Contents Einsteins proposed experiment Solving photoelectric problems Example 1 Example 2 Whiteboard Photon vs wave theory Waves Photons Color Brightness Energy per photon changes ID: 537953
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Slide1
The photoelectric effectContents:Einstein’s proposed experimentSolving photoelectric problemsExample 1 | Example 2WhiteboardPhoton vs wave theorySlide2
Waves Photons
Color
Brightness
Energy per photon changes
E = hf
Radio waves vs X-rays
Wavelength Changes
Amplitude Changes
# of photons changes
many = bright
few = dim
CCD Devices, High speed film
big
= red
small = blue
small = dim
big = bright
LightSlide3
Photo Electric EffectHow it worksEinstein’s idea
+
-
V
Reverse the voltage
Photon Energy = Work + Kinetic Energy
hf
=
+
E
max
- Work function
(Depends on material)
hf
– photon energy
E
max
– max KE of photoelectrons
(=
eV
s) – where V
s is the stopping potentialSlide4
Millikan’s setupSlide5
Photon Energy = Work + Kinetic Energyhf = + Emax 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?
117 nmSlide6
Photon Energy = Work + Kinetic Energyhf = + Emax 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?
12.4 VSlide7
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?Photon energy = Work + Kinetic EnergyPhoton Energy = 3.25 eV + 7.35 eV = 10.60 eVConvert to J: (10.60 eV)(1.602E-19 J/eV) = 1.69812E-18 JE = hf = hc/ , = hc/E = (6.626E-34 Js)(3.00E8 m/s)/(1.69812E-18 J
)
= 1.17059E-07 m = 117 nm
Photon Energy = Work + Kinetic Energy
hf =
+ E
max
hf =
hf
o + eV
- Work function (Depends on material)fo
- Lowest frequency that ejectse - Electron chargeV - The uh stopping potentialSlide8
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?E = hf = hc/ = (6.626E-34 Js)(3.00E8 m/s)/(70.9E-9) = 2.80367E-18 JConvert to eV: (2.80367E-18 J)/(1.602E-19 J/eV) = 17.5 eVPhoton Energy = Work + Kinetic Energy17.5 eV = 5.10 + Kinetic Energy
Kinetic Energy = 12.4 eV, so 12.4 V would stop the electrons.
Photon Energy = Work + Kinetic Energy
hf =
+ E
max
hf =
hf
o
+ eV
- Work function (Depends on material)f
o - Lowest frequency that ejectse - Electron chargeV - The uh stopping potentialSlide9
Ag (silver)
4.26
Au (gold)
5.1
Cs (cesium)
2.14
Cu (copper)
4.65
Li (lithium)
2.9
Pb (lead)
4.25
Sn (tin)
4.42
Chromium
4.6
Nickel
4.6
Metal
Work Function
Whiteboards:
Photoelectric effect
1
|
2
|
3
|
4Slide10
6.00 eVPhoton energy = Work function + Kinetic energy of electronPhoton energy = 2.15 eV + 3.85 eV = 6.00 eVPhotons 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?Slide11
147 nmE = hf = hc/,Photon energy = Work function + Kinetic energy of electronPhoton energy = 3.46 eV + 5.00 eV = 8.46 eVE = (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 10
8
m/s)/(1.3553 x 10
-18
J)
= 1.4667x 10-07 m = 147 nm
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)Slide12
6.67 VE = hf = hc/, 1 eV = 1.602 x 10-19 JPhoton energy = Work function + Kinetic energy of electronE = 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 electron11.079 eV = 4.41 eV + eVs
11.079 eV - 4.41 eV = 6.6688 eV = eVsVs = 6.67 V
112 nm light strikes a metal with a work function of 4.41 eV. What is the stopping potential of the ejected electrons? (2)Slide13
3.70 eVE = hf = hc/, 1 eV = 1.602 x 10-19 JPhoton energy = Work function + Kinetic energy of electronE = 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 electron4.847 eV = Work function + 1.15 eV11.079 eV - 1.15 eV = 3.70 eV
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)Slide14Slide15
Einstein’s Photon theory predicts:Photon energy = work function + Kinetic energy of electronhf = + Ekmax E
kmax
=
hf
-
(slope, y intercept)
Photon Theory predicts:
E
kmax
rises with frequencyIntensity of light ejects
moreWave Theory predicts:
Ekmax rises with Amplitude (Intensity)Frequency should not matterSurvey says
Millikan does the experiment
1915 conclusion1930 conclusion