1. Reverse the voltage Photo Electric Effect
Photon Energy = Work + Kinetic Energy
hf =  Emax
hf = hfo eV
 - Work function (Depends on material)
fo - Lowest frequency that ejects
e - Electron charge
V - The uh stopping potential + - V
Reverse the voltage

2. Example 1: A certain metal has a work function of 3. 25 eV
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?
( E-07 m = 117 nm)

3. Example 2: 70. 9 nm light strikes a metal with a work function of 5
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 V would stop the electrons.)

4. 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
Metal
Work Function

5. Photons of a certain energy strike a metal with a work function of 2
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 eV = 6.00 eV
6.00 eV

6. Another metal has a work function of 3. 46 eV
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 eV = 8.46 eV
E = (8.46 eV)(1.602 x J/eV) = x J
 = hc/E = (6.626 x Js)(3.00 x 108 m/s)/( x J)
 = x m = 147 nm
147 nm

7. 112 nm light strikes a metal with a work function of 4. 41 eV
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 = x J
Photon energy = Work function + Kinetic energy of electron
E = hf = hc/ = (6.626 x Js)(3.00 x 108 m/s)/(112 x 10-9 m)
E = x J
E = ( x J)/(1.602 x J/eV) = eV
eV = 4.41 eV eVs
eV eV = eV = eVs
Vs = 6.67 V
6.67 V

8. 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 = x J
Photon energy = Work function + Kinetic energy of electron
E = hf = hc/ = (6.626 x Js)(3.00 x 108 m/s)/(256 x 10-9 m)
E = x J
E = ( x J)/(1.602 x J/eV) = eV
4.847 eV = Work function eV
eV eV = 3.70 eV
3.70 eV