TheLAbsorption Discontinuities of Bismuth

The problem of obtaining a thin enough absorber was solved by using the bismuth in solution. Bismuth nitrate dissolved in dilute nitric acid was held in one cell, while in another like cell was put only the dilute acid. The difference in the absorption of these two cells was then due entirely to the bismuth atoms, since the absorption of the ${\left(\mathrm{N}{\mathrm{O}}_3\right)}_3$·5${\mathrm{H}}_2$O group added with each Bi atom almost exactly balanced that of the water and acid displaced. The cells had been balanced by trial and error, until their absorption was the same, both when they were filled with water and when they were empty. The solution was analyzed to find the amount of bismuth, and the thickness of the cell was measured with a microscope, so that $\frac{\mu }{\rho }$ could be found. This was measured at twenty wave-lengths between 0.56 and 1.54A. By plotting these values, the equations of the four branches of the curve were obtained. $\left\{\right(\frac{\mu }{\rho }\left)\right(L_{\mathrm{I}})=2.438\mathrm{}{\lambda }^{2.5};, (\frac{\mu }{\rho }\left)\right(L_{\mathrm{II}})=2.548\mathrm{}{\lambda }^4; (\frac{\mu }{\rho }\left)\right(L_{\mathrm{III}}\left)\right\}\{=2.51\mathrm{}{\lambda }^3; (\frac{\mu }{\rho }\left)\right(M)=1.860\mathrm{}{\lambda }^{2.715}.\}$ The magnitudes of the discontinuities, $r$, where $r(\mathrm{I}, \mathrm{II})$ is the ratio of $\frac{\mu }{\rho }$ on the short wave-length side of $L_{\mathrm{I}}$ to that on the long wave-length side, are $r(\mathrm{I}, \mathrm{II})$ 1.161; $r(\mathrm{II}, \mathrm{III})$ 1.572; $r(\mathrm{III}, M)$ 2.393) $r(\mathrm{I}, M)$ 4.37.