14. Photon Interactions with Matter I - Interaction Methods and Gamma Spectral Identification

[@mitocw]

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[@coldmayorelish6466]

I’m surprised this didn’t get as many views. Thank you so much for this

[@finsgutus]

I have a Master of Science in Engineering from the #1 technical university in my country and my physics lectures consisted of some senile dude literally reading their slides out loud. Most slides had a title and an equation on them. If someone had a question, he would stop for 10 seconds and then continue as if nothing happened. Needless to say, I was not charmed by university physics. I took the mandatory exams and did my best to forget that ordeal ever happened. I've now watched 15 lectures of this course and I've not had a boring moment yet. I'm almost in tears thinking how different my experience could have been if only they had a proper teacher. Or how different the experience of the brightest engineers of this entire country could be... Teachers make or break nations.

[@olfin88]

This is the coolest educational material i've ever seen. I wish I studied it in college instead of History/Economics.

[@runcycleskixc]

these youtube MIT courses are better than most universities' teaching (for which students pay 10-50K a year)

[@randallmcgrath9345]

But now we know why MIT alumni amd even current students get patents and inventions so often. I wish I could get into MIT, but the best I can do is try.

[@brainstormingsharing1309]

Absolutely well done and definitely keep it up!!! 👍👍👍👍👍

[@nurysantisteban]

At 18:55 you said that the maximum scattering angle is pi/2, which I think, needs to be corrected. the maximum value for (1 - cos\tetha) = 2 when \tetha=pi. Thus, the maximum Compton wavelength is twice the number you gave, and the minimum photon energy after the collision will be the initial energy minus (0.5*m_e*c^2)=0.511MeV/2.

[@natjimoEU]

woaw first reaction, hype.

[@colinpitrat8639]

32:00 I'm confused. Isn't the curve reconstructed from the number of electrons collected, so already accounting for these subtleties?

[@debajyotisg]

Yes, but those electrons required a minimum energy equivalent to the work function, to be produced; and so the pulse would be counted at energy with that amount of shift.

[@scipython]

10:32 I don't think H. B. Michaelson is the same person as A. A. Michelson because they have different names and A. A. Michaelson had been dead for 46 years by the time H. B. Michaelson wrote this paper on work functions

[@serialthreelah]

Thank you ☺️🙏🙋🏾‍♂️

[@GreatAuntKit]

so - are the electrons coming from the germanium?

[@spyhunter0066]

At 5.40minutes, you said by mistake increasing probability instead of decreasing for pair production as the gamma-ray energy increases. At 19.19minute, you call Delta lambda for both the wavelength differences of two photons and the compton wavelength. That was a little mistake. Also, at 21.11 minutes, you said by mistake "d theta/ d omega" for the differential cross section instead of "d sigma/ d omega". At 33.33 minutes, you said gamma undergoes a pair production by releasing 2 gammas, but indeed it's an annihilation of two particles instead. Note! Work function definition is not so accurate in books because they do not specify if the electron knocked out from a gas, liquid or a solid and the band where the electrons in those states of matters. You need to go and check how Einstein made his definition in his article. Remember solid has different band levels, but gases do not. I think at 25.48 mins, the gamma-ray energy in the equation should be 1460keV not 1332keV, so the Compton edge should appear at 1222keV as seen in the spectrum. It was K-40 peak not the Co-60 peak as you were confused at that time :)

[@ryandugal]

Higher wavelength? Or do you mean longer?

[@Kottam_Yallawa]

What is this cross section? Any lecture that links to this concept of cross section.

[@Spongman]

so, if a gamma is a photon emitted from the nucleus, why does an electron/positron annihilation produce a gamma and not just a regular photon.

[@TheErdnuss007]

in case you still wondering: a gamma is a specific photon with a high energy, the energy is determined by different factors (one of which is that usually gammes are found in nuclear decay), but essentially a electron positron annihilation produces a "regular photon" with an energy in the gamma spectrum technically a nucleus is giving of some amount of infra red photons aswell, bc it has an amount of temperature, and there arent called gammas bc they arent high energy

[@calebhaines3794]

Photon radiation constants

[@calebhaines3794]

So that quantum radiation is better understood

[@calebhaines3794]

Photon trade network

[@calebhaines3794]

Provide FOIA medical document request approval conditions.

[@danielmahmoudi6731]

@Ryan Wiley love the hall effect. Half the reason I passed electricity and magnetism. Lol