You said completely opposite johnsnow Kindly explore US university and courses(not only MIT) before passing such Bullcrap conclusion Stanford, MIT, UCB, Yale all have fantastic profs
Thanks to honourable Dr. HC Verma for his valuable lectures on nuclear physics. I m a student of bs.c. 6th sem; dptt. of Physics of North Lakhimpur College (autonomous). I find these lectures very much useful for better understanding of nuclear physics. These lectures can be understood easily if we concentrate a little. I hope the step taken by iits for better understanding and spreading of good knowledge will be a great corner stone in the educational history of india and the entire globe.
This was about the first time I felt I could grasp electron scattering on a nucleus. The one thing you have to keep in mid: the statements about the relation between input psi wave and output psi waves are only summary terms. Both input and output are considered plane waves, that is, we look at the experiment from far away. That H´, the integral of interaction energies is used as an operator (-density) in the QM treatment, or as a kernel. Then he writes down without much drama what we should expect for the probability distribution depending on input and output direction, k and k´. I was kind of surprised at first how this can be compared to light going through a small aperture but it really makes sense, when I tell myself that there is no such thing as solid little balls colliding. The nucleus is simply a region with high positive charge density. An electron can just go straight through it with a little deflection. The highest probability is that it will right go through the center without energy change in summary. When it approaches the nucleus it will actually accelerate and when it exits it will decelerate by the same amount. No energy gets lost of those 420 MeV. Remember: there is nothing crashing together. This type of Scattering is not crashing, only redirecting. Hope i can understand how you solve this complicated inverse problem to find the proton density distributions. I suppose that’s the next lecture. Now of course I wonder what it would be like if one could use positrons instead in the experiments. From the math it seems that nothing should be different ( as long as the positrons don’t decide to interact with the neutrons to do some weird stuff?)
A better teacher than the very honourable Dr. Pervez Hoodbhoy, who keeps his teaching level so high that average students like me find it impossible to catch up!
The nuclear radius formula, that I have recently derived in my research, is this: R = cube root of 4GKM/πc^2 where G is gravitational constant K is Coulomb constant M is the mass of nucleus (mass number A * 1.66054 * 10^-27). So I have demonstrated that in the nucleus does not operate the strong nuclear force, but the gravito-electric force (F = 4GKM^2/πR^4). Here you can find my paper where I demonstrate how to derive the nuclear radius formula: vixra.org/abs/1806.0433
thanks, i am simply amazed at a kind of maths that is required to model the nucleus ,the size of which is of the order of few femtometrs. this is wonderful exposition.
Prof.aquired explanatory skill in the grass root level and excels in all aspects.Tamil agathia muni is always great.Electricity is movement of electrons whereas light is emmision of electrons energy as photons when excited state so photons ie light energy has lower wavelength and energy unlike electrons.
in elastic collision energy is transfered as it is. so electrons encountered in detector must have high energy. and in inelastic collision energy must be less
@@vikasgupta8948 Apart from electron waves there are no other suitable form of radiation whose wavelength is comparable to size of a nucleus. light with wavelength much greater than size of nucleus will not show any diffraction pattern so it's size can not be measured. An important condition for diffraction.
The electron, approaching the proton, will have kinetic energy and potential energy. When it is far away, it will have a relatively huge amount of potential energy. As it moves towards the proton, it loses some of the potential energy. Some of it is radiated away, as electromagnetic energy. Some of it is converted to kinetic energy. Kinetic energy keeps an electron hopping, and keeps it from staying in a nucleus and combining with a proton.
sir at 54 min you were discussing about the charge distributions but how you intervened mass distribution into it ( you were saying how the mass is distributed based on charge) ? please kindly explain it sir.
while going to mass density from charge density we must know the number of total protons and neutrons(i.e. A and Z). At this stage how did we knew the exact numbers? More importantly how do we get any information about neutrons from experiments as it is neutral and should not respond to scattering?
Sir just a small doubt we know now that nuclear forces are a different category of forces so the hamiltonian for them should be different rather than electrostatic force at 39:19
sir you say that electrone of oxygen have same energgy as incident e have , and you are sying e scatterfrom nuclius also have energy abot 420 mev , then how we will differensiate between these two
43:49 he is writing the expectation value of the energy H' not the probability. Because where is the squared energy hamiltonian in the integral. Also he had to switch i and f since H' will act on psi i, not on the final state. And maybe alpha particles where a better choice than electrons. thanks anyway.
