This lecture provides information on nuclear force and nuclear models. The strong force is introduced through isospin. A comparison of exchange particles is provided. The use of mirror nuclei to examine the strong force is presented. An overview of nuclear potentials is provided and used to discuss the shell model. States of the shell model and their relationship to magic numbers are discussed. Use of the shell model is determine nuclide spin and parity is presented. The relationship between spin and parity with nuclear deformation is introduced with Nilsson diagrams. Additional information on Nilsson diagrams can be found in the Table of the Isotopes. An introduction of the Fermi model for energetic nuclei is given.
This was a great lecture. The section on the Nilsson Diagram and showing how to use it to determine if the nuclei is deformed based on the measured spin and parity was very helpful.
ReplyDeletethanks for the comments.
DeleteIt is interesting to see the somewhat conflicting models of the nucleus more fleshed out in this lecture, as the more general descriptions of the atom from high school chemistry leave the impression that the nucleus is a simple, well-understood concept compared to the electron which is usually described as the more exotic and complicated component of the atom.
ReplyDeleteThe nuclear models do change as the system becomes more complex. Earlier we discussed a simple liquid drop model. This had no shell configuration. Adding shells made the model more relevant. If the nuclei is non-spherical or at a higher energy then additions are needed.
DeleteGreat observation.
I do not understand how to calculate spin and parity using the shell model. Let's take for instance 7Li. I get the number of protons and neutrons and that spin and parity are dependent on the odd proton. I do not know what to do from there.
ReplyDeleteHi Daniel,
ReplyDeleteSo you're going to want to pull up the Shell Model from the Powerpoint (the one that appears on slides 6,7,9,10,11, and 12. Any will do) The spin and parity of 7Li will depend on that of its single unpaired proton, as you mentioned, so you will not need to consider the neutron shells as they are distinct from proton shells.
With 3 total protons, the first two will completely fill the 1s orbital. The final proton will reside in the next orbital above, 1p 3/2. The 3/2 is the spin of this proton, and therefore of 7Li. It is also in an odd shell, which corresponds to a negative parity giving a total spin and parity result of 3/2- for 7Li.
As you determine the spins and parities for the PDF quiz, however, keep in mind that this method assumes a spherical nucleus. Nuclei that are in reality deformed along either their polar or equatorial axis will have a completely different experimentally-derived spin and parity state. Consult the Chart of the Nuclides to compare the known states of each isotope to the state you theorized assuming a spherical nucleus. If there is a discrepancy, you can conclude that the nucleus in question is not a perfect sphere.
Hope that helps.
It helps a lot. Thank you Tyler.
DeleteAwesome Tyler!!
DeleteThis was a very good lecture. I am beginning to learn more about how the different types of information given in the Chart of the Nuclides pertain to an element. In the beginning, I was confused on what these type of information were. It is great learning where they all come from. The shell model diagram and the examples really helped me in figuring out how to calculate the spin and parity. The example that helped me the most was how to determine the spin and parity for odd-odd nuclei.
ReplyDeletethanks for your comments. I am glad to see the information is coming together.
DeleteThis lecture was one of the more difficult ones for me. Once again, it was really helpful to be able to go back through the lecture and work through the examples. Also, thanks to Tyler for his comment above. That helped as well!
ReplyDeletethanks for the comment. Let me know if more examples would be helpful.
DeleteThe examples for the shell model and blog comments were helpful in understanding the material in this lecture.
ReplyDeleteThAnka for your comments. I may add more examples to future lectures
DeleteThe lecture was interesting to learn but was a difficult one to grasp compared to the others, but the challenging a lecture is the more interesting it is to learn which I am thankful for.
ReplyDeleteThanks for the comment Ryan. You are correct this is a difficult topic but I am sure you can demonstrate some knowledge on the subjects
DeleteThis lecture seemed very easy on the surface, but was actually quite difficult to understand. I would love if you could explain how to calculate this a little bit more than the explanation above using unusual element examples. Thank you in advance!
ReplyDeleteIf you provide an example of what you would like calculated I will provide it.
DeleteI really enjoyed this lecture. It was very interesting to examine protons and neutrons using some of the same quantum numbers as I’ve used for electrons in the past. I have a couple of questions, though. Firstly, when filling shells in the shell model, is there anything akin to Hund’s rule for electrons that applies to the nucleons? Also, it was stated that the neutron potential in the Fermi Gas Model is a step function; can it be classified as a Heaviside step function? I’m trying to relate this material with some of the stuff I learned in my differential equations class, but it’s been a while.
ReplyDeleteThe nucleons will align in pairs. In electron configurations one can observe a number of unpaired electron. For a neutron or proton only one will be unpaired if odd.
DeleteThe lecture was quite good, the section on Nordheim numbers helped a lot.
ReplyDeleteThanks for your comment
Deleteexample from PDF quiz 8
ReplyDeleteAn example for 209Po is below.
Po has Z=84 and N=125. The 125th neutron is unpaired and will determine the spin and parity of the nucleus. Evaluating the shell model shows the 1i13/2 shell is filled with the 126th nucleon. This means the 125th, unpaired neutron sits in this shell. The spin is therefore 13/2. The parity from i is:
0 1 2 3 4 5 6
s p d f g h I
l=6, this is even so the parity is +
The expected spin is 13/2+. This should be compared to the actual value.
Dr. Czerwinski,
ReplyDeleteSorry for taking so long to respond on the blog.
I had interviews last week for medical school.
Just finished the lecture, and I will do the quiz.
Marko
no worries about the delay. I know you were going to do this. I hope it went well.
Deletethis was the most difficult lecture by far to grasp. Sorry for the delay in the response. I have been having a hard time posting on the blog. I keep getting error messages when trying to post.
ReplyDeleteSorry to hear you have had problems posting. If anyone else had this issue please let me know.
ReplyDeleteThe concept of nuclear models is by its nature somewhat difficult. You can use the PDF quiz to see how much you retained.
So I'm sorry about all the issues I'v had submitting things, I will comment on all these blogs though and email you the actual PDFs, because they don't seem to be sending
ReplyDeleteThis was one of my favorite lectures because I was able to add up to the knowledge that I acquired about shell models back in general chemistry. The practice problems in the quiz were fun to do by using the Nilsson diagram.
ReplyDelete