Michel you are brilliant!! I really like your unpretentious but thorough teaching style. I have been a teacher (mainly chemistry) for over 40 years. I can say that I enjoy your style immensely. Please don't change a thing!
@@MichelvanBiezen I have retired now (75 years old). But I cannot give up the science. I completed two BSc. (Hons) degrees, my first with a major in Biochemistry and later (began at age 55) with a double major in Mathematics and Physics. I have loved science since I was in primary school. To keep myself mentally active I am working through several textbooks at the moment including Lehninger (Biochemistry), Carrol and Ostlie (Astrophysics), Anton (Calculus), Serway and Jewett (Physics) and Serway, Moses and Moyer (Modern Physics). Your lectures have helped me on numerous occasions. I was particularly interested in the above lecture in relation to a problem I had in SMM's book. A few months back I followed your entire series on hyperbolic trig functions. It was (as is typical of your style) presented in a very logical and sequential manner which made it understandable, even for me. I know that my approach makes me a sort of "Jack of all trades" in science but a "master of none", but I am comfortable with that. You are doing an amazing job by posting these fabulous videos Michel. I hope that you can continue with this valuable work for many years. Best regards, Robert
DAMN Thanks Professor clear and concise. In my pr notes the formulas are every where and that just losses me everytime i glance at it. But with your modern physics notes I can seen a path and a clear way of approaching this kind of problems. I guess I am a pro now. Really thanks
Great learning videos. Your style of teaching and AK lecture's are the two most thorough methods on the Internet. You guys take each educational problem step by step for the viewer with great understanding. Thanks. "I salute you."
Thank you so much. Tried this problem on the textbook but found it challenging. Even after looking at the solutions it was still confusing but you shed light on it.
Thank you for these videos. The topics are very well explained they will be of great help for my exams. I just wish more people would know about them because you make physics easy to understand.
Why your videos don't show on UA-cam when I type the topics. The only way I can find you is to type your name with the topic!! I'm pretty sure students would love to see such physics topics well explained. Thank you so much you're the best.
Ali, Thanks for the feedback. Yes, it is hard to break through on the internet. The videos will become easier to be found when the viewership goes up. As people tell one another about these videos they will become more known.
Typically, when a photon collides with an atom, it will either be reflected or scattered, or its energy will be completely absorbed and cease to exist. Compton scattering is a strange exception to that.
In my view mass, charge, electric field & energy are different things. Mass,charge or electric field are not energy in themselves. Energy comes from motion of mass, charge or electric field. When a photon is absorbed by an atom, the photon transfers its KE to the electron and its electric field is absorbed by the atom. The KE & electric field is released when atom emits a photon. In compton scattering the photon transfer part of its KE but retains its electric field since electron cannot absorb its electric field. Electric field & mass are indestructible or convertible to energy. They can however carry & transfer energy. I think free space is an energy field with uniform energy density. The energy of this field comes from perpetual motion of sub microscopic particles which make up space.Imbalance in energy density induces force fields which cause the energy to flow from higher to lower density regions of space. Mass & charge change the energy density balance & induce gravitational & electric fields.Energy contained in gravitational and EM fields is energy borrowed from space. Energy extracted from these fields is ultimately returned back to space. My comments are not in line with current science.They are my conjectures.
Just one question why we wrote that the momentum of the photon is the same of the electron after the collision isn't it momentum of the system before equals momentum after the collision ?
Note that the equation indicates that the momentum in the y-direction is equal. Since the photon traveles in the x-direction, the momentum in the y direction is zero. Therefore after the collision the total momentum (in the y-direction) must be zero as well. That is why the momentum of the electron must equatl the momentum of the photon (in the y-direction) with opposite directions.
Hi sir, I have a question regarding Compton scattering. Why does intensity of graph of scatter wavelength decrease if we use heavier elements? Is it because there are more electrons which interact with x rays or because havy elements have much stronger bond of electrons within the atom? Thank you so much. Regards from Jelena
+Eric Golant When the velocities are 50% the speed of light, your answer will be off by about 15%. At 86.6% of the speed of light, your answers will be off by 100%. Typically at velocities greater than 10% the speed of light, you may want to start using relativistic equations.
