If an electromagnetic wave is shone on a piece of metal, electrons are emitted. Einstein explained this possibility using the quantum theory, that light behaves as particles. In 1905, Einstein proposed an explanation that was in line with Max Planck's theory.
The Photoelectric Effect
\(\large\mathsf{ KE_{max} = hf - \Phi }\)
... where KEmax is the maximum kinetic energy of the electron emitted, h is Planck's constant of 6.626 x 10-34 J·s, f is the frequency of the light shone on the surface, and Φ is the work function.
See if you can answer the questions on the tabs below using the work function equation for the Photoelectric Effect. Once you have answered the questions on your own, click the Answer button to check your work. Again, keep in mind that energy is often shown in electron-volts (eV). 1 eV is equal to 1.6 x 10-19 J, so you may have to convert to Joules from electron-volts and vice versa.
Problem 1
Problem 2
Problem 3
During a photoelectric effect experiment, green light possessing a frequency of 5.50 x 1014 Hz is incident on a photoelectric material. Electrons with a kinetic energy of 0.600 eV are observed to be emitted from the material. What is the work function for the material?
\(\mathsf{ KE_{max} = 0.600 \text{ eV} \cdot 1.60 \times 10^{-19} \text{ J/eV} = 9.60 \times 10^{-20} \text{ J} }\)
\(\mathsf{ KE_{max} = hf - \Phi }\)
\(\mathsf{ 9.60 \times 10^{-20} \text{ J} = 6.626 \times 10^{-34} \text{ J·s} \cdot 5.50 \times 10^{14} \text{ Hz} - \Phi }\)
\(\mathsf{ 9.60 \times 10^{-20} \text{ J} = 3.64 \times 10^{-19} \text{ J} - \Phi }\)
\(\mathsf{ \Phi = 3.64 \times 10^{-19} \text{ J} - 9.60 \times 10^{-20} \text{ J} }\)
\(\mathsf{ \Phi = 2.68 \times 10^{-19} \text{ J} }\)
OR
\(\mathsf{ \Phi = \frac{2.68 \times 10^{-19} \text{ J}}{1.60 \times 10^{-19} \text{ J/eV}} = 1.68 \text{ eV} }\)
During a photoelectric effect experiment, light possessing 3.10 eV of energy is incident on a photoelectric material. Electrons with a kinetic energy of 1.90 eV are observed to be emitted from the material. What is the work function for the material?
\(\mathsf{ KE_{max} = hf - \Phi }\)
\(\mathsf{ 1.90\text{ eV} = 3.10 \text{ eV} - \Phi }\)
\(\mathsf{ \Phi = 3.10 \text{ eV} - 1.90 \text{ eV} }\)
\(\mathsf{ \Phi = 1.2 \text{ eV} }\)
During a photoelectric effect experiment, light possessing 2.10 eV of energy is incident on a photoelectric material. The work function of the material is equal to 0.5 eV. What is the kinetic energy (in eV) of the electrons emitted?
\(\mathsf{ KE_{max} = hf - \Phi }\)
\(\mathsf{ KE_{max} = 2.10 \text{ eV} - 0.5 \text{ eV} }\)
\(\mathsf{ KE_{max} = 1.6 \text{ eV} }\)
Question
Do you always need to convert to Joules when doing these problems?
No, as long as all of the units in the problem match, you do not need to convert. However, if you have to do the calculation for energy of the photon using Planck's constant, you must convert to Joules.
