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Questions on the photoelectric effect and the photon
Light ejects electrons from a metal surface only above a threshold frequency, an observation no wave model of light accounts for. Einstein's photon explanation gives the photoelectric equation hf = φ + Ek(max), where the work function φ is the least energy an electron needs to escape and the stopping potential measures Ek(max).
27 original questions · 86 marks · the photoelectric effect and the photon notes · Quantum phenomena
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These questions cover both lessons on the photoelectric effect: The photoelectric effect and the photon and The photoelectric graph and measuring h.
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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 = 3.3 × 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 graph of the maximum kinetic energy of photoelectrons against the frequency of the incident light is a straight line. State what the intercept on the frequency axis represents, and what the intercept on the energy axis represents.
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The intercept on the frequency axis is the threshold frequency f0 of the metal (1). The intercept on the energy axis is at −φ, minus the work function of the metal (1).A student increases the pd across an LED of known wavelength until the LED is first seen to glow, and uses that pd to find a value for the Planck constant. Give two reasons why the value is described as an estimate rather than a measurement.
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The pd at which the glow first appears is judged by eye in a darkened room, so different observers take different readings (1). An LED emits a range of wavelengths about the value quoted on it, so λ is not exact (1). (Also accept: some of the energy an electron loses crossing the junction does not go to the photon, so eV overstates the photon energy.)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).
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E = (3.31 × 10−19)/(1.60 × 10−19) (1)
E = 2.07 eV (1)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).
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At the stopping potential the most energetic electrons are just turned back, so eVs = Ek(max) (1)
Vs = Ek(max)/e = (2.30 × 10−19)/(1.60 × 10−19) (1)
Vs = 1.44 V (1)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).
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The intercept on the frequency axis is the threshold frequency, so φ = hf0 (1). φ = 6.63 × 10−34 × 4.6 × 1014 = 3.05 × 10−19 J (1) = 1.9 eV (1).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).
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At the stopping potential eVs = Ek(max), so φ = hf − eVs (1). φ = 6.63 × 10−34 × 7.5 × 1014 − 1.60 × 10−19 × 0.60 = 4.01 × 10−19 J (1) = 2.5 eV (1).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).
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Photon energy E = hc/λ = (6.63 × 10−34 × 3.0 × 108)/(500 × 10−9) = 3.98 × 10−19 J = 2.49 eV (1). Ek(max) = E − φ = 2.49 − 1.9 = 0.59 eV (1) = 9.4 × 10−20 J (1).In an experiment to estimate the Planck constant, a blue-green LED of quoted wavelength 505 nm is found to just start glowing at a pd of 2.40 V. Using eV = hc/λ, calculate an estimate of the Planck constant, and find the percentage difference between this value and the accepted value 6.63 × 10−34 J s (c = 3.0 × 108 m s−1, e = 1.60 × 10−19 C).
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At the threshold pd, eV = hc/λ, so h = eVλ/c (1)
h = (1.60 × 10−19 × 2.40 × 505 × 10−9)/(3.0 × 108) (1)
h = 6.46 × 10−34 J s (1)
Percentage difference = (h − 6.63 × 10−34)/(6.63 × 10−34) × 100 ≈ 2.5% low (1)In a photoelectric experiment the graph of the maximum kinetic energy of the photoelectrons against the frequency of the incident light is a straight line. Extended backwards, the line meets the energy axis at −3.36 × 10−19 J. Determine the threshold frequency of the metal (h = 6.63 × 10−34 J s).
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The intercept on the energy axis is at −φ, so the work function is φ = 3.36 × 10−19 J (1)
At the threshold the photon supplies the work function exactly, hf0 = φ, so f0 = φ/h (1)
f0 = 5.07 × 1014 Hz (1)On one pair of axes, a student sketches the maximum kinetic energy of photoelectrons against the frequency of the incident light for two different metals, drawing a steeper line for the metal with the smaller work function. Explain why the sketch cannot be correct, and describe how the two lines should differ.
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The gradient of each line is the Planck constant, a universal constant that does not depend on the surface, so both metals give the same gradient (1). The two lines should therefore be parallel (1). The metal with the larger work function has its line shifted to the right: it meets the frequency axis at a higher threshold frequency and the energy axis at a more negative value (1).In an experiment to estimate the Planck constant, the pd at which each of five LEDs is first seen to glow is plotted against 1/λ, where λ is the wavelength quoted for that LED. The points lie on a straight line passing through (1.45 × 106 m−1, 1.80 V) and (2.60 × 106 m−1, 3.23 V). Determine the gradient of the line and hence estimate the Planck constant (c = 3.0 × 108 m s−1, e = 1.60 × 10−19 C).
