\question
{Yuki Tatsumi}
{01/16/12}
{tatsumi@flex.phys.tohoku.ac.jp}
{4-4}
%question
{Show that 1 eV corresponds to $8065 \, \rm{{cm}^{-1}}$. What is the energy in eV for the G-band Raman spectrum of graphite at $1580 \, \rm{cm^{-1}}$?}
%answer
{
The energy of a photon $E$ is given by

\begin{eqnarray}
E = h\omega = hck ,
\label{eq:Q4-4-1}
\end{eqnarray}

where $h$, $\omega$, $c$, and $k$ are Plank's constant, the angular frequency of light, the velocity of light and the wave number of light $(k = 1/\lambda \quad \lambda : \mbox{wavelength of right})$, respectively. Plank's constant and the velocity of light is given by

\begin{eqnarray}
h = 6.6262 \times 10^{-34} \, \rm{J{\cdot}s} , \\
\label{eq:Q4-4-2}
c = 2.9979 \times 10^{8} \, \rm{m/s} = 2.9979 \times 10^{10} \, \rm{cm/s} .
\label{eq:Q4-4-3}
\end{eqnarray}

Using $1 \, \rm{eV} = 1.602 \times 10^{-16} \, \rm{J}$, the wave number which corresponds to 1 eV is obtained from Eq.\,(\ref{eq:Q4-4-1}) as follows:

\begin{eqnarray}
E
& = &
hck
\nonumber \\
1.602 \times 10^{-16}
& = &
(6.6262 \times 10^{-34}) \times (2.9979 \times 10^{10}) \times k
\nonumber \\
k
& = &
8065 \, \rm{{cm}^{-1}} .
\label{eq:Q4-4-4}
\end{eqnarray}

Then it is shown that 1 eV corresponds to $8065 \, \rm{cm^{-1}}$. What wavenumber the G-band Raman spectrum of graphite at $1580 \, \rm{cm^{-1}}$ corresponds to is obtained as follows:

\begin{eqnarray}
1 : 8065
& = &
x : 1580
\nonumber \\
x
& = &
0.196 \, \rm{eV} .
\label{eq:Q4-4-5}
\end{eqnarray}

Thus the G-band Raman spectrum of graphite at $1580 \, \rm{cm^{-1}}$ corresponds to 0.196 eV.

}



