## Studio Session 11

### The photoelectric effect

In the last studio session you observed diffraction and interference.  You studied the diffraction and interference patters produced by single and multiple slits and verified that one can predict the positions of the maxima and minima in these patterns by assuming that light is an EM wave.  Where crests meet crests and troughs meet troughs we predict and observe maxima or bright regions, and where crests meet troughs we predict and observe minima or dark regions.

### So light is a wave, these experiments settle this?  Not so fast!

In this session you will simulate an experiment that suggests that light is a particle.  You will investigate the photoelectric effect.  To eject an electron from a metal surface a certain amount of energy Φ must be supplied to this electron.  Φ is called the work function of the metal.  (If no energy were required to free the electrons, they would just leave ordinary pieces of metal.)  In the wave picture the energy of the light beam does not depend on the frequency, but only on the intensity, which is proportional to the square of the amplitude.  Einstein explained the photoelectric effect by postulating that an electron can only receive the large amount of energy necessary to escape the metal from the EM wave by absorbing a single photon.  If this photon has enough energy, the electron is freed.  Excess energy appears as kinetic energy of the electron.  The maximum kinetic energy of the electron is given by E = hf - Φ.  If the photon does not have enough energy, then the electron cannot escape the metal.

In this session you will direct light with different wavelength onto a metal surface and measure the kinetic energy of the photoelectrons ejected from the metal as a function of the frequency of the light used to eject the electrons.  You will measure the work function of the metal and also determine the value of Planck's constant from your data.  This experiment reveals the "particle nature" of light.  A second experiment in the studio session will reveal the "wave nature" of electrons.  You will observe and analyze electron diffraction.

Equipment needed:

• Demonstration equipment

Open a Microsoft Word document to keep a log of your experimental procedures, results and discussions.  This log will become your lab report.  Address the points highlighted in blue.  Answer all questions.

Experiment 1

You will use an on-line simulation from the University of Colorado PhET group.

Explore the interface. There are some non-obvious controls.

• You can select Show photons in the Options menu to show the light beam as composed of individual photons.
• You can select Control photon number instead of intensity in the Options menu to change the Intensity slider to a Number of photons slider.
• You can use the camera icon to take a snapshot of the graphs so that you can compare graphs for different settings.
• You can Pause the simulation and then use Step to incrementally analyze.

In the simulation photons strike a metal cathode and eject electrons.  Electrons are ejected with a range of energies, up to a maximum energy.  The electrons are collected on the anode and then flow back from the anode to the cathode through a wire.  The current in the wire is measured.  Electrons are negatively charged.  If the anode is at a positive voltage compared to the cathode, electrons are attracted and gain energy.  If the anode is at a negative voltage compared to the cathode, electrons are repelled and loose energy.  Let Vca = Vcathode - Vanode.  If this potential difference Vca (energy per charge) becomes large enough, electrons will no longer reach the anode and current will no longer flow in the wire.  It is convenient to measure electron energy in units of electron volt (eV).  In SI units 1 eV = 1.6*10-19 J.
If the potential difference between the cathode and anode is x volts (Vca = x V), then electrons ejected from the cathode need an energy of at least x eV to overcome this potential difference and to reach the anode.  By determining voltage Vca needed to reduce the current in the wire to zero, we can determine the maximum energy of the ejected electrons.
The maximum energy E of electrons that reach the anode in eV has the same numerical value as the voltage in Vca.

(a)  Exploration:

For a Sodium target discuss:

• For a fixed number of photons and zero battery voltage, how does the number of photoelectrons ejected depend on the wavelength?  Does every photon eject an electron?  Does the probability of ejection change with wavelength?   Discuss!
• For a fixed wavelength and zero battery voltage, how does the current depend on the light intensity?  Discuss!
• For a fixed wavelength and light intensity, how does the current depend on the battery voltage?
• For a fixed wavelength and light intensity, do all ejected electrons have the same energy?   How can you measure the maximum energy of the ejected electrons.

(b)  Measurement:

Use a Sodium target.  Set the intensity to 100%.  For the wavelengths listed in the table below, find the maximum energy of the ejected electrons by finding the battery voltage Vca that just prevents the most energetic electrons from reaching the anode.
Note:  The text box under the battery displays -Vca.

