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nicedazed
Apr3-11, 04:51 PM
1. The problem statement, all variables and given/known data

Oxygen 17, work out j (nuclear spin)


Hi. Assume oxygen 17 atom. j is supposed to be 5/2. How does one work this out? from what i gather, l for protons is 1, and l for neutrons is 2. How does one work out j, and what would s be?

Partly what I'm confused about is for the protons l=1 but neutron l=2? what l do you use?

Thanks :devil:

2. Relevant equations


This is for the nucleus by the way. 8 protons and 9 neutrons


3. The attempt at a solution

ideasrule
Apr3-11, 06:07 PM
l doesn't have to be either 1 or 2. Rather, for all nucleons l can range, in increments of 2, from 0 to n for even n and 1 to n for odd n. For example, for n=5 l can be 1, 3, or 5.

However, what you really want is j=l+s. For this question, all of the protons pair off, so they contribute no angular momentum. All but one of the neutrons pair off, and the remaining neutron determines the angular momentum of the whole nucleus. So, what is j for that neutron?

nicedazed
Apr3-11, 06:24 PM
Parity= (-1) ^l

l=2, to parity=+.

s=5/2 for the neutron.

so is this is the spin? what is j?

as parity =+ then j=2+(5/2)?= 9/2?

I'm sure that's not right...

if anyone can tell me what I'm doing wrong it'd be massively (unlike a photon) appreciated! I know this is a very easy question, it's just that other links such as wikipedia are confusing the hell out of me.

thanks for the reply :)

ideasrule
Apr3-11, 07:15 PM
j is the total angular momentum, or "nuclear spin".

Where did you get that s=5/2 for the neutron? It's not; both the proton and the neutron has a spin of 1/2. Besides that, you have the right answer.

nicedazed
Apr3-11, 07:36 PM
Now it makes sense, thanks a lot. My lecture notes are slightly confusing.

There's just one thing I don't get:

Let's add a neutron: 18O. On Wikipedia, it says the spin is 0.

Once again, parity = +1. The additional spin is another +1/2.

Now, using the formula j=l+s, =2+1=3.

BUT, am I correct in thinking that because there's an even number of neutrons, they contribute no spin?

Thanks once again!

ideasrule
Apr3-11, 08:16 PM
Yes, you're correct. Any nucleus with an even number of protons and even number of neutrons has no spin.