Undergrad nuclear physics; spin parity rules

[grady]

Chapter 3 #17 (Krane) The spin-parity of ⁹Be and ⁹B are both 3/2^-. Assuming in both cases that the spin and parity are characteristic only of the odd nucleon, show how it is possible to obtain the observed spin-parity of ¹⁰B(3⁺). What other spin-parity combinations could also appear? (These are observed as excited states of ¹⁰B.)

Here's what I think I know. The parity is (-1)^l. So for ⁹Be and ⁹B, l = 1. It makes sense that these two nuclei have half-integer spin because A = 9 is odd. It makes sense that ¹⁰B has integer spin because A = 10 is even. The fact they're telling me to consider the spin and parity is charactersitic of the odd nucleon is supposed to be hinting at something, but I'm not sure what. I realize ⁹Be has an extra neutron and one less proton than ⁹B, but I don't know if I'm supposed to be getting any useful information from the fact that the unpaired particle is a proton in one case and a neutron in the other. If anyone can fill in any of these blanks for me, I would appreciate it.

Aside from this question, and in general, I'm confused about how nucleons pair off and have their spin cancel out with other nucleons. From what I've heard in class so far even numbers of nucleons should just pair off and leave nuclei with either 0 or 1/2 spin. :(

 

[quantumdude]

Originally posted by grady
I realize ⁹Be has an extra neutron and one less proton than ⁹B, but I don't know if I'm supposed to be getting any useful information from the fact that the unpaired particle is a proton in one case and a neutron in the other. If anyone can fill in any of these blanks for me, I would appreciate it.

I don't think there's anything to that, since J^P for both p and n are both (1/2)⁺.

Aside from this question, and in general, I'm confused about how nucleons pair off and have their spin cancel out with other nucleons. From what I've heard in class so far even numbers of nucleons should just pair off and leave nuclei with either 0 or 1/2 spin. :(

That would be true if only the nucleon spin were taken into account. However, the orbital angular momentum of the odd nucleon is seen from the outside as nuclear spin. So, for instance, if an odd proton is spin-up in the p-shell with m_l=1, then the whole nucleus is in a spin-3/2 state.

 

[M98Ranger]

The spin-parity rules in nuclear physics describe the allowed spin and parity values for a given nucleus based on its number of protons and neutrons. In the case of 9Be and 9B, both nuclei have an odd number of nucleons, which allows for half-integer spin values. The fact that the spin and parity are characteristic of the odd nucleon means that the spin and parity of the nucleus can be determined by looking at the odd nucleon alone.

In the case of 9Be and 9B, the odd nucleon is a proton in 9Be and a neutron in 9B. This means that the spin of 9Be is determined by the spin of the odd proton, which is 1/2. Similarly, the spin of 9B is determined by the spin of the odd neutron, which is also 1/2. In both cases, the parity is determined by the angular momentum of the odd nucleon, which is l = 1.

Now, let's consider the observed spin-parity of 10B(3+). This means that 10B has a spin of 3 and a positive parity, which corresponds to l = 2. Since 10B has an even number of nucleons, its spin and parity cannot be determined by looking at the odd nucleon alone. Instead, we need to consider the pairing of nucleons in the nucleus.

In 10B, there are four pairs of nucleons. Each pair consists of a proton and a neutron, which can combine to form a spin of either 1 or 0. In order to obtain a total spin of 3 for the nucleus, we need three of these pairs to have a total spin of 1 and one pair to have a total spin of 0. This results in a total spin of 3 and a positive parity.

Other spin-parity combinations are also possible for excited states of 10B. For example, 10B(2+) could have two pairs with a total spin of 1 and two pairs with a total spin of 0, resulting in a spin of 2 and a positive parity. 10B(1+) could have three pairs with a total spin of 1 and one pair with a total spin of 0, resulting in a spin of 1 and a positive parity.

In general, the spin and parity of a nucleus are determined by the combination of spins and

 

[M98Ranger]

To understand how it is possible to obtain the observed spin-parity of 10B(3+), we first need to understand the concept of nuclear spin and parity. In nuclear physics, spin is a property of nucleons (protons and neutrons) that describes their angular momentum. It can have a value of either half-integer (1/2, 3/2, 5/2, etc.) or integer (0, 1, 2, etc.). Parity, on the other hand, is a property that describes the symmetry of a nuclear state. It can have a value of either +1 (even parity) or -1 (odd parity).

In the case of 9Be and 9B, both nuclei have an odd number of nucleons (9). This means that they have a half-integer spin (3/2) as stated in the question. However, the parity of these nuclei is determined by the parity of the odd nucleon. In 9Be, the odd nucleon is a proton, which has a parity of +1. This gives 9Be a parity of +1. On the other hand, in 9B, the odd nucleon is a neutron, which has a parity of -1. This gives 9B a parity of -1.

Now, when we add an extra proton to 9Be to form 10B, the total number of nucleons becomes even (10). This means that the spin of 10B can have an integer value. In order to obtain a spin-parity of 10B(3+), we need to have a spin of 3 and a parity of +1. This can be achieved by pairing the extra proton with one of the protons in 9Be, resulting in a total spin of 3/2 + 1/2 = 2 and a parity of +1. This explains how it is possible to obtain the observed spin-parity of 10B(3+).

In addition to 10B(3+), there are other spin-parity combinations that can appear as excited states of 10B. For example, 10B can also have a spin-parity of 1+ or 2+. These can be obtained by pairing the extra proton with different nucleons in 9Be, resulting in different spin and parity combinations.

In general, the pairing of nucleons in a nucleus is