# Towards the Quantification of 5f Delocalization
## 3 Results and Discussion
### XES and 5f:6p Peak Ratios  
It is possible to calculate the electric dipole relative cross sections for the 6p \(\rightarrow\) 3d transitions, following the same procedures as discussed earlier for the 5f \(\rightarrow\) 3d transitions. The results are shown in Table 4 [30]. In this case, the 6 ps are completely full and, because the 6 ps should sit inside the 5f's [30], the electric dipole selection rules and cross sections should hold. One piece of information supporting this assertion is the absence of the 6p\({}_{1/2}\) peak in the M\({}_{5}\) spectrum in Figure 4, consistent with the cross section of zero in Table 4. For the 6p \(\rightarrow\) 3d transitions, the \(\Delta\)j = +1 transitions are the strongest and will be utilized for the normalization of the 5f intensities to the 6p intensities. At this point, it is possible to write out the relative intensity equations, including some constants for effects, such as the radial matrix elements. \[I_{5f}^{M4}=C_{5f}\bigg{(}\frac{n_{5/2}}{6}\left(\frac{56}{60}\right)\bigg{)} \tag{11}\]  
\[I_{5f}^{M5}=C_{5f}\bigg{\{}\frac{n_{5/2}}{6}\left(\frac{4}{90}\right)+\frac{n_ {7/2}}{8}\left(\frac{16}{18}\right)+\frac{6.4}{135}\bigg{\}} \tag{12}\]  
\[I_{6p_{1/2}}^{M4}=C_{6p}\bigg{(}\frac{1}{3}\bigg{)} \tag{13}\]  
\[I_{6p_{3/2}}^{M5}=C_{6p}\bigg{(}\frac{2}{5}\bigg{)} \tag{14}\]  
\[PR_{fp}^{M4}=\frac{I_{5f}^{M4}}{I_{6p_{1/2}}^{M4}}=\bigg{(}\frac{C_{5f}}{C_{6p }}\bigg{)}\bigg{\{}\frac{\left(\frac{n_{5/2}}{6}\right)\left(\frac{56}{60} \right)}{\left(\frac{1}{3}\right)}\bigg{\}} \tag{15}\]  
\[PR_{fp}^{M5}=\frac{I_{5f}^{M5}}{I_{6p_{3/2}}^{M5}}=\bigg{(}\frac{C_{5f}}{C_{6p }}\bigg{)}\bigg{\{}\frac{\frac{n_{5/2}}{6}\left(\frac{4}{90}\right)+\frac{n_ {7/2}}{8}\left(\frac{16}{18}\right)+\frac{6.4}{135}}{\left(\frac{5}{5}\right)} \bigg{\}} \tag{16}\]From the UF\({}_{4}\) experimental results [30], it is possible to determine (C\({}_{5\text{f}}\)/C\({}_{6\text{p}}\)). \[\left(PR_{fp}^{M4}\right)_{UF_{4}}^{Exp}=10.4\ \rightarrow\ \left(\frac{C_{5f}}{C_{6 \text{p}}}\right)=11.34 \tag{17}\]  
Finally, substituting, the equations become as follows:  
\[PR_{fp}^{M4}=\frac{I_{5f}^{M4}}{I_{6\text{p}_{1/2}}^{M4}}=5.31n_{5/2} \tag{18}\]  
\[PR_{fp}^{\text{MS}}=\frac{I_{5f}^{\text{MS}}}{I_{6\text{p}_{3/2}}^{\text{MS}}}= 0.2098n_{5/2}+3.150n_{7/2}+1.361 \tag{19}\]  
Setting n = 2 and applying Equations (16) and (17), the results in Figures 6 and 7 were generated. Here, the spectra are normalized to a value of one at the 6p maximum, as shown below. Note the strong agreement of the experiment and the n\({}_{5/2}\) = 2 curve in each, as expected. \begin{table}
\begin{tabular}{c c c c c c} \hline \hline  & & \begin{tabular}{c} 6P\({}_{\text{1/2}}\) \\ **Full** \\ \end{tabular} & \begin{tabular}{c} 6P\({}_{\text{3/2}}\) \\ **Full** \\ \end{tabular} & \
\begin{tabular}{c} 6P\({}_{\text{1/2}}\) \\ **Full** \\ \end{tabular} &
\begin{tabular}{c} 6P\({}_{\text{3/2}}\) \\ **Full** \\ \end{tabular} \\ \hline \multirow{2}{*}{M\({}_{5}\)} & \(\frac{3\text{d}_{5/2}}{\text{Empty}}\) & 0 & 12/5 & \(\frac{3\text{d}_{5/2}}{1\text{ Hole}}\) & 0 & 2/5 \\ \hline \multirow{2}{*}{M\({}_{4}\)} & \(\frac{3\text{d}_{3/2}}{\text{Empty}}\) & 4/3 & 4/15 & \(\frac{3\text{d}_{3/2}}{1\text{ Hole}}\) & 1/3 & 1/15 \\ \hline \multirow{2}{*}{M\({}_{5}\)} & \(\frac{I_{5f}^{\text{total}}}{I_{6\text{p}_{3/2}}^{\text{total},2}}=\left(1/3 \right)/\left(2/5\right)=0.833\) & & \\ \hline \hline \end{tabular}
\end{table} 
Table 4: Shown here are the 6p electric dipole. Note that (1) the total 6p cross section is 4; and (2) the 6p cross section per 3d hole is 2/5. Leftmost are the results for completely empty 3d states. Rightmost are the cross sections per 3d hole. See text for details. Figure 6: Shown here is the experimental M\({}_{4}\) XES spectrum of UF\({}_{4}\) and the simulated spectra for n = 2 and various values of n\({}_{5/2}\). The simulation used normalized Lorentzian line shapes, with a half-width at half-max of \(\Gamma\). For the 6p\({}_{\text{1/2}}\), \(\Gamma\) = 3.5 eV. For the 5f peak, \(\Gamma\) = 4 eV.