X-ray flux of 7 x 1013 and effective energy of 40 keV is allowed to pass through

Question

X-ray flux of 7 x 1013 and effective energy of 40 keV is allowed to pass through the metal window of a fluoroscopy intensifier. The window allows 82 % of the radiation which fell on the first phosphor (CsI:Na) whose conversion efficiency is 45 %. Each photon produced on the phosphor has energy of 2 eV. Only 6 % of the photons produce photoelectrons on the photocathode each of energy 1.4 eV. If the photoelectrons are accelerated through a potential difference of 40 kV and on the average each electron produces 0.10 photons on the second phosphor then find the Flux Gain of this fluoroscopy system.    

Explanation / Answer

The x-ray image intensifier converts the transmitted x rays into a brightened, visible light image. Within an image intensifier, the input phosphor converts the x-ray photons to light photons, which are then converted to photoelectrons within the photocathode

The ratio of the number of light photons created at the output phosphor to the number of x-ray photons striking the input phosphor is called flux gain.

The flux gain results from the acceleration of photoelectrons to a higher energy so that they generate more fluorescent photons at the output phosphor. Each light photon generated at the input phosphor will generate approximately 100 photons at the output phosphor, resulting in a flux or luminance gain of 100.

Let us analyse the given data

initial flux=7 x 1013

effective energy =40KeV

efficiency =45%

photon produced has energy =2eV

energy of photo electon =1.4 eV

potential difference = 40 kV

after calaculation the no of x-ray photons striking the input phosphor using the given conditions=100

output =0.10

therefore flux gain ratio 1000:1

flux gain =62%

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