No bro, its not the expectation value it is infact the probability, expectation value is calculated by taking the inner product of the same function, he has taken the product on different functions which gives the probability (hint- it is like calculating the coeficient in eigenvalue expansion), also since all H does is multiplication we can change its order whatever way we want.
@@viveksingh483 I agree with you, but I think what he actually wrote in integral form was the probability amplitude, not the probability, if he mod-squares that whole integral (like that coefficient sqaured), we'd get the probability
@@fatimaalishah1881 yahh, it's quite difficult to separate proton Or Neutron. But is this the only reason or there is something in charge that may play any role here?
#ALOK verma Actually it a empirical relation brought more correctly than just -lambda by D Therefore the factor 1.22 is added.more clearly it have value sin®=A¥/D-B(¥/D)`3+C(¥/D)`5..... Hence other factors neglected and coefficient A is given calculated value 1.22
Nice question, This is what I think about it bats use rwflection property of ultrasound waves and since reflection property of sound is similar for waves and particles (roughly) the bat uses it to find distance between it and its prey
![PASULOL รามเกียรติ์ ตอนที่ 4 นนทกพบรัก [Ramakien Ep.4: Nonthok in love]](http://i.ytimg.com/vi/tMUQgmZzf_A/mqdefault.jpg)
This man has been blessed by Almighty God with the gift of teaching. No one here in the US teaches with such perfect clarity.
True, unless you are from MIT
@@girishtripathy3354 yes there is a person named walter lewin unlike hc verma 😄🙏
You are Right. There is No
Spoon-feeding in the US.
You Read a Book & try to
Understand it on your Own. 🙂
Who is nothing but a sexual predator@@b_shakti_1375
You said completely opposite johnsnow
Kindly explore US university and courses(not only MIT) before passing such Bullcrap conclusion
Stanford, MIT, UCB, Yale all have fantastic profs
Thanks to honourable Dr. HC Verma for his valuable lectures on nuclear physics. I m a student of bs.c. 6th sem; dptt. of Physics of North Lakhimpur College (autonomous). I find these lectures very much useful for better understanding of nuclear physics. These lectures can be understood easily if we concentrate a little. I hope the step taken by iits for better understanding and spreading of good knowledge will be a great corner stone in the educational history of india and the entire globe.
why we are using the electron to estimate the size of nuclear? electron attract the nucleus
@@vikasgupta8948 nucleus attracts electron** coz of diffrence in 'rest' mass..
These are probably the best lectures for nuclear physics on TH-cam.
Is there any superior word than excellent. Sir deserves that.
#makemescientific
Sir we are too lucky to have a diamond like you
This was about the first time I felt I could grasp electron scattering on a nucleus. The one thing you have to keep in mid: the statements about the relation between input psi wave and output psi waves are only summary terms. Both input and output are considered plane waves, that is, we look at the experiment from far away. That H´, the integral of interaction energies is used as an operator (-density) in the QM treatment, or as a kernel. Then he writes down without much drama what we should expect for the probability distribution depending on input and output direction, k and k´. I was kind of surprised at first how this can be compared to light going through a small aperture but it really makes sense, when I tell myself that there is no such thing as solid little balls colliding. The nucleus is simply a region with high positive charge density. An electron can just go straight through it with a little deflection. The highest probability is that it will right go through the center without energy change in summary. When it approaches the nucleus it will actually accelerate and when it exits it will decelerate by the same amount. No energy gets lost of those 420 MeV. Remember: there is nothing crashing together. This type of Scattering is not crashing, only redirecting. Hope i can understand how you solve this complicated inverse problem to find the proton density distributions. I suppose that’s the next lecture. Now of course I wonder what it would be like if one could use positrons instead in the experiments. From the math it seems that nothing should be different ( as long as the positrons don’t decide to interact with the neutrons to do some weird stuff?)
Every concept of Kenneth's book is resolved with detailed.
best teacher I'm thankful
Sir very fabulous explanation every students will have work to make our country proud
Wow, this playlist is pure gold!
