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The Magnetic Moment of the Proton

1949; American Institute of Physics; Volume: 75; Issue: 10 Linguagem: Inglês

10.1103/physrev.75.1481

1536-6065

Herbert Taub, P. Kusch,

Particle accelerators and beam dynamics

The molecular beam magnetic resonance method has been employed to measure, in the same magnetic field, the frequency corresponding to a reorientation of the proton in the molecule NaOH and the frequency corresponding to a transition between certain of the h.f.s. levels of the ground state of both the atoms ${\mathrm{Cs}}^{133}$ and ${\mathrm{In}}^{115}$. From these data are calculated the ratio of the $g$ factor of the proton, ${g}_{H}$, to the $g$ factor of total electronic angular momentum, ${g}_{J}$, with the result $\frac{{g}_{H}}{{g}_{J}(\mathrm{Cs},^{2}S_{\frac{1}{2}})}=15.1911\ifmmode\times\else\texttimes\fi{}{10}^{\ensuremath{-}4}$ and $\frac{{g}_{H}}{{g}_{J}(\mathrm{In},^{2}P_{\frac{1}{2}})}=45.6877\ifmmode\times\else\texttimes\fi{}{10}^{\ensuremath{-}4}$. From a knowledge of the ratios $\frac{{g}_{J}({\mathrm{Cs}}^{133})}{{g}_{J}({\mathrm{Na}}^{23})}$ and $\frac{{g}_{J}({\mathrm{In}}^{115})}{{g}_{J}({\mathrm{Na}}^{23})}$ two entirely independent values of the ratio $\frac{{g}_{H}}{{g}_{J}({\mathrm{Na}}^{23})}$ have been obtained: From indium $\frac{{g}_{H}}{{g}_{J}(\mathrm{Na})}=15.1923\ifmmode\times\else\texttimes\fi{}{10}^{\ensuremath{-}4}.$From caesium $\frac{{g}_{H}}{{g}_{J}(\mathrm{Na})}=15.1931\ifmmode\times\else\texttimes\fi{}{10}^{\ensuremath{-}4}.$From the known value of $\frac{{g}_{s}}{{g}_{l}}=2(1.100116)$, where ${g}_{s}$ and ${g}_{l}$ are, respectively, the $g$ factors of electron spin and of orbital angular momentum, and on the basis of the assumptions that ${g}_{s}={g}_{J}$ and that ${g}_{l}=1$, we find, ${g}_{H}=30.4206\ifmmode\times\else\texttimes\fi{}{10}^{\ensuremath{-}4}\ifmmode\pm\else\textpm\fi{}0.005\mathrm{percent}.$Including a small diamagnetic correction, the magnetic moment of the proton, ${\ensuremath{\mu}}_{H}$, is ${\ensuremath{\mu}}_{H}=(15.2106\ifmmode\times\else\texttimes\fi{}{10}^{\ensuremath{-}4}\ifmmode\pm\else\textpm\fi{}0.005\mathrm{percent})$ Bohr magneton.