Exercise 6.2.12: Why sodium is transparent to ultraviolet

$N=\frac{968\times 6.022\times 10^{23}}{0.02299}=2.54\times 10^{28}{\mathrm{m}}^{-3}$,

$f=c/{\lambda }_0=6.0\times 10^{14}\mathrm{Hz}$, and ${\epsilon }_p/{\epsilon }_0=1-(1.4305/0.6{)}^2=-4.68$. Then $\gamma$ is real:

With the root $\sqrt{{\epsilon }_p/{\epsilon }_0}=-\mathrm{j}2.164$ that makes the field decay into the metal, as in Eq. (6.2.41),

It is purely imaginary, so $E$ and $H$ are in quadrature and $S_{av}=0$. No power enters a thick metal: the light is totally reflected, which is the metallic lustre.

$f=2.0\times 10^{15}\mathrm{Hz}$: ${\epsilon }_p/{\epsilon }_0=1-(1.4305/2{)}^2=0.488$ and $n=0.699$, so

$v_p{v}_g=4.29\times 2.10\times 10^{16}=9.0\times 10^{16}=c^2$. In (b), $|{\eta }_p|=376.99/2.164=174\Omega$ and $\alpha =k_0\sqrt{4.684}=1.2566\times 10^7\times 2.164=2.72\times 10^{7}Np/m$. The other root of ${\epsilon }_p/{\epsilon }_0$, $+\mathrm{j}2.164$, would give ${\eta }_p=-\mathrm{j}174\Omega$ and a field growing into the metal: the sign of the reactance follows from the direction of decay. R. W. Wood found in 1933 that thin films of the alkali metals become transparent in the ultraviolet, sodium near $210\mathrm{nm}$, in remarkable agreement with this free-electron estimate. For copper, silver and gold the bound d-electrons add to the permittivity and shift the edge into the visible or near ultraviolet, which gives copper and gold their colours.