Name(s) ____________________
PHOTOELECTRIC EFFECT
Figure 1
BACKGROUND
According to the photon theory of light, the maximum kinetic energy of photoelectrons depends
only on the frequency of the incident light and is independent of the intensity of light. The
prediction is that the higher the frequency of the light, the greater its energy. This differs from
the classical wave model which predicts that the maximum kinetic would depend on light
intensity. That would mean that the brighter the light, the greater the energy of the electrons.
Summarizing the differences between the classical view and what is seen in the photoelectric
effect:
1. The time lag between turning on the light and the ejection of the first electrons is
unaffected by the brightness or frequency of the light.
2. The effect is easy to observe with violet or ultraviolet light but not with red light.
3. The rate at which electrons are ejected (how MANY electrons per second) is proportional to
the brightness of the light.
4. The maximum energy of the ejected electrons is unaffected by the brightness of the light,
but does depend on the frequency of the light.
We will investigate the maximum energy of photoelectrons as a function of the frequency of light.
EQUATIONS:
These equations will be used to describe the photoelectric effect that we will see.
1) E = hf
Energy = Planks Constant x frequency
2) E = hf = KE + Wo
Energy = Kinetic Energy + Work Function
3) KE = Ve
Kinetic Energy = voltage x charge of electron
4) hf = Ve + Wo
Solve this for voltage:
V = (h/e)f - (Wo/e)
(Notice the similar form of the equation to y = mx + b)
APPARATUS : Pasco h/e apparatus as in Figures 1 and 2, digital voltmeter, banana leads.
Figure 2
PROCEDURE
1. When ready to take data, the room must be darkened.
2. Assemble the apparatus as shown in Figure 1 and tighten the thumbscrews. Turn on the mercury
light source and let it warm up for 5 minutes. Test the battery and make sure it is at least +/– 6 volts.
3. Rotate the detector apparatus with the coupling bar until you see the five colors of the mercury light
spectrum on the white reflective mask as in Figure 2. Use the bright, first order spectrum closest to the
central line from the source.
4. Connect the voltmeter to the red and black terminals on the detector. Set the scale to 2 Volts. Turn
on the detector.
5. Open the light shield on the detector (cylinder on detector behind the white reflective mask). Adjust
the apparatus so that only one of the colored bands of the mercury light spectrum falls on the opening of
the mask on the photodiode.
6. Press the zero button on the detector. After it has been reduced to zero volts on the voltmeter, release
the zero button. After it has stabilized, read the voltage (on voltmeter) produced by the light. Record
the light color and the voltage. Be sure to take all the decimal places you can read. Do not round off!
Use 0.613V instead of 0.6 V, for example.
7. Repeat this process for each of the five colors of the mercury light spectrum. Use yellow and green
filters respectively for each yellow and green color to block out the other, higher energy light. (You can
“see” the UV because the reflective mask is made of a fluorescent material. You can see the actual
colors of the light by holding a piece of non-fluorescent white material in front of the mask.)
DATA TABLE
Color
Yellow
Wavelength
578 nm
Frequency (Hz)
5.19x1014
Green
546 nm
5.49x1014
Blue
436 nm
6.88x1014
Violet
405 nm
7.41x1014
Ultraviolet
365 nm
8.20x1014
V (Volts)
ANALYSIS
8. In the table, the frequencies are given for the different colors of light in the spectrum. Plot Voltage
(y-axis) vs. frequency (x-axis) on Excel. Find the best fit line. Attach the graph.
9. From equation 4, we see that V = (h/e)f - (Wo/e) so when we plot the data this way, the slope will be
h/e. Why? Explain what the rest of the best fit line represents by comparing it to V = (h/e)f - (Wo/e).
10. From your graph, what is the experimental value for h/e (the slope)? (Take all decimal places
given.)
Experimental value from slope of line ______________
11. From the given values, compute the accepted value of h/e.
Accepted values:
h = 6.63 x 10-34 J۰s
e = 1.60 x 10-19 C
Accepted value of h/e ___________________
12. Compare your experimental value to the accepted value and find the % error.
% Difference =
Value actual  Valueexp erimental
Value actual
 100%
% Difference = _________________________
13. Explain how the photoelectric effect supports the idea of light as a particle instead of a wave.
14. What are some possible sources of experimental error?