Light knocks electrons out of a metal
When light shines on a clean metal surface in a vacuum, electrons are sometimes ejected from the metal. This is the photoelectric effect光电效应, and the ejected electrons are called photoelectrons; collecting them produces a current. Lesson 2.2.2 mentioned a puzzling result: even very bright red light ejects no electrons, while dim ultraviolet light ejects them immediately. The experimental pattern is:
- Each metal has a threshold frequency阈值频率, ν₀. Light below ν₀ ejects no electrons, however bright it is and however long it shines.
- Above ν₀, electrons are ejected immediately, even by very dim light.
- At a fixed frequency, brighter light ejects more electrons per second.
- The maximum kinetic energy of the electrons increases with the frequency of the light and doesn’t depend on its intensity.
If light were only a wave, brighter light would carry more energy, and bright enough red light should eventually eject electrons. Observations 1 and 4 show that what matters is the size of each packet of energy, not the total energy.
The photon model
Einstein explained the photoelectric effect with photons: light is made of photons, each with energy E = hν, and one photon gives all its energy to one electron. An electron needs a certain minimum energy to escape the metal, called the work function逸出功, Φ (also called the threshold energy).
- If hν < Φ, the photon doesn’t have enough energy, and no electron escapes. Brighter light only means more photons, each still too weak, and the energy doesn’t build up in an electron.
- If hν ≥ Φ, an electron is ejected, and the leftover energy becomes its kinetic energy:
- At the threshold, hν₀ = Φ, so the threshold frequency and threshold wavelength are:
Intensity changes only the number of photons arriving per second, so it changes only the number of electrons ejected. Photoelectron spectroscopy (U1.6) uses the same relationship: from the known photon energy and the measured kinetic energy of an electron, it finds the electron’s binding energy.
Work functions of some metals
The work functions below come from the table in Wikipedia, “Work function” (citing the CRC Handbook, 2008 edition); 1 eV = 1.602 × 10⁻¹⁹ J. A work function depends on the condition of the metal’s surface, so sources differ slightly.
| Metal | Φ (eV) | Φ (J) | Threshold wavelength λ₀ | λ₀ falls in |
|---|---|---|---|---|
| Cs | 1.95 | 3.12 × 10⁻¹⁹ | 636 nm | Red |
| K | 2.29 | 3.67 × 10⁻¹⁹ | 542 nm | Green |
| Na | 2.75 | 4.41 × 10⁻¹⁹ | 451 nm | Blue |
| Ca | 2.87 | 4.60 × 10⁻¹⁹ | 432 nm | Violet |
| Mg | 3.66 | 5.86 × 10⁻¹⁹ | 339 nm | Ultraviolet |
| Pb | 4.25 | 6.81 × 10⁻¹⁹ | 292 nm | Ultraviolet |
Only light with a wavelength shorter than λ₀ (a frequency higher than ν₀) ejects electrons. Potassium, for example, has λ₀ = 542 nm, so blue and violet light eject electrons from it, but red and yellow light don’t; magnesium and lead need ultraviolet light.
Choose a metal and drag the wavelength and intensity to see whether electrons are ejected and how much kinetic energy they have:
Work functions from the table in Wikipedia, “Work function” (citing the CRC Handbook, 2008 ed.); real values depend on the surface. On the graph, each metal is a straight line with slope h that meets the axis at its threshold frequency ν₀.
Graphing maximum kinetic energy against frequency
Plot KE_max = hν − Φ with the frequency of the light on the horizontal axis and the maximum kinetic energy of the electrons on the vertical axis:
- each metal gives a straight line with slope h, so the lines for different metals are parallel;
- each line meets the horizontal axis at the threshold frequency ν₀, which is farther right for a larger work function;
- below ν₀ there are no electrons, so there are no points.
Example 1: The threshold for sodium
The work function of sodium is 2.75 eV. Find the threshold frequency and threshold wavelength, and decide which of 650 nm red light, 532 nm green light, and 410 nm violet light can eject electrons from sodium.
