Questions on the Photoelectric Effect
Q9.
(a) Describe what occurs in the photoelectric effect.
Photons of light incident on the metal surface cause the emission of electrons. The electrons near the surface of the metalare the ones emitted.
[3 marks]
(b) Violet light of wavelength 380 nm is incident on a potassium surface. Deduce whether or not light of this wavelength can cause the photoelectric effect to occur when it is incident on the potassium surface.
Work function of potassium = 2.3 eV
Energy of the photon = hf = hc/λ
E = 6.63 x 10-34 x 3.0 x 108/(380 x 10-9)
E = 5.23 x 10-19 J
E = 5.23 x 10-19/1.6 x 10-19 eV
E = 3.3 eV
The energy of the photon is greater than the energy required to release a photoelectron (the work function), therefore photoelectrons will be emitted when violet light shines on a potassium surface.
Energy of the photon = hf = hc/λ
E = 6.63 x 10-34 x 3.0 x 108/(380 x 10-9)
E = 5.23 x 10-19 J
φ = 2.3 x 1.6 x 10-19 J
φ = 3.7 J
The energy of the photon is greater than the energy required to release a photoelectron (the work function), therefore photoelectrons will be emitted when violet light shines on a potassium surface.
[4 marks]
(c) The photoelectric effect provides evidence for light possessing particle properties. State and explain one piece of evidence that suggests that light also possesses wave properties.
Diffraction of light occurs when light is passed through a very narrow single slit
Interference patterns (light and dark fringes) are observed when light is passed through two slits or a diffraction grating
[2 marks]
(Total 9 marks)