\documentclass{article}
\usepackage {graphicx,subfigure,sidecap,array,tabularx}
\date{29. Juli 2005}
\author{Henning Schories and Philipp Spiekermann}
\title{THE FRANCK-HERTZ-EXPERIMENT (E 103)}


\begin{document}


\maketitle



The Franck-Hertz-Experiment shows that atoms can be excited by collision with electrons (or other particles). 
Thereby the electron confers just discrete energy-portions to the atom in an inelastic scattering. 
At lower kinetic energy of the electrons one observes elastic scattering. 
In this case there is no excitation of the atom up to higher energy levels.
For this experiment a so-called Franck-Hertz-tube is used. 
It is a tube filled with liquid mercury as well as mercury steam with a three-electrode-system in it.
The experimental setup is the following.
\begin {figure} [hbp]
\centering
\includegraphics[width=7cm]{Franck-Hertz-Rohr} \caption{Franck-Hertz tube} 
\end {figure}

In the first cathode electrons are set free by thermionic emission. 
In the next moment they accelerate in an electric field towards a positive charged grid, which provides as anode, 
in accordance with 
\begin {equation} 
E_{kin} = \frac {1}{2} m v^{2} \; .
\end {equation}
In the space between these two electrodes they scatter with mercury atoms. 
After passing the grid they are decelerated in a reverse poled field and, if their kinetic energy is high enough, 
they reach a third electrode. In the positive case a current is measured.
 
We measured this current for four different temperatures and two values of the deceleration-voltage, 
and plotted it against the acceleration-voltage. Our results are shown in figure 3 and 4.
\begin {figure} [hbtp]
\includegraphics[width=12cm]{1} \caption{anode-current for $164^{\circ}$\,C (blue), 
$180.5^{\circ}$\,C (red), $188,5^{\circ}$\,C (black), $196^{\circ}$\,C (green)} \label{anodetemp}
\includegraphics[width=12cm]{2} \caption{anode-current for acceleration-voltages 1\,V (blue) and 1.5V} \label{anodevolt}
\end {figure}

One sees the superposition of several effects. 
For the case of vanishing steam-density one would expect a quadratic increasing of the current with the voltage. 
This can be understood out of the relations 
\begin {displaymath} 
j = e n v \qquad \textnormal{and} \qquad v = \sqrt {\frac{2 e}{m_{e}} U} \;,
\end {displaymath}
where the second one follows from equation (1). Combining both leads to 
\begin {equation} \label {j}
j = e n \sqrt {\frac {2 e}{m_{e}} U} \;.
\end {equation} 
In the case of non-vanishing steam-density one can expect \emph{collisions} between electrons and Hg-atoms. 
The existence of minima of the current at certain voltages can be explained by the assumption 
that the Hg-atoms can be excited to a higher state by absorbing discrete energy-portions from the electrons which 
are corresponding to the height of the excited state over the ground state. 

If an electron has reached an energy which corresponds to the energy of the first excited state of the Hg-atom, 
it can, when hitting one, transfer his whole energy to that atom. After that the electron has no kinetic energy left, 
but if the acceleration-voltage is high enough, the electron can reach again his previous energy before passing the grid, 
and excite another atom. 
Those electrons who lost kinetic energy in such inelastic scatter processes and couldn't regain enough speed to 
override the decelerations-field aren't detected, and won't account for the current.

The observed equidistance of the minima proofs that only the first few states with nearly the same energy are excited. 
Thus the probability for an electron to reach an energy according to the next higher energy-level without 
colliding with an Hg-atom before seems to be very little.

A third effect we observed but cannot yet explain is the not-coincidence of the minima of the curves corresponding to 
different temperatures. They are shifted to higher voltages with increasing temperature. 
Nevertheless the distances of every two minima following each other seem to be quiet temperature-independent.
It is obvious that there is no dependency between the distances of the minima and the value of the deceleration-voltage, 
as one can see in figure \ref{anodevolt}. The decreasing of the current with higher temperature we will discuss later.

Regarding all this we proceed as follows to determine the voltage according to 
the first excited state(s) of the Hg-atom and the corresponding wavelength according to 
\begin {equation} \label{hmalf}
E   =   e U   =   h \frac{c}{\lambda} \;. %\quad \Rightarrow \quad \lambda = \frac {hc}{eU} \; .
\end {equation} 
We read off the distances of every two minima following each other and average over all temperatures and deceleration voltages. 
The results are shown in table~\ref {anodetable}.

