\question
{Yuki Tatsumi}
{01/24/12}
{tatsumi@flex.phys.tohoku.ac.jp}
{4-5}
%question
{Obtain the energy in eV of a laser with a wavelength of 633 nm. What is the conversion formula from nm to eV and from eV to nm?}
%answer
{
It is shown that 1 eV corresponds to $8065 \, {\rm cm}^{-1}$ in Q4-4. Then the units of energy is converted from eV to ${\rm {nm}^{-1}}$ as follows:

\begin{equation}
1 \, {\rm eV} = 8065 \, {\rm cm}^{-1} = 8065 \times 10^{-7} \, {\rm nm}^{-1}.
\label{eq:Q4-5-1}
\end{equation}

The relation between the wavenumber and the wavelength is given by

\begin{equation}
k = \frac{1}{\lambda} ,
\label{eq:Q4-5-2}
\end{equation}

where $\lambda$ and $k$ are the wavelength and the wavenumber, respectively. Using Eqs.\,(\ref{eq:Q4-5-1}) and (\ref{eq:Q4-5-2}), the energy in eV of a laser with a wavelength of 633 nm is written as follows:

\begin{equation}
E = \frac{1/633}{8065 \times 10^{-7}} = 1.96 \, {\rm eV} .
\label{eq:Q4-5-3}
\end{equation}

The conversion formula from $\lambda$ nm to $E$ eV and that from $E$ eV to $\lambda$ nm are written by 

\begin{eqnarray}
E = \frac{k}{8065 \times 10^{-7}} = \frac{10^7}{8065 \lambda} = \frac{1239.9}{\lambda} \, {\rm eV} ,
\label{eq:Q4-5-4}
\end{eqnarray}

or

\begin{eqnarray}
\lambda = \frac{10^7}{8065 E} = \frac{1239.9}{E} \, {\rm nm} .
\label{eq:4-5-5}
\end{eqnarray}

}