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Convert the work function to joules. 1 eV = 1.602 × 10⁻¹⁹ J:
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Threshold frequency. At the threshold the photon energy just equals the work function, hν₀ = Φ:
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Threshold wavelength.
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Decide. The shorter the wavelength, the more energetic the photon. Only light shorter than 451 nm ejects electrons: 410 nm violet light does; 532 nm green light and 650 nm red light don’t, however intense.
Example 2: What intensity and frequency each do
400 nm light shines on potassium (Φ = 3.67 × 10⁻¹⁹ J). Find the maximum kinetic energy of the ejected electrons. What happens if the intensity is doubled? What if 600 nm light is used instead?
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Photon energy.
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Maximum kinetic energy. Part of the photon’s energy overcomes the work function, and the rest becomes the electron’s kinetic energy:
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Double the intensity. Twice as many photons arrive each second, so twice as many electrons are ejected per second; each photon’s energy is unchanged, so the maximum kinetic energy stays 1.30 × 10⁻¹⁹ J.
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Switch to 600 nm. E = hc/λ = 3.31 × 10⁻¹⁹ J, less than Φ, so no electrons are ejected, however intense the light.
- Cesium (λ₀ = 636 nm) in 589 nm yellow light
589 nm is shorter than 636 nm, so the photon energy exceeds the work function.
- Potassium (λ₀ = 542 nm) in very bright 650 nm red light
Each photon has too little energy, and brightness can't make up for it.
- Potassium (λ₀ = 542 nm) in very dim 450 nm blue light
Each photon has enough energy; dim light just means fewer electrons.
- Sodium (λ₀ = 451 nm) in 532 nm green light
532 nm is longer than 451 nm.
- Sodium (λ₀ = 451 nm) in 405 nm violet light
405 nm is shorter than 451 nm.
- Magnesium (λ₀ = 339 nm) in any visible light
All visible light is longer than 339 nm.
- Magnesium (λ₀ = 339 nm) in 254 nm ultraviolet light
254 nm is shorter than 339 nm.
- Lead (λ₀ = 292 nm) in 300 nm ultraviolet light
300 nm is longer than 292 nm, so each photon falls just short.
- A metal with Φ = 3.0 × 10⁻¹⁹ J, photon energy 3.5 × 10⁻¹⁹ J
E > Φ; the maximum kinetic energy is 0.5 × 10⁻¹⁹ J.
- A metal with Φ = 4.0 × 10⁻¹⁹ J, photon energy 2.0 × 10⁻¹⁹ J, shining for a long time
Energy doesn't build up; if one photon isn't enough, no electron is ejected.
- Calcium (ν₀ = 6.94 × 10¹⁴ Hz) in light of frequency 8.0 × 10¹⁴ Hz
The frequency is above the threshold frequency.
Common mistakes
Mistake 1: “Brighter light ejects electrons with more energy.” Intensity only changes the number of photons, and so the number of electrons. The maximum kinetic energy of each electron, KE_max = hν − Φ, depends only on the frequency.
Mistake 2: “Given enough time, light below the threshold frequency will eject electrons.” An electron absorbs the energy of one photon at a time, and the energy doesn’t build up. If each photon has less energy than the work function, no electrons are ejected, however long the light shines.
Mistake 3: “Longer-wavelength photons carry more energy.” E = hc/λ, so the longer the wavelength, the less the energy. That makes the threshold wavelength λ₀ the longest wavelength that can eject electrons; the light must be shorter than it.
Quick check
Red, green, and violet light are each shone on magnesium (λ₀ = 339 nm). What happens?
Compare the photon energy with the work function. Violet photons have the most energy of the three, but violet light is still longer than 339 nm. C is another form of Mistake 2.
Light above the threshold frequency shines on a metal. If the intensity is doubled at the same frequency, what happens?
Twice the intensity means twice the photons and twice the electrons; each photon still has energy hν, so KE_max = hν − Φ is unchanged.
No matter how bright, red light can't eject electrons from sodium. Which explanation is best?
