Nuclear shell model, spin and parity predictions

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SUMMARY

The discussion centers on predicting the spin and parity of the ground states of isotopes ##^{89}Sr##, ##^{97}Zr##, and ##^{137}Ba## using the nuclear shell model. The user initially calculated the spins and parities as ##7/2^+## for both ##^{89}Sr## and ##^{97}Zr##, and ##11/2^-## for ##^{137}Ba##, which differed from established values of ##5/2^+##, ##1/2^+##, and ##3/2^+## respectively. Participants highlighted the complexities of the shell model, noting that while it serves as a foundational approximation, exceptions exist due to spin-orbit coupling and pairing energy effects. They emphasized the importance of considering the entire incomplete shell rather than solely the unpaired nucleon.

PREREQUISITES
  • Understanding of the nuclear shell model
  • Familiarity with parity calculations, specifically ##\pi = (-1)^l##
  • Knowledge of spin-orbit coupling in nuclear physics
  • Basic concepts of nuclear magnetic moments and their implications
NEXT STEPS
  • Study the effects of spin-orbit coupling on nuclear states
  • Learn about the role of pairing energy in nuclear structure
  • Research the "incomplete shell" model and its applications
  • Explore experimental observations of nuclear magnetic moments and their significance
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Nuclear physicists, students studying nuclear structure, and researchers interested in the complexities of the shell model and its predictions.

Incand
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Homework Statement


Not exactly a homework problem but I tried to predict the spin and parity of (the ground states of)
##^{89}Sr##,##^{97}Zr## and ##^{137}Ba##
using the shell model and my results seem to differ from the tables.

Homework Equations


Parity
##\pi = (-1)^l##

Figure 4 seems to have the same values as the diagram in my book (Krane)
http://courses.chem.indiana.edu/c460/documents/SEC9ShellModel.pdf

I also found another diagram where some of the energy levels had switched place:
https://www.eng.fsu.edu/~dommelen/quantum/style_a/ntsm.html

The Attempt at a Solution


##^{89}Sr## have ##51## neutrons. Looking at diagram the unpaired neutron is in the ##1g_{7/2}## state. The parity is then ##(-1)^4=1## so we have ##7/2^+##.

##^{97}Zr## have ##57## neutrons. The unpaired neutron is according to the diagram in an ##1g_{7/2}## state so we have ##7/2^+##.

##^{137}Ba## have ##81## neutrons. The unpaired neutron is then in a ##1h_{11/2}## state. We then have ##11/2^-##.

If however I look up the values in table I have:
##^{89}Sr \; \; 5/2^+##
##^{97}Zr \; \; 1/2^+##
##^{137}Ba \; \;3/2^+##.

I'm wondering why I get different values? I suspect I have the wrong diagram, so far every prediction I did with odd number of protons have worked out but not all the ones for neutrons. Is there a difference?
 
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Incand said:
I'm wondering why I get different values? I suspect I have the wrong diagram, so far every prediction I did with odd number of protons have worked out but not all the ones for neutrons. Is there a difference?

i think there are a bit different placements of spin orbit splitted levels as seen by neutron and as by protons-
further the rules are general and there are exceptions- moreover the shell model is an approximation and better ones do exist...
i have seen one assignment which you may also look up...ref is given below.

with that i get your odd neutron in N= 51 placed in 2d5/2 a 5/2+ state

N=57 goes to 1/2+ as it opts for 3s1/2 state - the particle tries to go to lowest j value state-leaving option of higher j-state (it has been observed)

N=81 goes to 2d3/2 state 3/2+

ref.
PDF]http://www2.ph.ed.ac.uk/~dwatts1/np3_Lecture8.pdf
www2.ph.ed.ac.uk/~dwatts1/np3_Lecture8.pdf

Shell Model predictions. & single-particle ... J = total angular momentum (nuclear spin) π = parity ... Determine spin and parity assignments for the ground states ... Fornuclei with an odd neutron or proton shell model predicts two possible.
 
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I think I have misunderstood the shell model. I thought the electrons filled up the lower energy states first in order. I get both the examples and exercise in your link correct. But I don't understand what you did with ##N=57## and ##N=81##. Is this is a general rule that they should fill the lowest ##j## state first or an exception? This doesn't seem to always be the case.

I thought they filled the lower energy levels first so the the ##N=57## nuclei would fill the ##2d5/2## level with ##6## neutrons and have one left for the ##1g7/2## level.
 
Incand said:
Is this is a general rule that they should fill the lowest jjj state first or an exception? This doesn't seem to always be the case.
Let me correct myself the N=57 (101Ru) state will be 5/2 rather than 3/2 which i wrote.

i do not think its a general rule but by experimental observation of nuclear magnetic moments such behavior is seen.'

if you analyze the magnetic moments one gets a pair of Schmidt's lines for magnetic moment variation with z or N number and most of the nucleons finds a place between those lines..

the pairing energy of the nucleons also plays the role that higher j value states get filled up first and the extra neutron/proton goes to lower j-value state.

e.g. at N=33 for 61Ni the spin is found to be 3/2 and not 5/2 which should be on the basis of diagram

for N=59 (105Pd) observed spin is 5/2 and for N=61 (109Cd) also it is 5/2.
though the higher j value state 1g7/2 is first being filled in pairs.

in certain regions the deformation of the nuclear shell becomes significant and residual inter nucleon interaction plays a role.
so the shell model (as the simplest) can not present the full explanation of experimental values
 
Last edited:
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Thanks for explaining! It was quite surprising to see that ##101Ru## have different spin from ##97Zr##. I seem to have lot more reading to do before I understand this. It seems what I've been using is only what my book calls the "extreme independent particle" model where only the unpaired nucleon is considered. Apparently I can get better results considering the entire "incomplete" shell.
 
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