The photoelectric effect
Contents:
•Einstein’s proposed experiment
•Solving photoelectric problems
•Example 1 | Example 2
•Whiteboard
•Photon vs wave theory
Light
Waves
Wavelength Changes
Color
small  = blue
Photons
Energy per photon changes
E = hf
X-Rays, UV, Gamma
big  = red
Amplitude Changes
Brightness
small = dim
big = bright
# of photons changes
many = bright
few = dim
CCD Devices, High
speed film
Einstein
•Proposes Photon theory
•Experiment:
Light ejects electrons
Ammeter detects
•Some Potential V
stops all electrons
•This is called the
“Stopping potential”
•(Pretty tough, huh?)
V
-
+
Reverse the voltage
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-
Electric Force
+
Consider an ejected electron hurtling toward the negative plate:
Remember V = W/q, so W = Vq
Definition of electron volt
Ekin turns to potential energy
If 5.12 V is the stopping potential, then Ekin = 5.12 eV
Einstein’s Photon theory predicts:
Photon energy = Work function + Kinetic energy of electron
hf
=

+
Ekmax
hf
=
hfo
+
eVs
 - Work function (Depends on material)
fo - Lowest frequency that ejects
e - Electron charge
Vs - The uh stopping potential
V
-
+
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Metal
Work Function
Ag (silver)
Au (gold)
Cs (cesium)
Cu (copper)
Li (lithium)
Pb (lead)
Sn (tin)
Chromium
Nickel
4.26
5.1
2.14
4.65
2.9
4.25
4.42
4.6
4.6
Photon energy = Work function + Kinetic energy of electron
hf
=

+
Ekmax
hf
=
hfo
+
eVs
e = 1.602 x 10-19 C
 - Work function
fo - Lowest frequency that ejects V = W/q
e - Electron charge
E = hf = hc/
Vs - The uh stopping potential
Example 1:
A certain metal has a work function of 3.25 eV. When
light of an unknown wavelength strikes it, the electrons
have a stopping potential of 7.35 V. What is the
wavelength of the light?
TOC
Photon energy = Work function + Kinetic energy of electron
hf
=

+
Ekmax
hf
=
hfo
+
eVs
e = 1.602 x 10-19 C
 - Work function
fo - Lowest frequency that ejects V = W/q
e - Electron charge
E = hf = hc/
Vs - The uh stopping potential
Example 2:
70.9 nm light strikes a metal with a work function of
5.10 eV. What is the maximum kinetic energy of the
ejected photons in eV? What is the stopping potential?
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Whiteboards:
Photoelectric effect
1|2|3|4
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Photons of a certain energy strike a metal
with a work function of 2.15 eV. The
ejected electrons have a kinetic energy of
3.85 eV. (A stopping potential of 3.85 V)
What is the energy of the incoming
photons in eV?
Photon energy = Work function + Kinetic energy of electron
Photon energy = 2.15 eV
+ 3.85 eV = 6.00 eV
6.00 eV
W
Another metal has a work function of
3.46 eV. What is the wavelength of
light that ejects electrons with a
stopping potential of 5.00 V? (2)
E = hf = hc/,
Photon energy = Work function + Kinetic energy of electron
Photon energy = 3.46 eV
+ 5.00 eV = 8.46 eV
E = (8.46 eV)(1.602 x 10-19 J/eV) = 1.3553 x 10-18 J
E = hf = hc/,
 = hc/E = (6.626 x 10-34 Js)(3.00 x 108 m/s)/(1.3553 x 10-18 J)
 = 1.4667x 10-07 m = 147 nm
147 nm
W
112 nm light strikes a metal with a
work function of 4.41 eV. What is the
stopping potential of the ejected
electrons? (2)
E = hf = hc/, 1 eV = 1.602 x 10-19 J
Photon energy = Work function + Kinetic energy of electron
E = hf = hc/ = (6.626 x 10-34 Js)(3.00 x 108 m/s)/(112 x 10-9 m)
E = 1.7748 x 10-18 J
E = (1.7748 x 10-18 J)/(1.602 x 10-19 J/eV) = 11.079 eV
Photon energy = Work function + Kinetic energy of electron
11.079 eV
= 4.41 eV
+ eVs
11.079 eV - 4.41 eV = 6.6688 eV = eVs
Vs = 6.67 V
6.67 V
W
256 nm light strikes a metal and the
ejected electrons have a stopping
potential of 1.15 V. What is the work
function of the metal in eV? (2)
E = hf = hc/, 1 eV = 1.602 x 10-19 J
Photon energy = Work function + Kinetic energy of electron
E = hf = hc/ = (6.626 x 10-34 Js)(3.00 x 108 m/s)/(256 x 10-9 m)
E = 7.7648 x 10-19 J
E = (7.7648 x 10-19 J)/(1.602 x 10-19 J/eV) = 4.847 eV
Photon energy = Work function + Kinetic energy of electron
4.847 eV
= Work function + 1.15 eV
11.079 eV - 1.15 eV = 3.70 eV
3.70 eV
W
Einstein’s Photon theory predicts:
Photon energy = work function + Kinetic energy of electron
hf =  + Ekmax
Ekmax = hf - 
Photon Theory predicts:
•Ekmax rises with frequency
•Intensity of light ejects more
Wave Theory predicts:
•Ekmax rises with Amplitude
(Intensity)
•Frequency should not matter
Survey says
Millikan does the experiment
1915 conclusion
1930 conclusion
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