Shell model determining spin and parity

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Homework Help Overview

The discussion revolves around determining the spin and parity of the calcium isotopes _{20}^{41}Ca and _{20}^{40}Ca using the shell model in nuclear physics.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants explore the implications of odd and even numbers of neutrons and protons in determining spin and parity. There is discussion about how the filled shells contribute to the overall spin and parity, with some questioning the calculations and assumptions made regarding the angular momentum of the particles involved.

Discussion Status

The conversation is ongoing, with participants sharing their thoughts on the contributions of filled shells to spin and parity. Some guidance on how to approach the addition of angular momenta is mentioned, but there is no explicit consensus reached on the final values for spin and parity.

Contextual Notes

Participants are considering the effects of closed shells and the rules for adding angular momenta, with some uncertainty about the parity calculations and the implications of odd versus even nucleon configurations.

Confundo
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shell model determining spin and parity - urgent

Homework Statement



Find the spin and parity of _{20}^{41}Ca and _{20}^{40}Ca

Homework Equations



http://hyperphysics.phy-astr.gsu.edu/hbase/Nuclear/shell.html


The Attempt at a Solution



For _{20}^{41}Cawe've an odd number of neutrons which will lie in 1f(7/2) level
so we have spin 7/2 and parity -

For _{20}^{40}Ca both the unpaired neutron and proon will be in the 1d(3/2) level which has parity +, do I need to add together the two spins in this level. So I'd get spin 3 with parity +.
 
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Thinking about it since the shell is full it can't contribute to spin.
 
Ca-40 has two closed shells, and since then it should have 0+ according to the shell model, right?
 
Confundo said:
Thinking about it since the shell is full it can't contribute to spin.

Yes, I think.

If we had something like _{21}^{42}X. Then we'd have an odd neutron and odd proton in the 1f_{7/2}, the spins would add and we'd have spin 7 and parity -1?
 
why negative parity?
 
Oh yeah. Should multiply the parities and thus get positive parity right? Basically the angular momentums are "parallel".
 
You can also have J_tot = 6,5,4,3,2,1,0.

Remember how to add two angular momenta.
 

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