Understanding Nuclear Spin: Odd Proton and Neutron Counts

[Gavroy]

hi

i have a question about nuclear spin

if the number of protons and neutrons is even, then my chemistry book says, that the nuclear spin is always zero.

i guess that the reason is, that there are always 2 protons and 2 neutrons in one state with opposite spin and therefore the overall spin is zero, so that's all right for me.

but then:

if the number of protons and neutrons is odd, then my chemistry book says, that the nuclear spin is some number like 1,2,3,4 and so on.

how so?

i mean, if i have let me say 19 protons and 9 neutrons, then there are 18 protons that give me zero spin(after the first rule) and 8 neutrons that give me zero spin(also after the first rule)

therefore i would say, that this proton and neutron could give me only 1, but not 2 or more...or are these rules not correct or where am i wrong?

 

[Bill_K]

Gavroy, For an even-even nucleus the spin is zero, without exception. For other nuclei you need to consider not just the spin of the unmatched nucleon but its orbital angular momentum as well, and their combination j = l + s.

For an even-odd or odd-even nucleus the spin of the nucleus is equal to the j-value of the unmatched nucleon.

For an odd-odd nucleus things are not so predictable. The j-values of the unmatched neutron and proton will couple to form the nuclear spin, which must therefore be in the range |j_p - j_n| ≤ J ≤ j_p + j_n. There is a set of guidelines called the Nordheim rules that often give the correct spin, but there are exceptions.

 

[bbbeard]

hi

i have a question about nuclear spin...

if the number of protons and neutrons is odd, then my chemistry book says, that the nuclear spin is some number like 1,2,3,4 and so on.

how so?

i mean, if i have let me say 19 protons and 9 neutrons, then there are 18 protons that give me zero spin(after the first rule) and 8 neutrons that give me zero spin(also after the first rule)

therefore i would say, that this proton and neutron could give me only 1, but not 2 or more...or are these rules not correct or where am i wrong?

First, keep in mind that almost all odd-odd nuclides are unstable. There are exactly four stable odd-odd nuclides; they are the lightest possible: [itex]^{2}_{1}H[/itex], [itex]^{6}_{3}Li[/itex], [itex]^{10}_{5}B[/itex], and [itex]^{14}_{7}N[/itex]. All four stable odd-odd nuclei have J=1. All other odd-odd nuclei are unstable.

Among the unstable odd-odd nuclides, there are a few that have J=0. Most of these have short half-lives (under a second). Thumbing through my http://www.nndc.bnl.gov/wallet/wallet05.pdf" , the longest half-life for an odd-odd nuclide with J=0 is [itex]^{170}_{71}Lu[/itex], which has a half-life right around 2 days.

As Bill K points out, these nuclides can have states with higher orbital angular momentum, so that J=1, 2, 3... are possible. But these excited states decay pretty quickly, I think...

BBB

 

[Astronuc]

Eu-156 (Z=63, N=93) has J^π = 0+ and half-life = 15.19 days.

Bi-208 (Z=83, N=125) has J^π = 5+, half-life = 3.68E+5 y.

La-138 (Z=57, N=81) has remarkable stability. J^π = 5+, half-life = 1.02E+11 y. Either side of La-138 are two stable nuclides, Ba-138 (stable) and Ce-138 (t_1/2 = ≥ 0.9E+14 y).

Ta-180m (Z=73, N=107) is also remarkably stable with J^π = 9-, half-life = > 1.2E+15 y, while the ground state, Ta-180 (J^π = 1+) has a short half-life of 8.154 h.

Thanks to bbbeard for pointing to the BNL pocket/wallet chart of nuclides.

 

[Vanadium 50]

Ta-180m is unusual because its decay has never been observed.

V-50 has a very long half life (over 10¹⁷ years). The last time I checked, its decay had never been detected directly and its lifetime is inferred from geochemistry.

 

[bbbeard]

Eu-156 (Z=63, N=93) has J^π = 0+ and half-life = 15.19 days.

Ah, yes, thanks. I had missed that when skimming the wallet cards...

BBB