Determination ofg-Factor Ratios for FreeRb85andRb87Atoms

Abstract

The ratio of the nuclear $g$ factor to the electronic $g$ factor in the ground electronic state of free ${\mathrm{Rb}}^{85,87}$ atoms has been determined using optical pumping techniques in a magnetic field of ≤50 G. Typical linewidths were ≤15 cps. The linewidth contribution due to magnetic field effects was less than 1 cps. The ratio of electronic $g$ factors for the Rb isotopes was also measured. The results are $-\frac{g_I}{g_J\left({\mathrm{Rb}}^{87}\right)}=4.969\mathrm{}9147\times (1\mathrm{}\pm 0.9\times {10}^{-6\mathrm{}})\times {10}^{-4\mathrm{}}$, $-\frac{g_I}{g_J\left({\mathrm{Rb}}^{85}\right)}=1.466\mathrm{}4908\times (1\mathrm{}\pm 2.1\times {10}^{-6\mathrm{}})\times {10}^{-4\mathrm{}}$, and $\frac{g_J\left({\mathrm{Rb}}^{87}\right)}{g_J\left({\mathrm{Rb}}^{85}\right)}=1.000\mathrm{}0000041\times (1\mathrm{}\pm 6.0\times {10}^{-9\mathrm{}})$. These results were obtained from evacuated wall-coated cells having the Lorentzian line shape. Cells filled with inert buffer gases exhibited small, uncontrollable, systematic error due to non-Lorentzian line shape. In both types of cells, the Zeeman resonances exhibited frequency shifts proportional to the pumping-light intensity. The ratio $\frac{g_I\left({\mathrm{Rb}}^{85}\right)}{g_I\left({\mathrm{Rb}}^{87}\right)}=0.295\mathrm{}0736\times (1\mathrm{}\pm 2.3\times {10}^{-6\mathrm{}})$ was obtained by combining the results above. The combination with results of other researchers yields the chemical shift of ${\mathrm{Rb}}^{+}$ in aqueous solution relative to the free atom as $\Delta \sigma \left(\frac{{{\mathrm{Rb}}^{+}}_{\mathrm{aq}}}{\mathrm{Rb}}\right)=-(211.6\mathrm{}\pm 1.2)\times {10}^{-6\mathrm{}}$. Absolute values in units of the Bohr magneton for the shielded nuclear moments are derived using only $g$-factor ratios for free atoms: $g_I\left({\mathrm{Rb}}^{87}\right)=-0.995\mathrm{}1414\times {10}^{-3\mathrm{}}\times (1\mathrm{}\pm 1.0\times {10}^{-6\mathrm{}})$ and $g_I\left({\mathrm{Rb}}^{85}\right)=-0.293\mathrm{}6400\times {10}^{-3\mathrm{}}\times (1\mathrm{}\pm 2.2\times {10}^{-6\mathrm{}})$.