In uranium ($Z=92$), the $K$ absorption edge is $0.107 \AA$ and the $K_{\alpha }$ line is $0.126 \AA$.Then the L wavelength absorption edge is-

$$\begin{gathered} \mathrm{E}\left(\mathrm{K}\right)=\frac{\mathrm{hc}}{{\mathrm{\lambda }}_{\mathrm{K}}}=\frac{12.4}{0.107}=115.9 \mathrm{keV} \\ \mathrm{E}\left({\mathrm{K}}_{\mathrm{\alpha }}\right)=\mathrm{E}\left(\mathrm{K}\right)-\mathrm{E}\left(\mathrm{L}\right)=\frac{\mathrm{hc}}{{\mathrm{\lambda }}_{\mathrm{\alpha }}}=\frac{12.4}{0.126}=98.4 \mathrm{keV} \\ {\mathrm{E}}_{\mathrm{L}}=\left({\mathrm{E}}_{\mathrm{K}}\right)-\mathrm{E}\left({\mathrm{K}}_{\mathrm{\alpha }}\right) \\ =115.9-98.4 \\ {\mathrm{E}}_{\mathrm{L}}=17.5 \mathrm{keV} \\ {\mathrm{\lambda }}_{\mathrm{L}}=\frac{\mathrm{hc}}{{\mathrm{E}}_{\mathrm{L}}}=\frac{12.4 \mathrm{keV} \mathrm{\AA }}{17.5 \mathrm{keV}}=0.709 \mathrm{\AA } \end{gathered}$$