\experiment{Photoelectric Effect}

\section{Background}
 
Around the turn of the century, Philipp von Lenard, studying a
phenomenon originally observed by Heinrich Hertz, showed that
ultraviolet light falling on a metal can result in the ejection of
electrons from the surface.\footnote{See, e.g., Robert Eisberg, Robert
  Resnick, \emph{Quantum Physics of Atoms, Molecules, Solids, Nuclei,
    and Particles}, 2nd ed. (Wiley, New York, 1985) Ch. 2.}  This
light-induced ejection of electrons is now known as the photoelectric
effect. Einstein's explanation of this effect in 1905 (the year he
also developed special relativity!)  is one of the cornerstones of
quantum physics.

According to the classical theory of electromagnetic fields, the
intensity of a light wave is directly proportional to the square of
the electric field of the wave. An electron in some material exposed
to this light wave should feel a force proportional to this electric
field. For an intense enough illuminating light, the electron should
be able to gain sufficient kinetic energy to escape the material. The
energy gained by the electron depends only on the intensity of the
light (and the nature of the material), not on the wavelength. 

That, however, is \emph{not} what is observed experimentally. In a series of
very careful experiments in the 1910\emph{s}, Robert Millikan showed that
the maximum kinetic energy $K_{\rm max}$ the ejected electron 
is independent of the intensity but linearly dependent on the
frequency $\nu$ of the incident light: 
\begin{equation}
	\label{eq:Photo1}
	K_{\rm max} = h \nu - W_{0}
\end{equation}
where $h$ is a constant and $W_{0}$ is the ``work function''
characteristic of the material. Millikan found experimentally that $h$
is numerically equal to the constant Max Planck introduced in his
explanation of blackbody radiation.

In fact, Einstein's theory of the photoelectric effect in 1905
(hypothesized before Millikan's experiments) predicted just such a
relationship, with $h$ being identical to Planck's constant. In this
theory, light exists in individual quanta, or photons. The energy of a
photon is given by its frequency, $E = h\nu$. In the photoelectric
effect a photon is absorbed by an electron, which then acquires the
energy lost by the photon. If the electron is right at the surface (so
it doesn't lose any energy in inelastic collisions on the way to the
surface), then the electron can escape, provided its kinetic energy is
greater than the work function $W_{0}$. Increasing the intensity of
the incident light of a given frequency would simply mean that
\emph{more} electrons are produced with sufficient kinetic energy to
escape; the maximum kinetic energy of the escaping electrons would
remain constant. However, if the frequency of the incident light is so
low that the photon energy is less than the work function then no
electrons will have sufficient energy to escape the
material.\footnote{For sufficiently intense illumination, it is, in
  fact, possible for two ``sub-threshold'' photons to be absorbed by a
  given electron, allowing it to escape the material, even though the
  individual incident photon energies are less than the work
  function. Such nonlinear effects require very intense laser beams.}
The simple linear relationship between photon frequency and energy
thus predicts Millikan's results. Two Nobel Prizes were awarded for
work done on the photoelectric effect---one in 1921 to Einstein for
his theoretical explanation, and one in 1923 to Millikan for his
experimental work on this effect and for his more famous experiments
establishing the charge of the electron.


\section{Experiment}

In this
experiment you will determine the maximum kinetic energy of electrons
photoejected from a metallic cathode in a vacuum tube under various
illuminations. The maximum kinetic energy is determined by measuring
the ``stopping potential,'' the minimum reverse potential $V$ between the
cathode and the anode which reduces the photoelectric current in the
tube to zero. In this case,
\begin{equation}
	\label{eq:Photo2}
	K_{\rm max} = e V
\end{equation}
where $e$ is the magnitude of the
electron charge. Substituting this expression for $K$ into
\eref{Photo1} and solving for the stopping potential V gives
\begin{equation}
	\label{eq:Photo3}
	V = \left( \frac{h}{e} \right) \nu - \frac{W_{0}}{e} 
\end{equation}
Thus a plot of $V$ vs.\ $\nu$ should give a straight line with a slope
of $h/e$ and an intercept of $-W_{0}/e$.

The experiment consists of two parts. In the first you will
study the effect of light intensity on the stopping potential and test
the predictions of the classical theory of electromagnetic
radiation. In the second you will look carefully at the effect of
light frequency on the stopping potential as a test of the quantum
theory.  

The experimental apparatus, made by PASCO Scientific, consists
basically of a mercury vapor light source, diffraction grating, and a
photodiode tube and associated electronics. The light
source/diffraction grating setup allows you to study five spectral
lines, from the near ultraviolet through yellow. Read quickly through
the PASCO lab manual to get familiar with the equipment and
procedures. You should assume that the basic alignment of the
apparatus has already been accomplished, so that you will only need to
properly locate the grating and photodiode detector for optimal
performance. \emph{Consult with your instructor before you make any
  other alterations.}

\caution{The mercury vapor lamp is a strong source of UV
light. Never look directly into the beam, and always use UV-absorbing
safety glasses when the lamp is on.}

Using the PASCO manual as a rough guide, study the dependence of the
stopping potential.on both the intensity and frequency of the
illuminating light. Your final analysis should include a determination
of Planck's constant and also the work function of the photocathode.%
\footnote{Note from \eref{Photo1} that $h/e$ has the dimensions of
  volt-sec (V\,s) and $W_{0}/e$ has the dimensions of volts (V). From these results you can
directly express $h$ in terms of eV\,s and $W_{0}$ in terms of
electron volts, where $\val{1}{eV} \equiv \text{(charge of electron)}
\times \pval{1}{volt}$.
If you had some independent determination of electron charge, you
could then give these results in terms of, say, joules rather than
electron volts, but that's not necessary here.}