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gradient = ΔV/Δ(1/λ) = (3.23 − 1.80)/(2.60 × 106 − 1.45 × 106) (1)
gradient = 1.24 × 10−6 V m (1)
At each threshold eV = hc/λ, so V = (hc/e) × (1/λ) and the gradient is hc/e (1)
h = gradient × e/c = 6.63 × 10−34 J s (1)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.
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φ = 2.0 × 1.60 × 10−19 = 3.20 × 10−19 J (1)
λ0 = hc/φ (1)
= (6.63 × 10−34 × 3.0 × 108)/(3.20 × 10−19) (1)
λ0 = 622 nm (1)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.
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Photon energy E = hc/λ = (6.63 × 10−34 × 3.0 × 108)/(400 × 10−9) (1)
E = 4.97 × 10−19 J (1)
(a) Ek(max) = E − φ = 4.97 × 10−19 − 3.20 × 10−19 = 1.77 × 10−19 J (1)
(b) Vs = Ek(max)/e = 1.11 V (1)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.
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Each electron is emitted by absorbing a single photon, and a photon's energy depends only on frequency (E = hf) (1). Increasing intensity delivers more photons per second, so more electrons are emitted (1), but each still carries the same energy, so the maximum kinetic energy is unchanged (1).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).
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The least energetic visible photon is at 700 nm: E = hc/λ = 2.84 × 10−19 J = 1.8 eV (1). To respond across the whole range, the work function must not exceed this photon energy (1). Y emits only for wavelengths below hc/φ = 565 nm, and Z only for ultraviolet, so neither covers the range (1). Coating X (1.4 eV < 1.8 eV) emits for all visible wavelengths, so the student should choose X (1).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).
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Photon energy E = hc/λ = (6.63 × 10−34 × 3.0 × 108)/(365 × 10−9) = 5.45 × 10−19 J (1). Photons per second = P/E = (2.4 × 10−3)/(5.45 × 10−19) = 4.4 × 1015 s−1 (1). Electrons per second = 4.4 × 1015/120 = 3.7 × 1013 s−1 (1). Current I = electrons per second × e (1) = 5.9 × 10−6 A, about 5.9 μA (1).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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An ultraviolet photon has energy hf greater than the work function of zinc (1), so each photon absorbed at the surface releases one electron from the plate (1). The plate gradually loses its negative charge, so the leaf falls (1). A visible photon has energy less than 6.9 × 10−19 J: its frequency is below the threshold frequency (1). At the intensities of an ordinary lamp an electron absorbs only one photon, so no electron gains enough energy to escape and energy cannot accumulate over time (1). Increasing the intensity increases the number of photons, not the energy of each, so no electrons are released and the leaf stays up (1). (Multiphoton absorption below the threshold is possible under an intense pulsed laser, but not with any ordinary light source.)A student plots the maximum kinetic energy of the photoelectrons against the frequency of the incident light for one metal and obtains a straight line. The line meets the frequency axis at 5.40 × 1014 Hz and passes through the point (9.50 × 1014 Hz, 2.72 × 10−19 J). Determine, from the graph alone, a value for the Planck constant and the work function of the metal in J and in eV (e = 1.60 × 10−19 C).
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gradient = ΔEk(max)/Δf = (2.72 × 10−19 − 0)/(9.50 × 1014 − 5.40 × 1014) (1)
The gradient is the Planck constant, h = 6.63 × 10−34 J s (1)
The intercept on the frequency axis is the threshold frequency, so φ = hf0 (1)
φ = 6.63 × 10−34 × 5.40 × 1014 = 3.58 × 10−19 J (1)
φ = (3.58 × 10−19)/(1.60 × 10−19) = 2.24 eV (1)A student estimates the Planck constant from a yellow LED of quoted wavelength 590 nm, taking the pd at which it is first seen to glow as 2.30 V. The voltmeter has been connected across the LED and its protective resistor in series, rather than across the LED alone. (a) Using eV = hc/λ, calculate the value of the Planck constant the student obtains. (b) Taking the accepted value as 6.63 × 10−34 J s, determine the pd across the resistor when the LED first glowed, and explain why the student's value is too large (c = 3.0 × 108 m s−1, e = 1.60 × 10−19 C).
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(a) h = eVλ/c = (1.60 × 10−19 × 2.30 × 590 × 10−9)/(3.0 × 108) (1)
h = 7.24 × 10−34 J s (1)
(b) The LED's own threshold pd is V = hc/(eλ) = 2.11 V (1)
so the resistor takes 2.30 − 2.11 = 0.19 V (1)
The voltmeter reads the pd across the resistor as well as across the LED, so the pd used is larger than the LED's threshold pd, and h is proportional to that pd, leaving the estimate about 9% high (1)
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