X-Ray Fluorescence Yields of Al, Cl, Ar, Sc, Ti, V, Mn, Fe, Co, Y, and Ag

The $K$-fluorescence yields of elements in the range $13\ensuremath{\le}Z\ensuremath{\le}27$ and the $L$-fluorescence yields of yttrium and silver have been measured. A monochromatic x-ray beam obtained through the Ross filter difference method produced the fluorescence in foil targets and argon gas, and some of the fluorescent x rays were detected by a flow counter mounted at right angles to the primary x rays. The incident x-ray flux was determined by a measurement in the same geometry of the x-ray flux scattered by helium. The fluorescence yields and standard deviations are ${\ensuremath{\omega}}_{K}(\mathrm{Al})=0.0379\ifmmode\pm\else\textpm\fi{}0.0023$, ${\ensuremath{\omega}}_{K}(\mathrm{Cl})=0.0970\ifmmode\pm\else\textpm\fi{}0.0054$, ${\ensuremath{\omega}}_{K}(\mathrm{Ar})=0.119\ifmmode\pm\else\textpm\fi{}0.007$, ${\ensuremath{\omega}}_{K}(\mathrm{Sc})=0.190\ifmmode\pm\else\textpm\fi{}0.010$, ${\ensuremath{\omega}}_{K}(\mathrm{Ti})=0.221\ifmmode\pm\else\textpm\fi{}0.012$, ${\ensuremath{\omega}}_{K}(\mathrm{V})=0.250\ifmmode\pm\else\textpm\fi{}0.012$, ${\ensuremath{\omega}}_{K}(\mathrm{Mn})=0.303\ifmmode\pm\else\textpm\fi{}0.017$, ${\ensuremath{\omega}}_{K}(\mathrm{Fe})=0.347\ifmmode\pm\else\textpm\fi{}0.022$, ${\ensuremath{\omega}}_{K}(\mathrm{Co})=0.366\ifmmode\pm\else\textpm\fi{}0.020$, ${\overline{\ensuremath{\omega}}}_{L}(\mathrm{Y})=0.0315\ifmmode\pm\else\textpm\fi{}0.0028$, and ${\overline{\ensuremath{\omega}}}_{L}(\mathrm{Ag})=0.0659\ifmmode\pm\else\textpm\fi{}0.0037$. The uncertainties include a 5% uncertainty in the helium-scattering cross section assumed for the data reduction. These $K$-fluorescence yields and the values for fluorescence yields for $Z\ensuremath{\ge}30$ considered best by Fink, Jopson, Mark, and Swift are fitted with the semiempirical formula ${[\frac{{\ensuremath{\omega}}_{K}}{(1\ensuremath{-}{\ensuremath{\omega}}_{K})}]}^{n}=A+BZ+C{Z}^{3}$, using $n=\frac{1}{4} \mathrm{and} \frac{1}{3}$. The best fit is given by ${[\frac{{\ensuremath{\omega}}_{K}}{(1\ensuremath{-}{\ensuremath{\omega}}_{K})}]}^{\frac{1}{3}}=\ensuremath{-}0.1019+0.03377Z+1.178\ifmmode\times\else\texttimes\fi{}{10}^{\ensuremath{-}6}{Z}^{3}$, with a standard deviation of 2.4%.