Right but by "non-relativistic velocities" you mean, velocities where the difference is non-negligible right? Technically the relativistic equation is more accurate no matter the velocity (above zero) correct?
the formula for calculating the scattered wavelength is derived using relativistic equations. doesnt that mean we should be using relativistic formulas for kinetic energy and momentum?
no actually, i was reading the derivation of the wavelength formula and there the book used relativistic equations. but then i checked the formula given for calculating the angle of scattered electron and the value matched with the one obtained here, so there must be something i am misunderstanding. thank you for replying :)
Hello sir, why the Total of E is E(initial) - E(final)? Rather than E(final) - E(initial)? From my previous experience it was always Final - initial, Thanks
The initial energy is scattered and separated into two different energy, the kinetic energy for the electron and the energy of the after collision photon. Both of these energy add up to the initial energy, making the initial energy bigger than them.
Michel you are brilliant!! I really like your unpretentious but thorough teaching style. I have been a teacher (mainly chemistry) for over 40 years. I can say that I enjoy your style immensely. Please don't change a thing!
40 years of teaching chemistry! Wow! That is a great service to the new generation! Thank you for your kind and encouraging words! 🙂
@@MichelvanBiezen I have retired now (75 years old). But I cannot give up the science.
I completed two BSc. (Hons) degrees, my first with a major in Biochemistry and later (began at age 55) with a double major in Mathematics and Physics. I have loved science since I was in primary school. To keep myself mentally active I am working through several textbooks at the moment including Lehninger (Biochemistry), Carrol and Ostlie (Astrophysics), Anton (Calculus), Serway and Jewett (Physics) and Serway, Moses and Moyer (Modern Physics). Your lectures have helped me on numerous occasions.
I was particularly interested in the above lecture in relation to a problem I had in SMM's book.
A few months back I followed your entire series on hyperbolic trig functions. It was (as is typical of your style) presented in a very logical and sequential manner which made it understandable, even for me.
I know that my approach makes me a sort of "Jack of all trades" in science but a "master of none", but I am comfortable with that.
You are doing an amazing job by posting these fabulous videos Michel. I hope that you can continue with this valuable work for many years.
Best regards,
Robert
DAMN Thanks Professor clear and concise. In my pr notes the formulas are every where and that just losses me everytime i glance at it. But with your modern physics notes I can seen a path and a clear way of approaching this kind of problems. I guess I am a pro now. Really thanks
Great learning videos. Your style of teaching and AK lecture's are the two most thorough methods on the Internet. You guys take each educational problem step by step for the viewer with great understanding. Thanks. "I salute you."
Thank you so much. Tried this problem on the textbook but found it challenging. Even after looking at the solutions it was still confusing but you shed light on it.
Thank you for these videos. The topics are very well explained they will be of great help for my exams. I just wish more people would know about them because you make physics easy to understand.
Thank you. This makes Comptom Scattering so much easier to understand.
তুই একটা জিনিস রে ভাই। পুরাই বস।
you are a boss,you r so much talented person 👨❤️👨
Why your videos don't show on UA-cam when I type the topics. The only way I can find you is to type your name with the topic!! I'm pretty sure students would love to see such physics topics well explained. Thank you so much you're the best.
Ali,
Thanks for the feedback.
Yes, it is hard to break through on the internet. The videos will become easier to be found when the viewership goes up. As people tell one another about these videos they will become more known.
THANK YOU VERY MUCH FOR YOUR WELL EXPLAINED EXAMPLES......GOD BLESS YOU......I'VE CLEARLY UNDERSTOOD. THANK YOU AND BE BLESSED.
well explained...breaking complex equations to a very simple physics
God bless your soul, professor
Thank you
Thanks to you I can finally understand what my professor is babbling about for the past few weeks
Thankkk you . Your videos has being helpful. God bless you 🙏
Thank you. Glad you find our videos helpful. 🙂
thank you sir..., either Issac Newton or Albert Einstein is your brother the way you presented physics concepts
😘
They were way smarter than me, but I do appreciate the work they did.
Photon is oscillating em field. When photon hits electron does it give away away some of its em field to electron, as its em field gets weaker.
Typically, when a photon collides with an atom, it will either be reflected or scattered, or its energy will be completely absorbed and cease to exist. Compton scattering is a strange exception to that.
Outstanding explanation
Glad it was helpful!
sir , I like you more then all things .
Wow, thanks
wonderful it made my concept really clear on compton effect
It is very useful thanks you sir
You are welcome.
Thank You! ❤️
Does the electron moving after collision have wavelength and frequency? Can we use De-broglie equation for momentum ?
