- #1

- 38

- 3

## Main Question or Discussion Point

I was trying to calculate the ##^{19}_{9}F_{10}## nuclear spin using the nuclear shell method. From what i know the nuclear spin for a nucleous with odd ##A## is the total angular momentum ##J## of the stand-alone nucleon:

P: ##(1d_{5 \backslash 2})^1## (stand-alone nucleon)

N: ##(1d_{5 \backslash 2})^2##

which in this case is a proton with ##J=\frac{5}{2}## so it leads to a nuclear spin of ##I=\frac{5}{2}##. But this isn't what i found in lecture, from the pdg (http://ie.lbl.gov/toi/nuclide.asp?iZA=90019) ##^{19}_{9}F_{10}## result to have ##I=\frac{1}{2}##.

I missed something?

P: ##(1d_{5 \backslash 2})^1## (stand-alone nucleon)

N: ##(1d_{5 \backslash 2})^2##

which in this case is a proton with ##J=\frac{5}{2}## so it leads to a nuclear spin of ##I=\frac{5}{2}##. But this isn't what i found in lecture, from the pdg (http://ie.lbl.gov/toi/nuclide.asp?iZA=90019) ##^{19}_{9}F_{10}## result to have ##I=\frac{1}{2}##.

I missed something?