Photoelectric Effect | X-ray interaction with matter | X-ray physics | Radiology Physics Course #23

[@skywater133]

I am a rad tech student studying for my boards exam and over my 2 year program, no one has explained this process better than you. Thank you SO much. You are concise and cover all the questions that come up for me.

[@JunkieMonkey321]

finally someone who explains how the photoelectric effect differs from characteristic radiation! was scratching my head for a while. thanks

[@lourenshenn867]

Hi I’m studying for my part one exams and I have found your channel to be of great value. Thank you for your hard work. Please keep posting videos on a regular basis.

[@radiologytutorials]

Yay! I’m glad it has been helpful 👍🏼 best of luck with the studying 📚

[@feruzarakhimova105]

I simply love the way the topics are so smoothly delivered! Incredibly helpful in understanding the concepts!!! Thank You so so much!!!

[@riio15]

So informative, i love how you answer the 'might arising questions on our minds' ,, thanks a lot ~

[@fazaltahir4216]

Superb explanation, very nicely explained, awesome

[@aiszelangeliligo2060]

I really appreciate your discussions, it helped me a lot. Im on my 3rd yr in residency training. I wish you also have a detailed discussion in mammography and CT physics. but overall, I really appreciate your discussions.

[@Jesus_peking]

Thanks for the knowledge as it was a great tool of understanding for me as I prepare for my radiation physics exam. I will further use your site as parallel working tool for my career improvement.

[@radiologytutorials]

Good luck for your exam - so glad you're finding the channel useful 👍

[@luigiGmalone1]

What happens if the energy of the incident X-ray is less than the binding energy of the electron (within the small range before the kedge energy)if the photoelectric effect is not occurring

[@radiologytutorials]

Hi Luigi. Great question! My answer might not be satisfactory but as with all things in physics nothing is as simple as it seams. This is the best way I have come to understand it without needing to get too much into quantum physics/mathematics. The issue comes when thinking of an X-ray and an electron as two solid balls colliding. It cognitively makes better sense when thinking of them as waves (easier to imagine waves passing through each another as opposed to solid balls passing through each other). The photoelectric effect requires x-rays to have a threshold wavelength (and therefore frequency) to occur. When x-ray wavelength is too long the photoelectric effect will not occur despite the x-ray coming into ‘direct contact’ with an electron. It will simply pass through the electron and not be attenuated. This is a gross oversimplification but is a good broad brushstroke way to think of it. Difficult to explain in a comment (and I don’t understand it in much more depth than that). Hope it helps!

[@ashutoshpani]

I might be totally wrong but what I inferred is this: For an electron in a certain shell (certain binding energy), if the incident xray photon energy is similar to or slightly more than the binding energy of the given electron, photoelectric effect occurs. But if the xray photon energy is way more than than the binding electron energy, Compton effect occurs (where the elctron is still ejected but the photon continues on with reduced energy having scattered or deviated). Now if the xray photon energy is less than that binding energy, it will not resulting in any electron ejection, instead as said by Sir earlier, it will pass through (albeit may be in a different direction). It will follow the coherent scattering phenomenon (i.e. Thompson or Rayleigh scattering). The electron will absorb all of the photon's energy, vibrate (as it is inadequate to eject the elctron) and simply re-emit the xray of the exact same energy as the original incident xray in a random direction. Mind you, this inadequate energy xray might eventually be stopped, probably by some other electron of a higher valency shell of some other atom which has a lower binding energy. This is why is it is important to use xrays of a certain energies spectrum. Too low - coherent Rayleigh scattering occurs which incurs unnecessary patient radiation dose and doesn't even contribute to the image. Too high - Compton scattering occurs, decreases contrast in image and also increases noise. Please, anyone correct me if I got something. Its just my understanding so far.

[@MENNAHABIB-r5b]

Immensely grateful for your videos🙏

[@hamidrezahatamishirkouhi3188]

If the photoelectric effect and the characteristic x-ray don't contribute to our image and it doesn't reach the detector then what does contribute? If it's just the difference between not attuned and attuned ones then we have only two grey colors and contrast?

[@pranavdev5691]

Thanks for explaining physics in its utmost beautiful form ❤️ No words to appreciate your efforts. ❤️❤️

[@radiologytutorials]

Thank you for such a lovely comment 😊 glad to hear that others find physics beautiful too!

[@sid17391]

this cleared up so many misconceptions that the popular books created. Thank you!!

