Understanding Bremsstrahlung Radiation - X ray Production

[@MrWilly-mm5rv]

The best explanation I found on KZbin. Thanks a lot❤.

[@user-gu7se5iw7b]

Nah

[@farxeennn_n]

im an Alevels student, it is so difficult to understand phys for A2. but you made it so easy.. Thankyou!!

[@SCCDentalAssisting]

Excellent explanation!

[@irwo]

Thx from medical college of iraq

[@AliFarhan-hm4ls]

عراقي ابو الغيرة

[@irwo]

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[@AliFarhan-hm4ls]

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[@irwo]

@@AliFarhan-hm4ls كوفة

[@AliFarhan-hm4ls]

@@irwo موفق ان شاء الله

[@zeph7691]

great video, very clear and the animations helped a lot! x

[@mbohhonorine8470]

This was so helpful. Thanks a lot

[@CloverLearning]

Glad it was helpful!

[@mohamedamr_1]

Thanks a lot . It was so helpful❤

[@Psychotherapeuticraphe]

Wonderful explanation. Thanks so much from Brazil

[@Dentalwali]

Thank-you so muchhhhh❤

[@محمدمحمد-ف9ز3ط]

شكرا جزيلا لك كان هذا مفيد جدا الكلية الطبية في سوريا ❤️❤️

[@aidanaarystanova7720]

Wonderful explanation

[@ashadasabzali4670]

Vet school from Trinidad 🙏👍

[@ScarlethAngeline25]

Thanks, it helps me alot. The best explanation i saw in youtube.😊

[@chamasuksansirikul9376]

keep up the good work!

[@TRFR-ye1mw]

You are great at physiology

[@leogoldn290]

Real good explanation 👍👍👍

[@annagerebo6561]

This is just what I needed! Thanx!!!! 😊

[@pavanajsridhar939]

superb! thanx a lot! best possible explanation !

[@alyssamaeamolopido1612]

Thank youuu ❤

[@fieryweasel]

5:10 What's up with the obvious spike on the right side of the Bell curve? Is that a resonance or something? It feels like that would be important.

[@mianascimben1858]

I think that could be the result of the excitation of some atomic electrons. The electron pass trough the atomic orbital can excite some atomic electrons whenever it has the right amount of energy. Then the excited atom relaxes and produces an x-ray with a specific frequency v = Ectitation_energy/h

[@fieryweasel]

@mianascimben1858 makes sense, thank you! It seemed clear that it wasn't random.

[@margheritamatarrese4038]

Hello! Is there a video of this series on characteristic radiation? Is the only key concept missing 😢

[@amegaviglory6979]

Thanks

[@mreatboom1314]

Why this energy is not transfered to the nucleus by moving it ?

[@blackalgae370]

I will preface this by saying i am not an expert, but; I think, it because atom generaly have a neutral charge. it has positvly charged protons and neutral neutrons in the nucleous and negativly charged eletrons in the electron shell. If an electron hits a neutraly charged atom and interacts with it then it will become negativly charged, and a negative charge will push away negatively charged particles. The same way a magnet will be atracted the metal and the negative end of a magnet will attract the positive side of another magnet but push the negative end away. On the atomic scale this would mean that if an electron shot into an atom and it locked into the electron shell it would knock another electron out of the shell and the too would produce an electromagnetic wave. If however it was the nucleus that recieved a transfer of energy and it recieved the impacting energy then the excited atom would release the energy by throwing out an electon or by altering the configuration of its electron shell also releasing a tiny busrt of energy in the form of heat or light. At least thats how i think it works.

[@AmeerHamza-el7un]

One video on characteristics radiation.

[@JavanKibet-lv7kw]

THANK U MONSIEUR

[@nawlifepeacefullife]

nice lecture

[@ienjoysandwiches]

Why is the x-ray beam that is produced fan shaped and in a perpendicular vector to the electron beam? Is the isotropic emission of Brehmstrahlung shaped into a fan by the glass/metal housing and aperture and anode shape? What I'm really asking is - are the shape of the x-ray beam and the shape of the anode mirror images? (ie. "enantiomers") Like how a tennis ball or pool ball hitting a surface will always form a complementary triangle? Not exactly because the surface would have to be "sticky" or allow the ball to sink in to it to really reflect what was happening, but just as a general explanation of why the beam width narrows in proportion to the focal spot width narrowing (theta increasing).

[@IkehMaryAnn]

Thank you for this ❤

[@lilianemccallin1143]

how can electron have an energy of 0 keV, as its rest mass is 500keV? Does this mean the electron has disappeared completely ?

[@miaehatamm8936]

i think it is only referring to the kinetic energy that is imparted by the voltage! the electron would still have rest energy

[@AliFarhan-hm4ls]

It is the kinetic energy

[@rahafalsultan4999]

Thank you ❤

[@ohanwenwachiamaka3078]

Thank you so much 💖

[@emilmohan1000]

any other way of x ray production?

[@cucciafr68]

You can get characteristic xrays from when the incoming electron knocks out an electron from one of the shells of the atom. To stabilize itself, an electron from a higher shell moves to that vacancy. To balance the energy difference the atom can release an xray of specific energy. If you look back at the graph it shows of the bell curve it has a large spike. That would be from the characteristic xray.

[@ch_arbaz_yt]

Salam very good concept

[@skitzo9917]

loving you

[@GM-uo5tq]

🙏🏻✨👏

[@YeshaniNethmi-q6o]

greattt

[@LinusFeynstein]

Sehr schöne Erklärung. However, please pronounce Bremsstrahlung as in „BRAMMS-STROAAA-LOONG“. The stress is on the second syllable.

[@cucciafr68]

I cringed trying to pronounce this in front of my German colleagues.

[@ienjoysandwiches]

Why is the x-ray beam that is produced fan shaped and in a perpendicular vector to the electron beam? Is the isotropic emission of Brehmstrahlung shaped into a fan by the glass/metal housing and aperture and anode shape? What I'm really asking is - are the shape of the x-ray beam and the shape of the anode mirror images? (ie. "enantiomers") Like how a tennis ball or pool ball hitting a surface will always form a complementary triangle? Not exactly because the surface would have to be "sticky" or allow the ball to sink in to it to really reflect what was happening, but just as a general explanation of why the beam width narrows in proportion to the focal spot width narrowing (theta decreasing).