\begin {table} \caption {anode-current metering and calculations \label {anodetable}
}{\small  
\begin{tabular} {l|rrrrrr|rrrrr|} 
           &      \multicolumn {5} {l} {blue curve ($164^{\circ}$\,C)}           &                   &              \multicolumn {5} {l} {red curve ($180.5^{\circ}$\,C)}     \\    
\hline
Minima [cm] &                 11.05 &      15.00 &      18.85 &      22.80 &      26.85 &            &                              15.80 &      19.75 &      23.55 &      27.45 &      31.30          \\

    $U_{N}$ [V] &                  15.64 &      21.23 &      26.67 &      32.26 &      37.99 &            &                              22.36 &      27.95 &      33.32 &      38.84 &      44.29        \\

$\Delta_{syst} U_{N}$  [V] &                   0.16 &       0.21 &       0.27 &       0.32 &       0.38 &                                    &       0.22 &       0.28 &       0.33 &       0.39 &       0.44    \\

$\Delta U_{N}$ [V] &                   0.23 &       0.28 &       0.34 &       0.39 &       0.45 &            &                               0.29 &       0.35 &       0.40 &       0.46 &       0.51       \\

$ U_{N} - U_{N-1}$ [V] &                        &       5.59 &       5.45 &       5.59 &       5.73 &            &                                   &       5.59 &       5.38 &       5.52 &       5.45         \\

$\Delta(U_{N} - U_{N-1})$ [V] &                        &       0.51 &       0.62 &       0.73 &       0.84 &            &                                   &       0.64 &       0.75 &       0.86 &       0.97        \\
\hline
\end{tabular} } \\  \\

{\small  
\begin{tabularx} {15,25cm} {l|Xrrrrr|XXrrr|}
\centering
           &       \multicolumn {4} {l} {black curve ($188,5^{\circ}$\,C)}    &&      & \multicolumn {4} {l} {green curve ($196^{\circ}$\,C)}  \\
\hline
Minima [cm] &          &  20.35 &      24.15 &      28.00 &      31.75 &            &      &      &      24.95 &      28.65 &      32.55 \\

    $U_{N}$ [V] &      & 28.80 &      34.17 &      39.62 &      44.93 &            &        &     &      35.30 &      40.54 &      46.06 \\

$\Delta_{syst} U_{N}$  [V] && 0.29 &       0.34 &       0.40 &       0.45 &            &     &    &       0.35 &       0.41 &       0.46 \\

$\Delta U_{N}$ [V] &     & 0.36 &       0.41 &       0.47 &       0.52 &            &       &     &       0.42 &       0.48 &       0.53 \\

$ U_{N} - U_{N-1}$ [V]&      &     &       5.38 &       5.45 &       5.31 &            &     &       &            &       5.24 &       5.52 \\

$\Delta(U_{N} - U_{N-1})$ [V] &   & &	   0.77 &       0.88 &       0.99 &            &      &      &            &       0.90 &       1.01 \\
\hline
\end{tabularx} } 
\end{table}

We assumed a metering-error of 0.5\,mm, corresponding to 0.7\,V, 
added to a systematic error caused by the impreciseness of the voltmeter of $ \Delta U_{syst}$ = 1\% \nolinebreak . 
Therewith the error of the \begin {math} (U_{N} - U_{N-1}) \end {math}\,'s is given by 
\begin {displaymath}
\Delta\,(U_{N} - U_{N-1}) \quad = \quad \Delta\,U_{N} + \:\Delta\,U_{N-1}
\end{displaymath} 
and the error for $\langle U\rangle$ by
\begin {displaymath}
\Delta\,\langle U \rangle \quad = \quad \frac{1}{N} \; \sum_{i = 1}^{N} \,\Delta\,(U_{N} - U_{N-1})
\end {displaymath}
Corresponding to this an error for $\lambda$ is given by (see equation (\ref{hmalf})) 
\begin {eqnarray*}
\Delta\lambda  & = & \biggl| \frac{\partial \lambda}{\partial U} \; \Delta U \biggr| \\*
& = & \frac{h c}{eU^{2}} \; \Delta U \;\; = \;\; \lambda \; \frac{\Delta U}{U}\;.
\end {eqnarray*}
Here we recapitulate the results: 
\begin {displaymath}
U = (5.47 \pm 0,81)\,\textnormal{V} \qquad \lambda = (2667 \pm 392)\,\textnormal{\AA} .
\end {displaymath}

Figure 4 shows the term-scheme of the lower states of mercury (a) and the values of the energy corresponding 
to the different states (b)
\begin{figure} [hbtp]
\centering
\subfigure[term-scheme]{
\label{term-scheme} \includegraphics [width=6cm]{term-schemeofHg}}
\hspace{.5cm}
\subfigure[energy-states]{\label{energy-states} \includegraphics{energy-states}}
\caption{characteristics of mercury}
\label{mercuryscheme}
\end{figure}

What we observed is, according to figure 4, the excitation of mainly the \textsuperscript{3}P~-~states. 
Their energy values listed in figure 4 (b) are within our error-bars. The next higher energy-level, 
the \textsuperscript{1}P$_{1}$ - state, 
seems to be quiet rarely excited. 

In figure \ref{crosssection} the theoretically derived cross-section for inelastic excitation of Hg is plotted over the electron-energy. 
\begin {figure}  [hbtp]
\centering
\includegraphics [width=9cm]{Cross-section2} \caption{cross-scattering for inelastic excitation of the mercury-atom} \label{crosssection}
\end {figure}
For 5 eV -- electrons one reads off a cross-section for the excitation of the \textsuperscript{3}P$_{0}$-state 
of 0.7 $\pi a_{0}^{2}$ . 
For this value the cross-section for the excitation of the \textsuperscript{3}P$_{2}$-states vanishes. 
So this level was very seldomly excited in the experiment, too. 

The Hg-atoms who are in the \textsuperscript{3}P$_{1}$-state can fall back to the ground 
state by optical transition, if j-j-coupling is realised. 
We can assume that because of the high atom-mass of mercury of 200.6 u. 
In this case the rules for optical transition are 
\begin {equation}
\Delta\,J = 0 ,\:\pm 1,
\end {equation}
whereas the transition $J = 0 $ to $J = 0$ is forbidden. 
Because of this the \textsuperscript{3}P$_{0}$- state cannot de-excite optically.

The next thing we wanted to determine was the energy corresponding to the \textsuperscript{3}P$_{1}$-state. 
For this we searched for emitted light. 
The wavelength corresponding to the excitation energy of the \textsuperscript{3}P-triplet-states we have already 
determined as $\lambda = 2667$\,\AA, that is why we expect ultra-violet light. 
As detector we used a photomultiplier with integrated interference-filter which is attuned to the wavelength we want to measure. 
The photomultiplier converts a light signal into a current (photo effect), which we plotted over the acceleration-voltage.
Our result for a temperature of $167^{\circ}$\,C is shown in figure \ref{undiffphoto}. 
\begin {figure} [hbtp]
\includegraphics[width=12cm]{3} \caption{photo-current for a temperature of $167^{\circ}$\,C} \label{undiffphoto}
\end {figure}
One can see here the current increase with the voltage 
with piecewise constant gradient. 
The differences of every two following values of the voltage pertaining to the sharp bends of 
this curve match quiet good with the ones pertaining to the minima of the anode-current-curves. 
The explanation for this is the following: If the kinetic energy of the electrons is less than ca. 5.5\,eV the Hg-atoms 
do not become excited. They remaining their ground state and do not emit photons. 
So we weren't able to observe any up to that voltage. 
When the electron energy reach that value of 5.5\,eV it is just enough to lift the Hg-atoms in the \textsuperscript{3}P-triplet states, 
especially in the \textsuperscript{3}P$_{1}$-state. This state has a lifetime of 117\,ns. 
After that time the atom will fall down while emitting a photon with a wavelength corresponding to its loss of potential energy. 
A nearly linear with the acceleration-voltage increasing \emph{light-current} can only be explained 
by a linear increasing of the number of electrons able to excite a Hg-atom in the \textsuperscript{3}P$_{1}$-state. 
The reason for that is the following. A linear increasing acceleration-voltage takes a linear increasing field-strength with it: 
In our case holds roughly 
\begin {equation}
U \approx E\;d .
\end{equation} 
In the case of sufficient electron density the field will be \emph{caught} by some electrons before reaching the cathode. 
To catch a stronger field more electrons are necessary. 
Therewith the emitted electrons that can be accelerated in the field increases with the voltage between cathode and grid.
 
