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:(adsbygoogle = window.adsbygoogle || []).push({});

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?

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Nuclear spin of Fluorine 19

Loading...

Similar Threads - Nuclear spin Fluorine | Date |
---|---|

A Nuclear spin of Fluorine 19 | Dec 20, 2017 |

Different stable nuclear spins for the same isotope | Dec 22, 2015 |

Nuclear spin of alkali atoms | Oct 9, 2014 |

Nuclear Spin | Sep 1, 2014 |

Nuclear spin | Nov 5, 2011 |

**Physics Forums - The Fusion of Science and Community**