Photoelectric Effect

The photoelectric effect is observed when light of a sufficiently high frequency is focused onto a polished metal surface, emitting "photo electrons" with a kinetic energy that is given by the difference betweent he photon energy and the work function of the metal.

Historically, this effect was very confusing. For most wave phenomina (and light was well known to behave as a wave phenomenon), one increases the energy carried by the wave by increasing the amplitude of the wave. In the case of light, this means increasing the ententisty. However, with increased intensity, the kinetic energy of the photoelectrons emitted does not increase - there are simple more electrons emitted!

Albert Einstein (1879-1955) offered a solution to this perplecing problem in hi 1905 paper1, "Über einen die Erzeugung und Verwandlung des Lichtes betreffenden heuristischen Gesichtspunkt", Einstein proposed the following relationship:

Ekinetic = hν - Ework function

in which Ekinetic is the kinetic energy of an emitted photoelectron, hν is the energy of a photon of frequency ν, and Ework function is the wrok function for the metal used in the experiment. The work function is the energy required to remove an electron from the surface of the metal.


References:

1Einstein, Albert (1905). "Über einen die Erzeugung und Verwandlung des Lichtes betreffenden heuristischen Gesichtspunkt". Annalen der Physik 17 (6): 132-148 (1905)


Questions

The work function of cesium is 2.1 eV. What is this energy in J?
3.36 x 10-19 J 6.36 x 10-19 J 4.29 x 10-19 J 1.96 x 10-19 J
Hint: 1 eV = 1.602 x 10-19 J

What is the energy of a photon with a frequency of 6.12 x 1014 s-1?
2.05 x 10-19 J 4.05 x 10-19 J 6.24 x 10-19 J 6.12 x 10-19 J
Hint: h = 6.626 x 10-34 J s

Will a photon with a frequency of 6.12 x 1014 s-1 have sufficient energy to eject a photoelectron from the surface of cesium?
Yes, it will! No, it will not.

What will be the kinetic energy of a photon ejected from the surface of cesium by a photon with a frequency of 6.12 x 1014 s-1?
6.12 x 10-19 J 4.05 x 10-19 J 6.90 x 10-20 J 1.09 x 10-18 J

What will be the velocity of a photon ejected from the surface of cesium by a photon with a frequency of 6.12 x 1014 s-1?
7.25 x 105 m/s 4.84 x 105 m/s 3.58 x 105 m/s 2.75 x 105 m/s
Hint: Ekin = 1/2 * m * v2
me = 9.11 x 10-31 kg


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Patrick E. Fleming
Department of Chemistry and Biochemistry
California State University, East Bay
patrick.fleming@csueastbay.edu