Compton Scattering
Вставка
- Опубліковано 4 лип 2017
- MIT 8.04 Quantum Physics I, Spring 2016
View the complete course: ocw.mit.edu/8-04S16
Instructor: Barton Zwiebach
License: Creative Commons BY-NC-SA
More information at ocw.mit.edu/terms
More courses at ocw.mit.edu
Words can’t express my gratitude to whoever made these courses available to the public for free. Thank you. ❤️
Well said. In my studying years (70s/80s) there was nothing like this (of course). We had to make a reservation to make use of a computer, haha! Btw, I think this man is just a brilliant lecturer. Good luck!
I think it’s adorable how he says:
Photons: phot’ns
Electrons: electr’ns
The way they ended this video in mystery makes the viewer want to continue studying the next video 😂
absolutely flawless professor Zwiebach!
I can't believe it, MIT is cliffhanger us with PHYSICS CLASSES
Wow he made equations look simple by explaining everything simply like it's natural things to happen.
A mark of a good lecture and professor. :)
Llevo viendo estos videos toda la semana. joder que guapo es tener plata para pagarte un buen maestro
Fantastic lectures. No wonder MIT is world's number 1.
Hello very interesting lecture it is exactly what i studied here in my country in france, in 1980 on my last school year before going for my first year at the university, we studied everything shown here, photoelectric effect and compton scatering plus the special relativity including all the equations writen on the blackboard, we even had to learn the demonstration of the final compton scatering formula i still have this in my book that i kept all these years, i think the level was high in those years , we were only 17 years old. best regards
Thank you for sharing this lecture
Compton Scattering reminds me of the way Rutherford detected the nucleus of an atom!
My best guess for the reading of λi is that they are the reflected x-ray particles that did not interact with a virtually free electron. The prepared crystalline structure of the metal might provide places where the x-ray particle can oscillate on the surface without resistance and also without an electron in affecting proximity and then could be reflected by a following x-ray particle with trivial energy loss.
Great!
Boy if there ever was a cliffhanger... It does seem L3.4 doesn't continue where this one stops. Anyone have an explanation to why the first bump is there?
Nvm, I found the correct answer. The first bump is present because some of the electrons which scatter the light are very closely bound to the nucleus of the atom. This means that the m in the Compton equation is not the mass of the electron, but the mass of electron+nucleus which is much larger. This makes Δλ very small.
@stijn D'hondt thanks
Thank you for the answer, that makes sense. So, for the Carbon nucleus is accelerated by a photon?
can't a photon scatter off the nucleus directly?
because of the photon is reflecting from nuclei instead of the electron.
great!
Ermahgerd, the squeaky eraser.
David Battle Right? You’d think MIT could afford the non-squeaky kind.
Thanks very good ❤❤
I also think photons are just particles that look like a wave when it bounces off other particles. I though it was my idea....last time I checked it was considered wave/particle duality...
mind blowing 😘
can we use gamma ray photon for compton and photoelectric effect
Where can I find the first homework he mentioned?
By diagonalizing the matrix M = (cos alpha, sin alpha; cos beta, sin beta)^T, in the new basis B ={v1, v2}, we see that if the beam splitter's lines are set to v1 and v2, the eigenvectors of M with two distinct eigenvalues a1 and a2, then the phase factors become simply a1*v1 and a2* v2. So the calculations become very simple when the light rays travel along lines v1 and v2 instead of 45 degrees up and 45 degrees down.
Small edit: Change 'the phase factors become" to "the new phases are".
Sorry. This comment should be in Lecture 2.3 of the professor's 8.04 course, not in this course. He has so many I got confused.
Photon must have repelling with some other X particle which makes high energy photons bounce off a lot....
What a cliffhanger!
why is the intensity of the second beak is bigger?
did the Professor say "collision"? So it wasnt like the electron absorbed the entire photon and then re-emit it?
if electron absorb photon.photoelectric happen
Now I'm curious. Why is there the other peak? :D
It is due to interaction of photon with the nucleus of the atom.
According to the notes from MIT open courseware, the other peak is caused by the collision between the photon and the whole atom. Because of the negligible Compton wavelength of the atom, the wavelength shift is very small.
The peak at λi represents a process
in which an electron receives some momentum from the photon but still remains bound. This is not
very unlikely: the typical momentum of a bound electron is actually comparable to the momentum of
the photon. In this case the photon scatters at 90◦ and the recoil momentum is carried by the whole
atom. The relevant Compton wavelength is therefore that of the atom. Since the mass of the carbon
atom is several thousands of times larger than the mass of the electron, the Compton wavelength of
the atom is much smaller than the electron Compton wavelength and there should be no detectable
change in the wavelength of the photon
@@ryanyi8900 Thx a lot😀
The Compton Effect UA-cam video link: ua-cam.com/video/jY58YHTbzXs/v-deo.html
Si atunci care este mecanismul prin care din fotonul gama incident de mare energie, de mare frecventa, ar aparea fotonul gama imprastiat, de frecventa si energie mai mica decat a fotonului gama incident.?
And then what is the mechanism by which the incident gamma photon of high energy, of high frequency, would appear the scattered gamma photon, of lower frequency and energy than the incident gamma photon.?
