Before you move on, make sure you can use Planck's constant in problem solving.
Energy In a Photon
\(\large\mathsf{ E = hf }\)
... where E is the energy, h is Planck's energy constant of 6.626 x 10-34 J·s, and f is the frequency of the light.
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.
Use these principles to solve each problem in the multiple-choice activity below. If you have difficulties with the problems, revisit the example problems provided in the lesson and read the feedback for some hints.
A photon has an energy of 8.9 eV. What is the frequency of the photon?
- 2.15 x 1015 Hz
- 2.27 x 1015 Hz
- 1.34 x 1034 Hz
- 5.90 x 10-33 Hz
First, convert the eV to Joules, then use \(\small\mathsf{ E = hf }\) to solve for f.
First, convert the eV to Joules, then use \(\small\mathsf{ E = hf }\) to solve for f.
First, convert the eV to Joules, then use \(\small\mathsf{ E = hf }\) to solve for f.
First, convert the eV to Joules, then use \(\small\mathsf{ E = hf }\) to solve for f.
An electromagnetic wave has a frequency of 2.27 x 1014 Hz. How much energy does a photon at this frequency have?
- 0.45 eV
- 0.78 eV
- 0.94 eV
- 1.2 eV
Use \(\small\mathsf{ E = hf }\) to solve for the energy. Then convert to eV.
Use \(\small\mathsf{ E = hf }\) to solve for the energy. Then convert to eV.
Use \(\small\mathsf{ E = hf }\) to solve for the energy. Then convert to eV.
Use \(\small\mathsf{ E = hf }\) to solve for the energy. Then convert to eV.
A photon with frequency of 4.83 x 1014 is incident on a photoelectric surface. The work function of the material is 0.500 eV. What is the kinetic energy of the electrons emitted from that surface?
- 8.0 x 10-20 J
- 3.2 x 10-19 J
- 1.5 x 10-20 J
- 2.4 x 10-19 J
First, convert the work function to Joules, then plug the known values into the equation \(\small\mathsf{ KE_{max} = hf - \Phi }\) to solve for KE.
First, convert the work function to Joules, then plug the known values into the equation \(\small\mathsf{ KE_{max} = hf - \Phi }\) to solve for KE.
First, convert the work function to Joules, then plug the known values into the equation \(\small\mathsf{ KE_{max} = hf - \Phi }\) to solve for KE.
First, convert the work function to Joules, then plug the known values into the equation \(\small\mathsf{ KE_{max} = hf - \Phi }\) to solve for KE.
An incident photon of 2.1 eV is incident on a photoelectric surface that emits electrons with 0.5 eV of kinetic energy. What is the work function of the surface?
- 1.4 eV
- 1.6 eV
- 1.9 eV
- 2.6 eV
Use the photoelectric effect equation to solve for the work funciton. There is no need to convert the units since everything in the problem uses eV.
Use the photoelectric effect equation to solve for the work funciton. There is no need to convert the units since everything in the problem uses eV.
Use the photoelectric effect equation to solve for the work funciton. There is no need to convert the units since everything in the problem uses eV.
Use the photoelectric effect equation to solve for the work funciton. There is no need to convert the units since everything in the problem uses eV.
Summary
Questions answered correctly:
Questions answered incorrectly:
