%\chapter*{Photoelectric Effect}
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\chapter*{Photoelectric Effect}
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 \textbf{Experiment objectives}: measure the ratio of Planck's constant to the electron charge $h/e$ using the photoelectric effect.

\section*{History}


    The photoelectric effect and its understanding was an important step in development
    of quantum mechanics. You probably know that Max Planck was the first to postulate that
    light was made up of discrete packages of energy. At that time it was a proposed hypothetical light property that
    allowed for the proper description of black body radiation. Several years after Planck made this
    suggestion, Albert Einstein refined the idea to explain the
    strange features of electrons which are emitted from metal
    surfaces after absorbing energy from light. Later he received a Nobel prize for this work.

\section*{Theory}

\begin{figure}[h]
\centering \includegraphics[height=3in]{./pdf_figs/pefig1} \caption{\label{pefig1} A simplified schematic of the photo-electric apparatus.}
\end{figure}
 Consider an apparatus as outlined in Figure \ref{pefig1} (it is the
apparatus which Heinrich Hertz used to inadvertently discover the ``photoelectric effect''). Light of a
frequency $\nu$ strikes a cathode, causing electrons to be emitted with velocity $v$. A positive voltage applied
between the anode and the cathode can accelerate emitted electrons towards the positive anode, producing an
electrical current in the circuit. A reverse bias potential applied between the anode and cathode will slow down
the electrons, and even stop them from reaching the anode by matching their kinetic energy. This is the way to
carefully measure the kinetic energies of the photo electrons.
\begin{equation}
KE_{max} = (\frac{1}{2}mv^2)_{max} = eV_0
\end{equation}
%
where $KE_{max}$ is the maximum kinetic energy, $m$ and $e$ are the mass and the charge of an electron, and
$V_0$ is a potential required to stop the electrons (known as the stopping potential).

From the point of view of a wave theory of light, the energy carried by the light wave is proportional to its
intensity, and independent of light frequency. Thus, it was logical to expect that stronger light should
increase the energy of photoelectrons. However, the experiments of Philipp Lenard in 1900 showed something
different: although the maximum current increased with light intensity, the stopping potential $V_0$ was
independent of light intensity. This meant that increasing the rate of energy falling onto the cathode does not
increase the maximum energy of the electrons. This was quite a surprising result from the classical point of
view.

This ``controversy'' was elegantly resolved five years later by Albert Einstein, who postulated that light
arrived in discrete quanta known as ``photons'', and each photon led to an emission of a single electron. The
energy of each photon is determined by the frequency of light - $h\nu$. Thus the energy of an emitted
photoelectron (and therefore the value of the stopping potential $V_0$) is determined by the frequency of an
individual incident photon:
\begin{equation}
\label{eq:pe}
 eV_0= \frac{1}{2}mv^2 = h\nu-\phi = \frac{hc}{\lambda} - \phi
\end{equation}
where $\phi$ is known as the ``work function''  - the amount of energy needed to free the electron from the
cathode. The value of the work energy is a property of the cathode material. Also, it is now clear why the
intensity of light does not affect the stopping potential: more intense light has higher photon flux and thus
increase the number of emitted photoelectrons, but the energy of each electron is determined only on the energy
of a single photon, and that depends only by light frequency  $\nu$.

\begin{boxedminipage}{\linewidth}
\textbf{Where do all those electrons come from?} \\
%
At Jefferson Lab in nearby Newport News, high-energy electrons (GeV range) are
aimed at targets to probe the fundamental properties of quarks. But where do
all those electrons come from? From the photoelectric effect, of course! A
laser beam is aimed at a cathode consisting of a material like copper or GaAs.
The frequency of the laser is such that electrons will be emitted from the
cathode. In this way, high current electron beams are produced. One benefit of
using a laser and the photoelectric effect is that if the laser is "pulsed" in
time, the electron beam will be also (this allows synchronized timing in the
experiments). Also, the polarization of the laser can be manipulated to allow
for the emission of electrons with particular spin. See: www.jlab.org
\end{boxedminipage}

\section*{Procedure}

\textbf{Equipment needed}: Pasco photo-electric apparatus, Hg lamp, digital
voltmeter. The Pasco apparatus contains a circuit which automatically determines the stopping potential, which you measure off of a voltmeter, so there is no need to adjust the stopping potential yourself or measuring the current (lucky you!). Read the brief description of its operation in the appendix.

\begin{figure}
\centering \includegraphics[width=\linewidth]{./pdf_figs/pefig3} \caption{\label{pefig3}
The Pasco photoelectric effect setup.}
\end{figure}

Set up the equipment as shown in Fig. \ref{pefig3}. First, place a lens/grating assembly in front of the Mercury
lamp, and observe a dispersed spectrum on a sheet of paper, as shown in Fig.~\ref{pefig4}. Identify spectral
lines in both the first and the second diffraction orders on both sides. Keep in mind that the color
``assignment'' is fairly relative, and make sure you find all lines mentioned in the table in Fig.~\ref{pefig4}.
Often the first/second order lines on one side are brighter than on the other - check your apparatus and
determine what orders you will be using in your experiment.

After that instal the $h/e$ Apparatus and focus the light from the Mercury Vapor Light Source onto the slot in
the white reflective mask on the $h/e$ Apparatus. Tilt the Light shield of the Apparatus out of the way to
reveal the white photodiode mask inside the Apparatus. Slide the Lens/Grating assembly forward and back on its
support rods until you achieve the sharpest image of the aperture centered on the hole in the photodiode mask.
Secure the Lens/Grating by tightening the thumbscrew. Align the system by rotating the $h/e$ Apparatus on its
support base so that the same color light that falls on the opening of the light screen falls on the widow in
the photodiode mask, with no overlap of color from other spectral
 lines. Return the Light Shield to its closed position.

Check the polarity of the leads from your digital voltmeter (DVM), and connect them to the OUTPUT terminals on
the $h/e$ Apparatus.


\section*{Experimental procedure}



\section*{Part A: The dependence of the stopping
    potential on the intensity of light}
\begin{enumerate}
\item Adjust the $h/e$ Apparatus so that one of the blue first order spectral lines falls upon the opening of the mask of the photodiode. 
\item Press the instrument discharge button, release it, and observe how much time\footnote{Use the stopwatch feature of your cell phone? You don't need a precise measurement.} is required to achieve a stable voltage.
\item Place the variable transmission filter in front of the white reflective mask (and over the colored filter, if one is used) so that the light passes through the section marked 100\% and reaches the photodiode. Record the DVM voltage reading in the table below.
\item Do the exact same thing for another blue spectral line.
\item It's important to check our data early on to make sure we are not off in crazyland. This procedure is technically known as ``sanity checking''. According to equation Eq.~\ref{eq:pe} the frequency (or, better, wavelength) of light can be found from the measured voltage with knowledge of the work function $\phi$. Unfortunately, we don't know the work function. You'll measure it in the second part of the lab. For now, compute $\Delta V_0$ between the two measurements and use it to compute $\Delta \lambda$ or $\Delta \nu$. Does your data agree with the accepted value?  For this computation, it might be helpful to realize that $eV_0$ is the electron charge times the voltage. If you measured $V_0=\unit[1]{V}$ then $eV_0 =1$ ``electron-volt'' or ``eV''.  This is why electron-volt units are useful. Also, $hc=\unit[1240]{nm~eV}$.
\item Go back to the original spectral line.
\item Move the Variable Transmission Filter so that the next section is directly in front of the incoming light. Record the new DVM reading, and time to recharge after the discharge button has been pressed and released.
\item Move the Variable Transmission Filter so that the next section is
directly in front of the incoming light. Record the new DVM reading and
approximate time to recharge after the discharge button has been pressed and
released.

Repeat step 3 until you have tested all five sections of the filter.

Repeat the procedure with a second color from the spectrum.

\end{enumerate}

\begin{figure}[h]
\centering \includegraphics[width=\linewidth]{./pdf_figs/pefig4} \caption{\label{pefig4}
The Pasco photoelectric effect setup.}
\end{figure}

\section*{Part B: The dependence of the stopping potential on the frequency
    of light}
\begin{enumerate}
\item You can easily see five brightest colors in the mercury light spectrum. Adjust the $h/e$ Apparatus so that the 1st order yellow colored band falls upon the opening of the mask of the photodiode. Take a quick measurement with the lights on and no yellow filter and record the DVM voltage. Do the same for the green line and one of the blue ones. 

\item Repeat the measurements with the lights out and record them (this will require coordinating with other groups and the instructor).  Are the two sets of measurements the same?  Form a hypothesis for why or why not.
\item Now, with the lights on, repeat the yellow and green measurements with the yellow and green filters attached to the h/e detector. What do you see now?  Can you explain it? (Hint: hold one of the filters close to the diffraction grating and look at the screen).
\item Repeat the process for each color using the second order lines.  Be sure to use the the green and yellow filters when you are using the green and yellow spectral lines.

\end{enumerate}

\section*{Analysis}
\section*{Classical vs. Quantum model of light}
\begin{enumerate}
\item Describe the effect that passing different amounts of the same
colored light through the Variable Transmission Filter has on the stopping
potential and thus the maximum energy of the photoelectrons, as well as the
charging time after pressing the discharge button.
\item Describe the effect that different colors of light had on the stopping
potential and thus the maximum energy of the photoelectrons.
\item Defend whether this experiment supports a classical wave or a
quantum model of light based on your lab results.
\end{enumerate}
Read the theory of the detector operation in the Appendix, and explain why
there is a slight drop in the measured stopping potential as the light
intensity is reduced. \\{\bf NOTE:} While the impedance of the unity gain
amplifier is very high ($10^{13}~\Omega$), it is not infinite and some charge
leaks off. 

\section*{The relationship between Energy, Wavelength and Frequency}

\begin{enumerate}
\item
Use  the table in Fig.~\ref{pefig4} to find the exact frequencies and wavelengths of the spectral lines you used and plot the measured stopping potential values versus light frequency for of measurements of the first and second order lines (can be on same graph).

\item Fit the plots according to $eV_0 = h\nu-\phi$, extracting values for slopes and intercepts. Find average value for slope and its uncertainty. From the slope, determine  $h$ counting $e=1.6\cdot10^{-19}$~C. Do your measured values agree with the accepted value of  $h=2\pi\cdot 10^{-34}$J$\cdot$s within experimental uncertainty?

\item From the intercepts, find the average value and uncertainty of the work function $\phi$. Look up some values of work functions for typical metals. Is it likely that the detector material is a simple metal?

\end{enumerate}

\section*{Appendix: Operation principle of the stopping potential detector}

The schematics of the apparatus used to measure the stopping potential is shown in Fig.~\ref{pefig5}. Monochromatic light falls on the cathode plate of a vacuum photodiode tube that has a low work function $\phi$. Photoelectrons ejected from the cathode collect on the anode. The photodiode tube and its associated electronics have a small capacitance which becomes charged by the photoelectric current. When the potential on this capacitance reaches the stopping potential of the photoelectrons, the current decreases to zero, and the anode-to-cathode voltage stabilizes. This final voltage between the anode and cathode is therefore the stopping potential of the photoelectrons.
\begin{figure}[h]
\centering \includegraphics[width=0.7\linewidth]{./pdf_figs/pe_det}
\caption{\label{pefig5} The electronic schematic diagram of the $h/e$
apparatus.}
\end{figure}

To let you measure the stopping potential, the anode is connected to a built-in
differential amplifier with an ultrahigh input impedance ($> 10^{13}~\Omega$),
and the output from this amplifier is connected to the output jacks on the
front panel of the apparatus. This high impedance, unity gain ($V_{out}/V_{in}
= 1$) amplifier lets you measure the stopping potential with a digital
voltmeter.

Due to the ultra high input impedance, once the capacitor has been charged from
the photodiode current it takes a long time to discharge this potential through
some leakage. Therefore a shorting switch labeled ``PUSH TO Zero'' enables the
user to quickly bleed off the charge.
\newpage
\section*{Sample data tables:}
{\large
%\begin{tabular}{|p{27mm}|p{27mm}|p{27mm}|p{27mm}|}
\begin{tabular}{|c|c|c|c|}
\hline
 Color  & \%Transmission & Stopping Potential &
Approx. Charge Time \\
\hline
&100&&\\\hline
&80&&\\ \hline
&60&&\\ \hline
&40&&\\ \hline
&20&&\\ \hline
\hline
 Color & \%Transmission & Stopping Potential &
Approx. Charge Time \\
\hline
&100&&\\\hline
&80&&\\ \hline
&60&&\\ \hline
&40&&\\ \hline
&20&&\\ \hline
\end{tabular}
}
\vskip .1in
%
%
%{\large
%\begin{tabular}{|p{27mm}|p{27mm}|}
%\hline
% Light Color  & Stopping Potential \\\hline
%Yellow&\\\hline
%Green&\\ \hline
%Blue&\\ \hline
%Violet&\\ \hline
%Ultraviolet&\\ \hline
%\end{tabular}
%}

{\large
\begin{tabular}{|c|c|c|c|}
\hline
 1st Order Color&$\lambda$ (nm) &$\nu$  ($10^{14}Hz$) &
Stopping Potential (V) \\
\hline Yellow&&&\\\hline Green&&&\\ \hline Blue&&&\\ \hline Violet&&&\\
\hline Ultraviolet&&&\\ \hline \hline
2nd Order Color&$\lambda$ (nm) &$\nu$  ($10^{14}Hz$) &
Stopping Potential (V) \\
\hline Yellow&&&\\\hline Green&&&\\ \hline Blue&&&\\ \hline Violet&&&\\
\hline Ultraviolet&&&\\ \hline \hline

\end{tabular}
}
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