Moving electrons do indeed exhibit wavelenght and frequency. That is not to be confused with Compton scattering.
In my view mass, charge, electric field & energy are different things. Mass,charge or electric field are not energy in themselves. Energy comes from motion of mass, charge or electric field. When a photon is absorbed by an atom, the photon transfers its KE to the electron and its electric field is absorbed by the atom. The KE & electric field is released when atom emits a photon. In compton scattering the photon transfer part of its KE but retains its electric field since electron cannot absorb its electric field. Electric field & mass are indestructible or convertible to energy. They can however carry & transfer energy. I think free space is an energy field with uniform energy density. The energy of this field comes from perpetual motion of sub microscopic particles which make up space.Imbalance in energy density induces force fields which cause the energy to flow from higher to lower density regions of space. Mass & charge change the energy density balance & induce gravitational & electric fields.Energy contained in gravitational and EM fields is energy borrowed from space. Energy extracted from these fields is ultimately returned back to space. My comments are not in line with current science.They are my conjectures.
What specifically are you disagreeing with? I couldn't determine that from your comment.
amazing videos
Lifesaver
You're welcome
thank you
Just one question why we wrote that the momentum of the photon is the same of the electron after the collision isn't it momentum of the system before equals momentum after the collision ?
Note that the equation indicates that the momentum in the y-direction is equal. Since the photon traveles in the x-direction, the momentum in the y direction is zero. Therefore after the collision the total momentum (in the y-direction) must be zero as well. That is why the momentum of the electron must equatl the momentum of the photon (in the y-direction) with opposite directions.
flawless, I don't know what to say
Hi sir, I have a question regarding Compton scattering. Why does intensity of graph of scatter wavelength decrease if we use heavier elements?
Is it because there are more electrons which interact with x rays or because havy elements have much stronger bond of electrons within the atom?
Thank you so much.
Regards from Jelena
Thanks man.
For what velocities would it me appropriate to use the relativistic equations of kinetic energy ? (gamma-1)mc^2
+Eric Golant
When the velocities are 50% the speed of light, your answer will be off by about 15%. At 86.6% of the speed of light, your answers will be off by 100%. Typically at velocities greater than 10% the speed of light, you may want to start using relativistic equations.
+Eric Golant v>0.1 c
Thank you so much for your help :)
awesome! very helpful :)
Hello! How come you used sinθ when finding the angle of the scattered electron?
We just used the equation: p (photon in the y direction) = p (electron in the y direction)
Oh right! Thank you
Would this be a lvl or uni lvl? Anyone knows?
The velocity calculated is only valid in classical physics right? Even if the relativistic effects at .1C would be "negligible"?
+Tommy Juszczyk
Yes, these methods are only valid for non-relativistic velocities. We must employ different methods for relativistic velocities.
Right but by "non-relativistic velocities" you mean, velocities where the difference is non-negligible right? Technically the relativistic equation is more accurate no matter the velocity (above zero) correct?
the formula for calculating the scattered wavelength is derived using relativistic equations. doesnt that mean we should be using relativistic formulas for kinetic energy and momentum?
The wavelengths are those of the photon. No relativistic equations needed for that.
no actually, i was reading the derivation of the wavelength formula and there the book used relativistic equations. but then i checked the formula given for calculating the angle of scattered electron and the value matched with the one obtained here, so there must be something i am misunderstanding. thank you for replying :)
Only if the electron moves at relativistic speeds ( > 10% the speed of light) do you need to worry about using the relativistic equations.
Hello sir,
why the Total of E is E(initial) - E(final)? Rather than E(final) - E(initial)? From my previous experience it was always Final - initial, Thanks
The initial energy is greater than the final energy, therefore to make the delta energy positive it is written that way.
The initial energy is scattered and separated into two different energy, the kinetic energy for the electron and the energy of the after collision photon. Both of these energy add up to the initial energy, making the initial energy bigger than them.
I tried to do it relativistically and got 74.9 degrees, but maybe thats partially due to rounding
That is relatively close.
What if you need to find the relativistic velocity? Can somebody either direct me to another video, text or let me know by replying on how to find it?
from which book these problms are ??
We make up many problems ourselves and we use many references for ideas.
i like your bow tie
Why i got the theta 73.05°? Am i wrong??😓
You are close. (I got 73.5)
Michel van Biezen im just realised that im using the value of the momentum of electron is 1.704×10^-23 kgm/s..TQ