Thanks to Mr. H C Verma for lecture of nuclear physics. I am preparing for lecturer of physics
A better teacher than the very honourable Dr. Pervez Hoodbhoy, who keeps his teaching level so high that average students like me find it impossible to catch up!
The nuclear radius formula, that I have recently derived in my research, is this:
R = cube root of 4GKM/πc^2
where G is gravitational constant
K is Coulomb constant
M is the mass of nucleus (mass number A * 1.66054 * 10^-27).
So I have demonstrated that in the nucleus does not operate the strong nuclear force, but the gravito-electric force (F = 4GKM^2/πR^4).
Here you can find my paper where I demonstrate how to derive the nuclear radius formula: vixra.org/abs/1806.0433
This sounds brilliant. I will read the entire 250 pages and how should I contact you, sir?
Thankyou sir. I will look into it
thanks, i am simply amazed at a kind of maths that is required to model the nucleus ,the size of which is of the order of few femtometrs. this is wonderful exposition.
This is a wonderful lecture on nuclear physics. One of the best teachers of physics.
Thank you Dr. HC Varma sir for your excellent lecture on nuclear physics
High clarity in depth of science session sir. Thank you for your wonderful lecture.
A very informative and wonderful lecture. Thanks to Hon. Verma Sir
perfect clarity of subject
Best lecture in nuclear physics. I have found lot of excitement to listen further lecture
Thank you sir for reminding me the important concepts in Nuclear Physics clearly.
Excellent Lectures on Nuclear Physics. Thanks Sir.
I am very happy to be listening to be this lecture given by you sir. You are the best teacher I have ever had, Sir. Thank you very much Sir.
Prof.aquired explanatory skill in the grass root level and excels in all aspects.Tamil agathia muni is always great.Electricity is movement of electrons whereas light is emmision of electrons energy as photons when excited state so photons ie light energy has lower wavelength and energy unlike electrons.
very much benefited in terms of fundamental ideas
Dr. HC Verma, thank you.
Thank you for such a wonderful explanation
Really very informative lecture on nuclear physics
Thank u so much sir for this marvelous series of nuclear physics 🌸🌸🌸💯💯🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🌹🌹🌹🌹🌹🌹
Thanks sir🎉❤
Great way to teach nuclear physics. The subject is explained at the grass root level.
Nice lecture from the eminent professor H C verma
Very informative and useful session on nuclear physics thank you sir
Detailed lecture. Thanks Professor.
Thank u sir for giving nice presentation Verma sir
Legend of physics ❤️❤️
U r great , sir...
Thank u so much, sir
It's awesome
Very clear explanation 🙂
Thank you for the wonderful presentation
very nice and informative session. Thank you sir
Crystal clear explanation
Very well explained!
Thank you sir for your excellent lecture on nuclear physics
Excellent explanaion. Thank you Sir.
Really very informative and useful lecture
Very nice presentation. Thanks
Very much important session
Best lecture, thanks sir
Very nice presentation Sir
धन्यवाद
These are really very nice lectures thanks a lot sir :)...
Excellent Explanation.
Thank you sir for for the wonderful lecture.
Excellent presentation
in elastic collision energy is transfered as it is. so electrons encountered in detector must have high energy. and in inelastic collision energy must be less
why we are using the electron to estimate the size of nuclear? electron attract the nucleus
@@vikasgupta8948 Apart from electron waves there are no other suitable form of radiation whose wavelength is comparable to size of a nucleus. light with wavelength much greater than size of nucleus will not show any diffraction pattern so it's size can not be measured. An important condition for diffraction.
Very informative talk sir
Thank u sir your excellent lecture
Thank you sir for wonderful session
You are amazing
very clearly explained sir
NICE PRESENTATION THANKS
Nice lecture sir.
Thank You Sir
How an electron beam being-vely charged can be scattered by a nucleus of +ve charge.
Fruitful
Very helpful lecture 👍
Very nice lecture
Thankyou sir, sir why doesnot the electrons get attracted by the nucleus rather than scattering.
Informative
You are great sir.....
Why are the high energy electrons not being absorbed by the nucleus? Nucleus is positively charged and electrons are negatively charged.