[@radiologytutorials]

I'm so glad!

[@AfiOye]

Hey. I'm in xray school and scored pretty low in the interactions with matter portion of a mock boards. So question, if the body is absorbing these xrays, how is it hitting the IR to create an image?

[@jenifer.s170]

I have one question: which contribute to image formation in photoelectric effect....Is characteristic X-rays ? But u told it didn't have sufficient energy to reach the detector..?

[@noodlesthe1st]

So the difference between characteristic radiation and the photoelectric effect is that characteristic radiation happens in the anode and the photoelectric effect happens in tissue?

[@anilchuhan1236]

Sir,what do you exactly mean by dose to the patient tissues?

[@sowmiyasiva7390]

Great thanks for your lecture series in Radiology physics I have a doubt The characteristic xray released from tissue will have much much lower energy and wont be able to reach the detector -this makes sense .What will happen to the characteristic xray in case of iodine , since atomic number is 53 and energy of the characteristic xray would be around 28KeV?

[@Rupesh99]

Love from India 🇮🇳

[@radiologytutorials]

Thanks Rupesh. Love from South Africa 🇿🇦

[@irishummel2983]

The BEST!

[@LilMissSqueak]

Why is the line for iodine plotted before the k-edge? Surely if the k-edge is the KeV at which the photon energy is greater than the K shell binding energy which is what allows the photoelectric effect to occur then there should be nothing happening before the K-edge as no photoelectric effect can occur for iodine before this KeV?

[@robeenaanjum6166]

In the strict definition of photoelectric effect there is a threshold energy below which X-rays will not release a photoelectron. However,the photoelectric effect can occur with outer shell electrons (the reason we focus on inner shell is that these interactions are more common) this accounts for the pre K edge attenuation.(i have copy pasted the answer of Dr Michael to the same question by another viewer.hope that would answer yours)

[@chad969]

Great video! I have a question about the K edge. You said that the photon energies to the left of the K edge on the graph are insufficient to release a K shell electron. So shouldn't the probability of the photoelectric effect be zero at those energy levels? If I understood correctly, the incident X-ray needs to be powerful enough to release an electron from its orbit in order for the photoelectric effect to occur. But this graph shows that there's a non-zero probability of the photoelectric effect occurring, even at photon energies that are to the left of the K edge (below the K shell binding energy). This suggests that the photoelectric effect can still occur even if the incident X-ray is too weak to release an electron. How is that possible?

[@radiologytutorials]

Excellent question. You are correct. In the strict definition of photoelectric effect there is a threshold energy below which X-rays will not release a photoelectron. The photoelectric effect can occur with outer shell electrons (the reason we focus on inner shell is that these interactions are more common) this accounts for the pre K edge attenuation.

[@chad969]

@@radiologytutorials Oh I see, that makes sense. Thanks for answering my question

[@radiologytutorials]

No worries. Most people don’t pick up the subtleties that you do. Shows you’re really engaging with the content at a high level 🙌🏼

[@karlwagner8905]

Why do interactions with the electrons (= photoelectric effect) play such an important role here? do the xrays also interact with the protons/nuclei just like the elektrons do in the anode?

[@radiologytutorials]

Great thought. I hadn’t actually considered this. My initial thought is that the atomic number in tissue is very low (as opposed to the anode) therefore with high energy electrons the electrostatic force from the nucleus will be negligible (no significant Bremmstrahlung radiation produced). That would be my guess. Perhaps some one reading this knows better. Let me know if you find out anything more 🙂

[@LolyCabrera19]

Thank you so much

[@rakhisandhu9339]

Thank you sooooo much sir 🙏😊

[@radiologytutorials]

An absolute pleasure!

[@steves3948]

Thank you.

[@radiologytutorials]

You're welcome!

[@balasubramanian6680]

B

[@kennyx8482]

one word...IMMACULATE

[@radiologytutorials]

Thank you mate!

[@emiliasalas1447]

[@NoorCamelia]

Man you are a miracle im a freshman and i cant understand anything thank you so so much you are amazing human being 🩷

[@radiologytutorials]

Yay! So glad it’s helpful 🙂

[@mia_kayy]

which selection does photoelectric falls under in your question bank questions

[@radiologytutorials]

Predominantly in section 5. Although it does come up at other points (ie when talking about filtration). Hope the studying is going well Mia 🙂