, know its efficiency curve for different energy, source detector distance, the solid angle between the source
and the sensitive area detector, the attenuation in the air and on the windows of the tube and the detector.
I recommend you see our paper: X-Ray Spectrometry
Volume 44, Issue 2, pages 69–74, March/April 2015
Hi Rodolfo,
I am using a photodiode type AXUV20A with a known responsivity to the wavelength emitted from the the source. If I am right so I am using the formula
No. of Photons = Power / photon energy
where P = Voltage value/ (responsivity x oscilloscopes impedance).
Power is taken in J/s and photons energy is taken in joules.
Hi All,
Many thanks for your answers.
So if I have a source that emits x-ray photons from two elements, say Fe of 6.404 eV and Cr of 5.415eV (k alpha). Which energy would I use when counting the no. of photons or shall I take the average of the two?
I think, average is not enough. It depends on the weight percentage of the individual materials, surface area (if any, if not a homogenous mixture), K X-ray yield of each elements etc.
Hi Mallick,
Many thanks for your answer. Its hard, my source is a non homogeneous as I have a combination of 28.9% Fe and 6.1% Cr, and 17% C. The measured no. of Photons / s is 5x10^12. So the source is emitting a range of energies.
Best,
Rad
hope you have corrected the window transmission factor say 90% transmittance
Hi Iman,
A common recommended practice if you are limited by your detector ' s saturation limit would be the use of filters, naturally dependent on your source emission range (energy, wavelength ) so you can choose the proper filter. Once you get a non saturated signal then substitute by the absorption factor of your filter.
Hope this helps.
Best,
Radhwan
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