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Faq 017
Dear Photomultiplier Doctor,
What is the relationship between the energy of the x-rays and the number of photons or photoelectrons produced?
Mr. Derek Tam Sing
Rapiscan
Take NaI(Tl) for example. We know that the light output is about 30 photons per keV of x-ray energy
deposited in the crystal. As the photomultiplier has a QE of about 25% at 400 nm, the peak emission
wavelength of NaI(Tl), 1 keV of x-ray energy will give 30 x 0.25 ~ 7.5 photoelectrons per keV. The
crystal is essentially linear so 100 keV x-ray will give 750 photoelectrons.
At 50 A/lm the anode signal will be Q = 1.6 x 10-19 x 750 x 0.7 x 106 = 8.4 x 10-11 Coulombs
If you have a load resisitor of 100k then you will get a voltage signal of peak amplitude
Alternatively if this charge is injected into a charge sensitive amplifier with feedback capacitor of 10 pF then the peak voltage is
n p = Q/C = 8.4 x 10-11 / (10-11) = 8.4 volts
I hope this gives you some feel for the sort of signal levels you will find with NaI(Tl). In this example, the photomultiplier gain is clearly too high by about a factor of 10 because no application I know of calls for signals of 34 volts.
Regards,
Photomultiplier Doctor.