Wavelength (nm) Frequency (s-1) Maximum Electron Energy (eV)
150 2.00e15
200 1.50e15
300 1.00e15
400 0.75e15
500 0.60e15

Prepare a graph of the maximum electron energy versus the frequency of the light.  Choose an X-Y Scatter plot.  Plot maximum electron energy on the vertical axis and frequency on the horizontal axis.
Prediction: E = hf - Φ
This equation is of the form y = ax + b, with a being the slope and b being the y-intercept.  The slope of the plot of E versus f will yield Planck's constant and the y-intercept will yield the work function of the metal cathode.
Add a trendline and insert the equation for the trendline into your plot.  Use the slope of your trendline to find Planck's constant h and the intercept value to find the work function Φ of Sodium.  If there are data points that seem to deviate to far from the trendline, repeat those measurements until you are satisfied that you have done your best.
Since the electron energy is measured in eV and the frequency in 1/seconds, Planck's constant h will have units of eV s and the work function will have units of eV.  Convert Planck's constant to SI units (J s) by multiplying the value you obtained from the slope by 1.6*10-19J/eV.

• What value did you obtain for h in units of eV s and J s?
• How does this value compare with the accepted value h = 6.626*10−34 J s = 4.136*10−15 eV s?
• Describe how the maximum energy of the photoelectrons depends on the wavelength of the incident light.
• Defend whether this experiment supports a wave or a quantum model of light based on your lab results.

Experiment 2

How does matter behave on a scale of a few nanometers or smaller?  Is its behavior governed by Newton's laws or by a wave equation?  The de Broglie relations associate a wavelength λ = h/p = h/√(2mE) with each particle of momentum p.  For an electron which has been accelerated through a potential difference of 5 kV and therefore has a kinetic energy of 5000 eV = 8*10-16 J, this wavelength is λ = 1.74*10-11 m.  Can we measure λ?

If a beam of accelerated electrons passes through a thin crystal, the crystal planes can act like a diffraction grating.  Different crystal planes can produce different diffraction patterns.  Planes that are spaced farther apart produce a narrower patterns.  You will observe the diffraction pattern produced when 5 keV electrons pass through graphite.

For light normally incident on a grating with slit spacings d, we find diffraction maxima at angles θ away from the normal such that dsinθ = mλ.  If we observe the diffraction pattern on a screen a distance L away from the grating, then we can write dz/(mL) = λ if θ is a small angle.  For electron diffraction from crystal planes at small angles away from the forward direction, we can use the same formula to find the diffraction maxima.

Graphite has the crystal structure shown below.

In this orientation, the d100 planes  produce a horizontal pattern.
d100 = 2.10*10-10 m

.

In this orientation, the d110 planes produce a horizontal pattern.
d110 = 1.21*10-10 m.

The graphite in our apparatus is not a single crystal, but a polycrystalline powder.  The orientation of the various crystals is random.   Powder diffraction produces a pattern of concentric rings. The powder diffraction patters is just the superposition of the patterns produced by the individual crystals with random orientations.

Assume for a single crystal with a fixed orientation we observe the pattern below.  The diffraction pattern is rotated by the same angle as the crystal.

Then for 4 crystals with 4 different orientations, we also observe 4 different orientations of the diffraction patterns.  The individual diffraction patterns plotted in the same color as the corresponding crystal start to add up to rings.

For 40 randomly oriented crystals, powder rings become clearly visible.

In our experiment accelerated electron with 5 keV kinetic energy pass through a graphite target in an evacuated tube and hit a fluorescent screen.  We observe the ring pattern on this screen.
Two rings are clearly visible.  These correspond to the first-order maxima produced by the d100 and the d110 planes.  All other rings are too dim or at angles too large to observe with our apparatus.  You will measure the diameter of the two visible rings and calculate the angles θ, given the distance L = 13.5 cm from the target to the screen.
Using the known plane spacings d100 = 2.10*10-10 m and d110 = 1.21*10-10 m you will then use dz/L = λ to experimentally determine the de Broglie wavelength λ of the 5 keV electrons.

Schematic sketch off the apparatus:
L = 13.5 cm (distance between graphite foil and screen)
D = diameter of a diffraction ring observed on the screen
z = D/2

Make the measurements and fill in the table below.  Measure the center-to-center distance for each bright ring.  The scale on the picture is a mm scale.

1st ring (d100) 2nd ring (d110)
d (m) 2.10E-10 1.21E-10
L (m) 0.135 0.135
D (m)
z
λ = dz/L