CHAPTER 6
PHOTOELECTRIC EFFECT When we apply the quantum theory to the photoelectric effect, we must realize that the KE of the ejected electron is directly related to the energy it receives from the incident photon, minus the energy (W) required to remove the electron from the surface of the metal (work function). KE = hf –W 1/2 mv2 = hf–W We see from the equation above that no electrons can be emitted from the surface of the metal unless the product of the frequency and Planck’s constant is greater than the work function (W). In addition, the kinetic energy of the ejected electron depends upon the frequency of the photon and the work function of the metal. As we have seen from the chart of the color, frequency, and photon energy, the energy of the emitted photon is very small. A more convenient method to measure the energy of the electron (whether it absorbs or emits the photon) is to use an energy scale that is of the same magnitude as the electron: the electron volt (eV). The electron volt is defined as the quantity of work required to move an electron through a potential difference of 1 volt. (Remember, 1 volt is equal to 1 Joule per coulomb of charge.) Work = QV 1ev = (1.6 × 10 −19 C )(1V ) 1eV = 1.6 × 10 −19 J = 6.25 × 1018
eV J
Using this relationship, we will go back and recalculate the energy of a photon of yellow light in eV.
E = hf eV = (6.6 × 10 −34 J • s )(5.2 × 1014 Hz ) 6.25 × 10 25 J
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