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The photoelectric effect questions
Shine light on a metal and electrons can leap out, but the details do not match the predictions of a wave model. One experiment forced physics to accept photons, and its equation reads as a three-term energy budget you can balance in a single line.
19 original questions · 58 marks · the the photoelectric effect notes · Quantum phenomena
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State two observations of the photoelectric effect that the wave model of light cannot explain.
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There is a threshold frequency below which no electrons are emitted whatever the intensity (1). The maximum kinetic energy of the emitted electrons depends on the frequency of the light, not on its intensity (1). (Also accept: emission is instantaneous even in very dim light.)A photon has a frequency of 5.0 × 1014 Hz. Taking h = 6.63 × 10−34 J s, calculate its energy.
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E = hf = 6.63 × 10−34 × 5.0 × 1014 (1)
E = 3.31 × 10−19 J (1)Define the work function of a metal.
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The work function is the minimum energy needed (1) to release an electron from the surface of the metal (1).A student shines light of frequency 9.0 × 1014 Hz onto a clean gold surface. Gold has a work function of 8.2 × 10−19 J. Explain, with a calculation, why no electrons are emitted, however intense the light (h = 6.63 × 10−34 J s).
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Photon energy E = hf = 6.63 × 10−34 × 9.0 × 1014 = 5.97 × 10−19 J (1). This is less than the work function, and at ordinary intensities an electron absorbs only one photon, so no photon supplies enough energy to release an electron; greater intensity means more photons, not more energetic ones (1).A graph of the maximum kinetic energy of photoelectrons against the frequency of the incident light is a straight line. State what the gradient of this line represents.
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The Planck constant h (1).Light of a fixed frequency, above the threshold frequency, falls on a metal surface. The intensity of the light is doubled. State and explain the effect of this change on the number of electrons emitted per second.
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The number emitted per second doubles (1), because the number of photons arriving per second doubles and each emission needs one photon (1).A metal has a work function of 3.2 × 10−19 J. Calculate its threshold frequency (h = 6.63 × 10−34 J s).
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At the threshold, hf0 = φ, so f0 = φ/h (1)
= (3.2 × 10−19)/(6.63 × 10−34) (1)
f0 = 4.83 × 1014 Hz (1)Light of frequency 8.0 × 1014 Hz falls on a metal of work function 3.0 × 10−19 J. Calculate (a) the maximum kinetic energy of the emitted electrons and (b) their maximum speed (me = 9.11 × 10−31 kg).
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(a) Ek(max) = hf − φ (1)
= 6.63 × 10−34 × 8.0 × 1014 − 3.0 × 10−19 = 2.30 × 10−19 J (1)
(b) v = √(2Ek/m) (1)
v = 7.11 × 105 m s−1 (1)A photon has an energy of 3.31 × 10−19 J. Calculate the photon energy in electronvolts (e = 1.60 × 10−19 C).
In a photoelectric experiment, electrons leave a metal surface with a maximum kinetic energy of 2.30 × 10−19 J. Calculate the stopping potential, the minimum potential difference across the cell that reduces the photoelectric current to zero (e = 1.60 × 10−19 C).
In a photoelectric experiment a student varies the frequency of the light falling on a metal cathode and plots the maximum kinetic energy of the photoelectrons against frequency. The straight line obtained meets the frequency axis at 4.6 × 1014 Hz. Determine the work function of the metal, in J and in eV (h = 6.63 × 10−34 J s, e = 1.60 × 10−19 C).
Light of frequency 7.5 × 1014 Hz falls on the cathode of a vacuum photocell. The photoelectric current falls to zero when the stopping potential is 0.60 V. Determine the work function of the metal, in eV (h = 6.63 × 10−34 J s, e = 1.60 × 10−19 C).
A green laser pointer emits light of wavelength 500 nm. The beam falls on a caesium surface of work function 1.9 eV. Calculate the maximum kinetic energy of the emitted electrons, in J (h = 6.63 × 10−34 J s, c = 3.0 × 108 m s−1, e = 1.60 × 10−19 C).
A metal has a work function of 2.0 eV. Taking h = 6.63 × 10−34 J s, c = 3.0 × 108 m s−1 and e = 1.60 × 10−19 C, calculate the longest wavelength of light that can release an electron.
Light of wavelength 400 nm falls on the same metal (work function 2.0 eV). Calculate (a) the maximum kinetic energy of the emitted electrons and (b) the stopping potential.
Explain, using the photon model, why increasing the intensity of light of a fixed frequency does not increase the maximum kinetic energy of the emitted electrons.
A student is designing a light sensor that must respond to every wavelength of visible light, from 400 nm to 700 nm. Three photocathode coatings are available: coating X has a work function of 1.4 eV, coating Y 2.2 eV and coating Z 4.3 eV. Deduce which coating the student should choose (h = 6.63 × 10−34 J s, c = 3.0 × 108 m s−1, e = 1.60 × 10−19 C).
In a vacuum photocell, ultraviolet radiation of wavelength 365 nm and power 2.4 mW falls on the cathode. On average, one photoelectron is emitted for every 120 photons striking the cathode. Calculate the photoelectric current (h = 6.63 × 10−34 J s, c = 3.0 × 108 m s−1, e = 1.60 × 10−19 C).
A zinc plate is given a negative charge and placed on the cap of a gold-leaf electroscope, so that the leaf rises. When ultraviolet radiation is shone on the plate, the leaf slowly falls. When very bright visible light is used instead, the leaf stays up, however long the light shines. The work function of zinc is 6.9 × 10−19 J. Explain both observations.
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