In the photoelectric effect, we know that the material absorbs light by quanta. But in Compton scattering, the electrons partially *absorbs*(collide) the energy of high energy photon. So are the two (the photoelectric effect and Compton scattering) contradictory?
I would say NO. Here the photon is a particle with some momentum. In the photoelectric effect the photon gives off ALL of it's momentum, while in Compton scattering the photon only gives off A PART of it's momentum.
Jacob van Dijk Thank you!
@@ryanyi8900 You're (always) welcome.
@@jacobvandijk6525 isn't that the question? If a photon carries a certain quantum of energy, then how is it possible for it to lose only a part of it?
Consider the photoelectric effect as an example, the photoelectrons don't eject for as long as we provide photons with a frequency lower than the threshold frequency. If it were possible for photons to lose part of their energy, then it would require a higher intensity of light for photocurrent to flow; which is totally against the experimental data.
So, what's going on here?
@@tanfeexulhaqq4616 Nothing wrong with the picture you describe. In the photo-electric effect the emphasis is indeed on photon-absorption. But when you shine light an a material photons can be reflected (or scattered) too. They don't enter the material; think of a mirror. And if the scattering is inelastic, the photon gives up some of its energy to the material (which is heated a bit). Absorption always kills the photon, in collisions it can survive to some degree.
What a cliffhanger
Does anyone know why the peaks aren't sharper? What causes the smaller variation in wavelength of the scattered photons that makes the peaks broad rather than pointy? Is it a detector issue? Some other quantum effect?
yeah good question. collision should be instanteneous
This is due to Doppler broadening
@@AnshulSharma1997This, plus all detectors have some finite resolution, in angle and in energy, etc.
wave particle duality
why when theta=0 there is no collision ? can someone explain?
@Thejas CS I know you 🤔
If a photon carries a certain quantum of energy, then how is it possible for it to lose only a part of it?
Consider the photoelectric effect as an example, the photoelectrons don't eject for as long as we provide photons with a frequency lower than the threshold frequency. If it were possible for photons to lose part of their energy, then it would require a higher intensity of light for photocurrent to flow; which is totally against the experimental data.
So, what's going on here?
how can a photon loses its energy?
As it has packets of energy how can it be lost after hitting to electron.There is something more which is till Now unnoticable which changes some of photon's property.
it changes wavelength
It is just an inelastic collision and conservation of energy
@@summadayze733 Yes you are right it is inelastic collision but in what terms it is losing energy?
@@shivamkakkar4027 the lost energy is the kinetic energy of the recoiled electron which was before almost motionless
guys, someone please explain to me why is that extra peak there in the experiment described at the end of the video? next video doesn't cover this. as the angle is 90, it can't be that some of the photon are detected without being scattered at all.
Photons are scattered not only by electrons but also by Carbon nuclei. And there is a negligible Compton shift for those nuclear part of scattered photons wavelength
"Thomson," not "Thompson."
how does electron and photon decide in wich direction (angle) to scater if photon hits electron directly? I understand formula and experime so we take fact final result but if i want to predict direction?
In aggregate, all possibilities happen, with the x-section formula governing the probability distribution. In a single event, a single random one.
Sir, homework done-
If electron at rest absorbs photon, the velocity of photon had to decrease i.e change which is against relativity or electron has to move with the speed of light, that makes its mass infinite which is against the law of conservation of momentum.
Is linear algebra essential for this course
no
10:08 I think you have to correct the formula of cross-section. And after correction unit becomes "per unit area per unit solid angle" rather than "area per solid angle".
No he's right
It is counterintuitive to have no mass but still have momentum. Can this be explained in layman’s terms?
E ²=(mc ²) ²+(pc) ² is the full equation. So when m=0, then E ²=(pc) ² or p=E/C
The idea that mass is required for momentum is a simplification (oversimplification) from highschool physics. At low speeds p = mv. But at high speeds (relativistic regime) it's more complicated, see the formula given in the other response.
The other pick, why is it there??
On guy up says it is because it is not only an electron detected, but also the entire nucleus for the Carbon, so, if you use the equation for initial lenghtwave, it makes sense
I am from in Iraq
عراقي... #طالب سادس
منور يستااااا 😂😂
But why was the first peak there?
because of the photon is reflecting from nuclei instead of the electron.
Static/uniform particle distributed throughout universe???
Matrix...lol...this is crazy
mass of photon=0 ??
Rest mass of photon is 0
Could be better a little bit,feels something is missing
It's unfortunate that the teaching of this subject hasn't changed much since I was a post-grad at a lower ranked university 20+ years ago. Even back then I thought to myself that I wasn't there to take a class in history along with some interesting qualitative ideas with not enough detail in the reference to experiments. Back then I put in down to the professor/university I attended but if MIT is teaching it pretty much the same way 15+ years later, there's a wider problem. Almost no other topic in physics or any other science is taught in this way.
MFTI or MGU are much better!
What is this?
@@valentinsarmagal Moscow PhysTech and Moscow State University
@@javierhernandez1390 do they have open courses in English?
@@valentinsarmagal nope. Learn Russian!
They've got a different style. Whatever suits you best
.