How spin orbit coupling affects the nuclear potential
Пікірлер: 50
@ashleighnaysmith3618 жыл бұрын
I love how you sound a little bit like a kids tv presenter - much more engaging than regular tutorials! Even for undergrads! :D
@JenniferHL310 жыл бұрын
This is great stuff. Would have been very useful during my degree.
@LadyShaydex5 жыл бұрын
Thanks man real easy to follow and clear, helped a lot to work it out in my head
@uhgar43277 жыл бұрын
Sir you are awesome, please keep making videos... ur knowledge is much needed, thanks alot :)
@drdeveshprasadbhatt32137 жыл бұрын
Thanks for this nicely presented lecturer.
@sarahkanawati40716 жыл бұрын
I really should thank you sir, you helped me to pass my exam.
@DrPhysicsA6 жыл бұрын
Congratulations. Well done!
@alibachir67799 жыл бұрын
Brilliant explination!
@robinashaheen17138 жыл бұрын
great explanation, thanks
@animeshmondal12088 жыл бұрын
What a beauty!
@hannah2843 Жыл бұрын
thank you for this great video!
@getachewsolomon75796 жыл бұрын
at 19:05, i think, the anti-parallel alignment(l-1/2) is more “favourable”, has a lower energy. That is why l+1/2(maximum J) appears at higher binding energy in photoemission spectrum (XPS).
@ankushyerawar28487 жыл бұрын
thanks for the video
@baharasingh38405 жыл бұрын
I always watch your videos
@faizahussein63265 жыл бұрын
really wonderfull thank you so much
@baharasingh38405 жыл бұрын
Best ever explanation
@benmenanaazdine2782 Жыл бұрын
Thank you very much
@chowonjang22204 жыл бұрын
hey , finaly i understand what spin-orbit-coupling is . thanks
@PiranhaFlip6 жыл бұрын
im a little confused, why have you said that s(s+1) is always 3/4? when s could be +/- 1/2?
@savosia4997 жыл бұрын
What about the jj coupling? why it occurs? and in wich way is it different from the LS one?
@howardhall23009 жыл бұрын
Thank you very much for these excellent videos. At about 18.08 you explain how the positive LS term depresses the energy level. I,m confused why the positive LS term reacts this way with the negative values of the Potential Well . Again, many thanks.
@thenorup7 жыл бұрын
When LS is positive there is some energy "trapped inside" the coupling, and when it is negative it has "released the energy". In other words, when LS is positive it takes more energy to rip off the electron.
@fuzzylumpkin80304 жыл бұрын
What happens if you raise an electron and fill it’s previous energy level with a new electron
@crystalpeng44857 жыл бұрын
Thank you for the lecture. It helps me a lot! But I still have some questions, hope that someone could answer me, please. 1. Why does the j term have only one value when L=0? Don't we take account to the other j value when s=-1/2? 2. I don't quite understand how does the LS term work on the binding energy of the levels. What does the the value in LS term mean? And why the negative term means less contribution to the potential well? I have watched another video (shell model video) mentioned in this video, but still can't find these answers. Please answer my questions if you could. Thank you very much for your kindness!
@debajyotisg7 жыл бұрын
Actually, S has only one value (1/2); it's ms(components of s) which takes positive or negative values and make +1/2,-1/2. So j would still have only one value if l is zero.
@brendawilliams80622 жыл бұрын
It’s harder to locate the math on one side as a crystal.
@dipanshueminem9 жыл бұрын
Brilliant video. I wanted to ask, for l=2, what are the possible values of j?
@kaiserdostuff9 жыл бұрын
Dipanshu Gupta 5/2
@TheImpressionable8 жыл бұрын
+aknelkaiser Could it b 3/2 as well?
@kaiserdostuff8 жыл бұрын
+Stefan Epler - Snow Yes.
@kaiserdostuff8 жыл бұрын
+Stefan Epler - Snow however OP didnt said anything if n=3 or not. So i assume, he knew that much. If n=3, l = 2,1,0 and each j is equal l+s. so, there is 3/2.
@verumINscientia9 жыл бұрын
for the operators on these vectors wouldnt it be more proper (or more easily destinguished from other values) for those tems to have a "hat" or ( ^ ) above them?
@kaiserdostuff9 жыл бұрын
verumINscientia A lot of book don't even bother about putting those '^' above the operator, as people assume once you able to reach to this video, you already well known about operator.
@ramysaad10473 жыл бұрын
your always the best
@igorantoniazzi5 жыл бұрын
Awesome!!! thanks!
@brendawilliams80622 жыл бұрын
Thankyou
@baharasingh38405 жыл бұрын
Please keep it up
@GlodChip3 жыл бұрын
Are you sure +1/2 decreases the energy level and -1/2 increases it? Because everywhere else I look it is the other way round. Or maybe I get mixed up with notations? Does j-j coupling and LS coupling lead to different energy directions?
@AlirezaNabavian-eu6fz7 ай бұрын
I feel the same way..when it's positive it should reduce the energy level since it's negative therefore less energy is needed to release the electrons not more
@dpcon19948 жыл бұрын
Wait, if s = +- 1/2, why is s(s+1) = 3/4? s(s+1) = 3/4 is only the case for when s = 1/2, what about s = -1/2??
@Richard-rx8uo8 жыл бұрын
s is always 1/2, there is a point where he says that s=+1/2 or -1/2 in explaining why j can be l+s or l-s, but this is because j can take values from modulus(l-s) up to l+s in integer steps. As l=1 and s=1/2, j can be either 1/2 or 3/2. Hope that helps :)
@djangogeek6 жыл бұрын
quantum number s for the electron is always a half. But the z component projection m_s may be plus or minus a half. the s^2 operator only acts on the quantum number s and has eigenvalue s(s+1), the s^2 operator does not act on m_s which is why we do not consider the case for m_s equal to minus a half.
@EvaPev9 жыл бұрын
Your voice resembles that of Richard Dawkins. (in my ears at least) Good lecture.
@anweshaguha73663 жыл бұрын
Ikr I couldn't stop thinking about that
@AbhishekMahajan7 жыл бұрын
19.47 spin is -1/2 so there should be 1/4
@muntazerahmed5261 Жыл бұрын
I need someone who know how to use it CCFULL programe for the japanese scientists [ K. Hagino , N. Rowley and A. T. Kruppa ] called : A FORTRAN77 program for coupled-channels calculations with all order couplings for heavy-ion fusion reactions .
@ganquan19905 жыл бұрын
when s=1/2 the term of -s(s+1)=-3/4 and when s=-1/2 the term of -s(s+1)=+1/4,then you calculate LS term in l=0 will give you both 0. then when calculate l=1 LS term should be l*(h bar sqr/2) and -l*(h bar sqr/2). That should be right answer. I think professor make a small mistake here.
@arpitchoudhary64754 жыл бұрын
Sometimes you sound like Kane Williamson!
@haniefsofi4 жыл бұрын
Not a neat lecture. I think there is a mix up.
@henryjackson70509 жыл бұрын
Why call the video spin orbit coupling then go through really basic physics at the beginning that anyone who's watching this video will already know? and you don't even explain the physics of spin orbit coupling. what a waste of my time