Nuclear Spin
Each nuclear state is assigned a unique “spin” quantum number 𝑰, representing the total
angular momentum (orbital plus intrinsic) of all the nucleons in the nucleus. The vector 𝑰
can be considered the sum of the orbital and intrinsic contributions to the angular
momentum
𝑰=
𝒍𝑖 +
𝑖
𝒔𝑖
𝑖
=𝑳+𝑺=
𝒋𝑖 = 𝑱 𝑜𝑓 𝑛𝑢𝑐𝑙𝑒𝑢𝑠
𝑖
the quantum number 𝐼 has the usual connection the vector 𝑰 :
𝑰=
𝐼(𝐼 + 1) ℏ
For nuclear ground states, there are several rules for determining the spins :
1- All even−𝑍, even−𝑁 nuclei have 𝐼 = 0. This results from the strong tendency of
nucleons to couple pairwise to zero spin.
2- In odd−𝐴 nuclei, the net spin is almost always determined by the 𝑗 of the last odd
particle, with the remaining 𝐴 − 1 nucleons (having even numbers of protons and
neutrons) pairing to zero spin as above.
3- In odd−𝑍, odd−𝑁 nuclei, the spin is determined by the vector coupling of the 𝑗 of the
odd proton and neutron, 𝑰 = 𝒋𝒑 + 𝒋𝒏 , and thus any of several values are possible. To
determine which of these possible couplings will be the ground state, we use the empirical
rule that the ground state is usually the coupling with the proton and neutron intrinsic spins
𝑠𝑝 and 𝑠𝑛 parallel. As an example, consider 38Cl , which consists of a 𝑑3/2 proton coupled
to an 𝑓7/2 neutron. For the proton, 𝑙𝑝 = 2 and thus 𝑠𝑝 is opposite to 𝑗𝑝 . For the neutron,
𝑙𝑛 = 3 and 𝑠𝑛 is parallel to 𝑗𝑛 as shown in figure (2.9). If we putting the occupation
number of nucleons on Fig. (2.9) we get the Fig. (2.10) ,
we get 𝑰 = 𝒋𝒑 − 𝒋𝒏
momentum) of
38𝐶𝑙 .
or 𝑰 =2, which is in fact the ground-state spin (total angular
(The first excited state is 𝑰 = 5, corresponding to 𝑰 = 𝒋𝒑 + 𝒋𝒏 )
On the other hand, consider 50𝑆𝑐 , resulting from an 𝑓7/2 proton coupled to a 𝑝3/2 neutron
as shown in Fig. (2.11).
Here making 𝑠𝑝 and 𝑠𝑛 parallel also makes 𝑗𝑝 and 𝑗𝑛 parallel, and thus
𝑰 = 𝒋𝒑 + 𝒋𝒏 = 5, in agreement with observation. (The state 𝑰 = 𝒋𝒑 − 𝒋𝒏 = 2 is a low
excited state of 50Sc).