If the kinetic energy of the electrons reaches ca. 10\,eV one electron becomes abruptly able to excite two Hg-atoms. 
This causes another sharp bend in the photo-current-curve. 
As one can see in the graph the exact values of the voltage according to the sharp bends are hard to read off. 

To overcome this problem we integrated a differentiator in our experimental setup, 
with the effect that we now have a roughly piece wise constant curve. 
For this rearrangement holds 
\begin {equation}
U \to -RC \frac{dU}{dt} \;.
\end {equation}
With this setup plotted light-curves for four different temperatures, like we have done for the anode-current. 
The graphs and the metering for which we proceed as previous are shown in figure \ref {diffphoto} and table \ref {phototable}.
\begin {figure}
\includegraphics[width=12cm]{4} \caption{differentiated photo-current for $167^{\circ}$\,C (blue), 
$179^{\circ}$\,C (red), $189^{\circ}$\,C~(black), $198.5^{\circ}$\,C (green)} \label{diffphoto}
\end {figure}
\begin {table} \caption {photo-current metering and calculations} \label {phototable}
{\small  
\begin{tabularx} {5cm} {l|rrrrXr|rrrrr|} 
           &      \multicolumn {4} {l} {blue curve ($167^{\circ}$\,C)}    &           &         &              \multicolumn {4} {l} {red curve ($179^{\circ}$\,C)}     \\    
\hline
Minima [cm] &                 14.20 &      17.90 &      21.90 &      25.90 &      &            &             14.55 &      18.00 &      21.95 &      25.55 &      30.65          \\

    $U_{N}$ [V] &                  17.18 &      21.66 &      26.50 &      31.34 &   &            &           17.61 &      21.78 &      26.56 &      30.92 &      37.09        \\

$\Delta_{syst} U_{N}$  [V] &        0.17 &       0.22 &       0.26 &       0.31 &    &                &       0.18 &       0.22 &       0.27 &       0.31 &       0.37    \\

$\Delta U_{N}$ [V] &                   0.35 &       0.40 &       0.44 &       0.49 &    &            &        0.36 &       0.40 &       0.45 &       0.49 &       0.55       \\

$ U_{N} - U_{N-1}$ [V] &                        &       4.48 &       4.84 &       4.84 &      &      &             &       4.17 &       4.78 &       4.36 &       6.17         \\

$\Delta(U_{N} - U_{N-1})$ [V] &                        &     0.75 &       0.84 &       0.94 &      &         &     &       0.75 &       0.84 &       0.93 &       1.04        \\
\hline 
\end{tabularx} }  \\  \\
 
{\small  
\begin{tabularx} {14,8cm} {l|rrrrr|XXrrr|}
\centering
           &       \multicolumn {4} {l} {black curve ($189^{\circ}$\,C)}    &      & \multicolumn {4} {l} {green curve ($198.5^{\circ}$\,C)}  \\
\hline
Minima [cm] &            15.50 &      19.70 &      22.70 &      26.95 &   30.65        &      &      &      23.65 &      27.65 &      30.95 \\

    $U_{N}$ [V] &       18.76 &      23.84  &      27.47 &      32.61 &    37.09       &        &     &      28.62 &      33.46 &      37.45 \\

$\Delta_{syst} U_{N}$  [V] & 0.19 &       0.24 &       0.27 &       0.33 &   0.37         &     &    &       0.29 &       0.33 &       0.37 \\

$\Delta U_{N}$ [V] &      0.37 &       0.42 &       0.45 &       0.51 &     0.55       &       &     &       0.47 &       0.51 &       0.55 \\

$ U_{N} - U_{N-1}$ [V]&           &       5.08 &       3.63 &       5.14 &     4.48       &     &       &            &       4.84 &       3.99 \\

$\Delta(U_{N} - U_{N-1})$ [V] &    &	   0.79 &       0.87 &       0.96 &      1.06      &      &      &            &       0.98 &       1.07 \\
\hline
\end{tabularx} } \\ \\
\end{table} 
The results are: 
\begin {displaymath}
U = (4.68 \pm 0.91)\,\textnormal{V} \quad ; \quad \lambda = (2649\pm515)\,\textnormal{\AA}.
\end {displaymath} 
The value for the \textsuperscript{3}P$_{1}$-state quoted in \cite{Ph}, 
$U$ = 4.89\,eV, lies between the error bars. The errors were calculated the same way as we did for the anode-current-curves.
 
