Photoelectric absorption in x-ray imaging

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Characteristic x-rays emitted during photoelectric absorption occur when an outer electron fills a vacancy left by an inner electron that has escaped after absorbing a photon. These emitted photons are released in random directions and typically do not contribute to the image formation in x-ray imaging. Instead, the image is primarily formed by photons that pass through the object without interaction or those that are scattered. The few characteristic x-rays that do reach the detector are considered noise and can degrade image quality, but their impact is usually minimal. Understanding this process clarifies the role of absorbed photons in x-ray imaging.
curleysue
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I'm sure I'm overlooking something simple here.
I am trying to understand characteristic x-rays given off during photoelectric absorption of a photon. The energy from the photon is given to an inner orbital electron and the photon no longer exists. The electron uses the energy to escape, and any remaining is converted to kinetic energy. So far so good.
An outer electron drops down to fill the hole, and emits a photon in the form of a characteristic x-ray. What happens to this photon? Does it form part of the image? My understanding was that for x-ray imaging, what gets detected are photons that pass through without any interaction, and photons which get scattered (compton scattering). I thought absorbed photons were ones that got stopped, resulting in a shadow. So what is the story with these photons, where do they fit in the picture?

Thanks
 
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You're right that the image is formed by the photons that pass through the object being imaged. The photons emitted by the excited atoms are emitted randomly in all directions, so they don't form an image. Most of them miss the detector, and so they don't play a role. The ones that do hit the detector are a form of noise or background level, and degrade the quality of the image. However, there usually aren't enough of these to be a problem.
 
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