Physics - Modern Physics (6 of 26) X-Rays

[@marcelstieger7639]

Hey Micheal, I just watched some of your videos. Your explanations help me quite a lot. Couldn't find nearly anything as good as your videos in german. Seeing a brighter future now for my physics exam

[@MichelvanBiezen]

Marcel, Good luck on your exam. Study hard.

[@feelingzhakkaas]

Is the wavelength calculated is Lamda -min or Lamda-max ? if it is min then larger wavelength will tend towards UV not Xray ranges....

[@deepeshpadala5329]

Shouldn't the negative terminal go to the cathode? cuz then it'll get -ve charge, and accelerate the electrons?? (I'm not trynna criticize, I'm genuinely confused) Great video BTW, thankyouuu so much

[@MichelvanBiezen]

We always appreciate all comments, because that is how we learn and solidify our understanding, AND discover when we are wrong. The confusing part is that the definitions reverse depending on the system is charging or discharging. When discharging, the anode is the negative terminal from which electrons leave and the cathode is the positive terminal at which electrons arrive. In such a system, reduction occurs at the cathode. But when forced and the system is reversed (the system is being charged which required the input of energy, the anode is the terminal at which electrons arrive and the cathode is the terminal from which the electrons leave.

[@Hydrogen1972]

Just a small fraction of the electrons will cause x-rays (bremstralung) Increasing KA voltage and the current mA will result in a increasing x-rays

[@srishtisadhu]

Thank you so much..it really helped me to solve the numericals based on minimum wavelength.. This video has given me concepts based on X-rays.. :)

[@Yashtackle]

Thank you ......... i would be delighted if you explain a few little things more like why the heater is used....... is it a thermionic emission ? and few more depths in places like Why the X-rays are emitted?? you explained what is the experiment but not the phenomenon! please....

[@rasikaprabhu8583]

Xrays are emitted as per wavelength... (depends on energy) ifyou change potential difference, you will get different kind of radiation.

[@nickbrady1153]

Hey Professor, Two things: 1)In the video You have the wanted Emax as max energy from photons when you actually find the Emax of electrons. also Im curious as to how we can just label lambda as min when you get to the equation of Emax=hc/lamda

[@dvdplayer128]

For the first part of your question, as I understand it: The Emax of electrons is fully converted into the Emax of photons. That's why the situation gives the maximum Energy for photons

[@Jarrod_C]

Not very clear on how it works. Furthermore so that also means that some of the photons will be UV photons? So when you go into an x-ray machine some of the light hitting you will be UV?

[@karhukivi]

Four years late (!) and I suppose you know by now that the lower wavelengths won't get through the beryllium window, only x-rays will.

[@MikeH-ho7lm]

I'm having trouble getting the results on my TI30x calculator. Please tell me where I can get help punching in the calculations properly.

[@Shubham.Kumar.Mishra]

sir I can not understand how continuous x ray are produced.. I read that e without colliding when passes near by nucleus it produces continuous x ray but I can not understand that 1. why doesn't it fall in nucleus? and why it produces radiation?

[@MichelvanBiezen]

If the source oscillates continuously, then the radiation produced is continuous.

[@drrx4308]

Thanks alot sir this was very helpful

[@MichelvanBiezen]

Glad it helped

[@jakubkusmierczak695]

Amazing.

[@MichelvanBiezen]

Thank you. Glad you found our videos!

[@jakubkusmierczak695]

I would love to have such a teacher. Getting so sophisticated knowledge never was so simple and pleasant. I can not stop being impressed. Thank You Mr. Professor I am very garateful, it is like science adventure. Best greetings.

[@gautamchettiar4150]

I think the battery terminals are wrong

[@shayanmoosavi9139]

Cathode isn't always the negative terminal if that's what you mean.