By comparing both values for the acceleration-voltage one finds that the \emph{first} one is higher than the \emph{second} one. 
This shows that even higher states than the \textsuperscript{3}P$_{1}$ have been excited by collision. 

The amplitude of the anode-current-curve as well as the amplitude of the photo-current-curve decreases with higher temperatures. 
For the explanation we calculate the mean free path-length of electrons and the mean time between Hg-Hg-collisions. 
The mean free path length is given by 
\begin{equation}
\Lambda = \frac{1}{n\sigma}\;,
\end{equation}
where $n$ follows from the ideal gas law: 
\begin{equation}
n = \frac{N}{V} = \frac {p}{kT}\;,
\end {equation}
thus holds
\begin {equation}
\Lambda = \frac{kT}{p\sigma}\;.
\end {equation}
The pressure in Torr is in our case given by 
\begin{equation}
\log p = 10.55 - 3333/T - 0.85\log T.
\end{equation} 
For a temperature of $180^{\circ}$\,C one finds for the free path-length of an electron $\Lambda_{e^{-}\to Hg}$ = 0.44\,mm, 
where $\sigma_{e^{-}\to Hg}$ = 1.4 $\pi a_{0}^{2}$ is read off of figure \ref {crosssection}. 
For the calculation of the free path-length of a Hg-atom we take as cross-section for Hg-Hg-collisions
\begin {equation}
\sigma = \pi (r_{1} + r_{2})^{2} = \pi d^{2},
\end {equation} 
where $d$ = 0.44\,nm is the gaskinetic diameter of the Hg-atom. By assuming a temperature of $180^{\circ}$\,C, too, one finds:
\begin{displaymath}
\Lambda_{Hg \to Hg} = 8.91\,\mu m.
\end{displaymath}
To calculate the mean time between Hg-Hg-collisions we still need the average velocity of an Hg-atom. 
This we get from the \emph{Maxwell-Boltzmann}-distribution
\begin{equation}
f(v) = \Bigl( \frac{m}{2 \pi k T}\Bigr)^{3/2} 4 \pi v^{2} e^{-mv^{2}/2kT} .
\end{equation}
This leads us to 
\begin{equation}
\bar v = \int_{0}^{\infty} v \;f(v)\;dv = \sqrt{\frac{8kT}{\pi m}} .
\end {equation}
 we put in the mass of a Hg-atom which is 200.6\,u , we get 
\begin{displaymath}
\bar v_{Hg} = 218.83\,m/s .
\end {displaymath} 
The mean time between Hg-Hg-collisions is therewith 
\begin{displaymath}
t_{Hg \to Hg} = \frac{\Lambda_{Hg \to Hg}}{\bar v_{Hg}} = 40.7\,ns ,
\end {displaymath} 
which is much smaller than 117\,ns what corresponds to the life-time of the \textsuperscript{3}P$_{1}$-~state. 
The values for different temperatures one finds in table %\ref {statetable}.

%\begin {tabular} {l|rrrrrr} %\caption {several energy-states of mercury} \label {statetable}
%states & \textsuperscript{1}S$_{0}$&\textsuperscript{3}P$_{0}$&\textsuperscript{3}P$_{1}$&\textsuperscript{3}P$_{2}$&\textsuperscript{1}P$_{1}$&\textsuperscript{3}S$_{1}$\\
%\hline
%energy [eV]& 0.00&4.67&4.89&5.46&6.70&7.73\\
%\end {tabular}
 
This shows that the de-excitation of the Hg-atoms becomes with increasing temperature more and more realized 
by collisions than by optical transition. 
This explains the decreasing of the amplitude of the light-current-curves with increasing temperature. 
The decreasing of the amplitude of the anode-current-curves results out of the shorter becoming free-path-length 
of the electrons. 
Thus they have after a collision less and less length to get an energy sufficient to reach the third 
electrode after passing the grid and to be detected.

\begin{thebibliography}{9}
\bibitem{Ph}
Physikalisches Praktikum f\"ur Fortgeschrittene, Teil 1\\ Universit\"at Bonn; BONN-MS-2005-01NA; Juli 2005
\end{thebibliography}


\end {document}