The electron, approaching the proton, will have kinetic energy and potential energy. When it is far away, it will have a relatively huge amount of potential energy. As it moves towards the proton, it loses some of the potential energy. Some of it is radiated away, as electromagnetic energy. Some of it is converted to kinetic energy. Kinetic energy keeps an electron hopping, and keeps it from staying in a nucleus and combining with a proton.
sir at 54 min you were discussing about the charge distributions but how you intervened mass distribution into it ( you were saying how the mass is distributed based on charge) ? please kindly explain it sir.
THANK YOUY SIR
Very interesting lecture
Love it
while going to mass density from charge density we must know the number of total protons and neutrons(i.e. A and Z). At this stage how did we knew the exact numbers? More importantly how do we get any information about neutrons from experiments as it is neutral and should not respond to scattering?
Thanks Sir
Collision between electron and electron is elastic or inelastic??
H.C. Verma is god
If I get sound waves over a wall which is more than a meter then can I detect the size of the wall by just hearing the sound ???
How can I find pdf of ur vedio lecture?
Sir just a small doubt we know now that nuclear forces are a different category of forces so the hamiltonian for them should be different rather than electrostatic force at 39:19
sir you say that electrone of oxygen have same energgy as incident e have , and you are sying e scatterfrom nuclius also have energy abot 420 mev , then how we will differensiate between these two
My question too
48:08
43:49 he is writing the expectation value of the energy H' not the probability. Because where is the squared energy hamiltonian in the integral. Also he had to switch i and f since H' will act on psi i, not on the final state. And maybe alpha particles where a better choice than electrons. thanks anyway.
No bro, its not the expectation value it is infact the probability, expectation value is calculated by taking the inner product of the same function, he has taken the product on different functions which gives the probability (hint- it is like calculating the coeficient in eigenvalue expansion), also since all H does is multiplication we can change its order whatever way we want.
@@viveksingh483 I agree with you, but I think what he actually wrote in integral form was the probability amplitude, not the probability, if he mod-squares that whole integral (like that coefficient sqaured), we'd get the probability
Sir why we are using only electrons for this? We Can Use proton Or neutron.
How could you segregate proton or neutron from nucleus ? By nuclear fission?
@@fatimaalishah1881 yahh, it's quite difficult to separate proton Or Neutron. But is this the only reason or there is something in charge that may play any role here?
@@fatimaalishah1881
Don't worry we have nice source of nutrons that is 11Na23
But we can't allign them
The estimate for the charge density came out of the assumption that the interaction is entirely coloumbic?
is there any relation between atomic density and nucleus size in nuclear and other elements
6:00.....i put this time stamp bcz i stop it here ...i could play when I visit here that's y
nice music.
🙏🙏🙏
lecture is fine but one doubt is there how to come 1.22 factor in relation given by sin theta = 1.22 LAMBDA BY D
#ALOK verma Actually it a empirical relation brought more correctly than just -lambda by D
Therefore the factor 1.22 is added.more clearly it have value
sin®=A¥/D-B(¥/D)`3+C(¥/D)`5.....
Hence other factors neglected and coefficient A is given calculated value 1.22
Is nuclear act as a aperture or opaque for high energy electron wave?
Opaque obviously! And diffraction would still occur quantum mechanically. Not sure will it have the same form as the the classical aperture or not.
@@mohammedsrivastava5917 I did not understand what you want to say. Can you please elaborate?
i dont understand, how interference pattern is formed by a single slit.???? two slits are necessary for the interference of waves......
I think it's a big slit (size of b). So, it's like lots of small slits.
Can one even tell the width of a object around 34cm from sound waves?
that's what the SONAR does..which detects the size distance and depth
17:19 Hofstadter expt?
But can't a bat tell the difference from the sound wave?
Nice question, This is what I think about it
bats use rwflection property of ultrasound waves and since reflection property of sound is similar for waves and particles (roughly) the bat uses it to find distance between it and its prey
49:00 there r three elephant taking selfie
Hello
Shouldn't the energy be E =pc
Why is it E=√(p^2c^2+m^2c^2)
E=pc is for only particles having no rest mass. Also 2nd expression is wrong written by u.
attendance in 2023
Very informative lecture