\question
{Shoichi Takahata}
{02/03/12}
{physics2070@yahoo.co.jp}
{1-8}
%question
{
In spectroscopy, a wave vector is defined by 1/$\lambda$ (where $\lambda$ is the wavelength)
while in solid state physics, the definition of a wave vector is 2$\pi$/$\lambda$. 
Show that a 1eV photon corresponds to 8065 ${\rm cm}^{-1}$ (wavenumber). 
In Raman spectroscopy, the difference between the wave vectors for the incident and scattered light 
is called the Raman shift whose units are generally given in ${\rm cm}^{-1}$ .
}
%answer
{
The Energy of a photon is given by

\begin{equation}
E=\hbar c \,\frac{2\pi}{\lambda} \,,
\end{equation}

where $\hbar=(1.054571596\pm0.000000082)\times{10}^{-34} \,{\rm J}\cdot {\rm s}$ is the Planck constant 
and $c=2.99792458\times{10}^{10} \,{\rm cm} \cdot {\rm s}^{-1}$ is the velocity of light.
Using Eq. (1-8-1), we can obtain the corresponding value of the wave vector 
for a $1eV=(1.602176462\pm0.000000063)\times {10}^{-19}\,{\rm J}$ photon 
in the units of ${\rm cm}^{-1}$ as follows:

\begin{eqnarray}
\frac{1}{\lambda} 
&=&
\frac{1}{2\pi}\,\frac{E}{\hbar c} 
\nonumber \\
&=&
\frac{1}{2\pi}\,
\frac{(1.602176462\pm0.000000063)\times {10}^{-19} \,{\rm J}} 
{(1.054571596\pm0.000000082)\times{10}^{-34}\,{\rm J} \cdot s \,
\times 2.99792458\times{10}^{10}\,{\rm cm}\cdot {\rm s}^{-1}}
\nonumber \\
&=&
8065.54477\pm0.00094\,{\rm cm}^{-1} \,.
\end{eqnarray}
}