What decides whether electrons are ejected is the energy of each photon. A contradicts observation 1; C has it backwards, since red light has a long wavelength; D has nothing to do with the photoelectric effect.
A metal has a work function of 4.41 × 10⁻¹⁹ J. What is its threshold frequency?
ν₀ = Φ/h = 4.41 × 10⁻¹⁹ ÷ 6.626 × 10⁻³⁴ = 6.65 × 10¹⁴ Hz. A is h/Φ; B multiplies the two; C is the threshold wavelength (in meters).
For metals A and B, graphs of the maximum kinetic energy of ejected electrons against the frequency of light are straight lines. A's line meets the frequency axis at 5.5 × 10¹⁴ Hz and B's at 8.8 × 10¹⁴ Hz. Which statement is correct?
KE_max = hν − Φ, so every slope is h; each line meets the axis at ν₀ = Φ/h, so the farther right it meets the axis, the larger the work function.
250 nm ultraviolet light shines on lead (Φ = 6.81 × 10⁻¹⁹ J). What is the maximum kinetic energy of the ejected electrons?
E = hc/λ = 7.95 × 10⁻¹⁹ J, so KE_max = 7.95 × 10⁻¹⁹ − 6.81 × 10⁻¹⁹ = 1.14 × 10⁻¹⁹ J. B forgets to subtract the work function; C adds it; and 250 nm is shorter than lead's λ₀ (292 nm), so electrons are ejected.
AP-style free response
The work function of calcium is 2.87 eV (1 eV = 1.602 × 10⁻¹⁹ J).
(a) Convert the work function to joules.
(b) Calculate the threshold wavelength of calcium, and identify the color of light it falls in (violet is about 380–450 nm).
(c) Ultraviolet light of wavelength 350 nm shines on calcium. Calculate the maximum kinetic energy of the ejected electrons.
(d) The wavelength stays at 350 nm, but the intensity is halved. How do the number of electrons ejected per second and their maximum kinetic energy change? Justify your answers.
(e) A student predicts, “If 500 nm light is bright enough, it will eject electrons from calcium.” Evaluate the prediction.
Model answer and scoring
(a) Φ = 2.87 × 1.602 × 10⁻¹⁹ J = 4.60 × 10⁻¹⁹ J.
(b)
This is violet light.
(c) E = hc/λ = (6.626 × 10⁻³⁴)(3.00 × 10⁸) ÷ (3.50 × 10⁻⁷) J = 5.68 × 10⁻¹⁹ J; KE_max = 5.68 × 10⁻¹⁹ J − 4.60 × 10⁻¹⁹ J = 1.08 × 10⁻¹⁹ J.
(d) Half as many electrons are ejected per second, because half the intensity delivers half as many photons per second. The maximum kinetic energy is unchanged, because each photon still has energy hν, so KE_max = hν − Φ stays the same.
(e) The prediction is wrong. A 500 nm photon has E = hc/λ = 3.98 × 10⁻¹⁹ J, less than the work function of 4.60 × 10⁻¹⁹ J (equivalently, 500 nm is longer than the 432 nm threshold wavelength). An electron absorbs one photon at a time, so brighter light only adds more photons that each have too little energy, and no electrons are ejected.
Scoring: (a) 4.60 × 10⁻¹⁹ J; (b) 432 nm, violet; (c) 1.08 × 10⁻¹⁹ J; (d) number halved and maximum kinetic energy unchanged, each with a reason; (e) photon energy compared with the work function, noting that intensity can’t make up for it.
Summary
- Photoelectric effect: only light above a metal’s threshold frequency ν₀ ejects electrons; intensity sets the number of electrons, and frequency sets their maximum kinetic energy.
- Photon model: one photon gives its energy hν to one electron; KE_max = hν − Φ, ν₀ = Φ/h, and λ₀ = hc/Φ.
- Light must be shorter than λ₀ (photon energy greater than Φ) to eject electrons; energy doesn’t build up.
- A graph of KE_max against ν is a straight line with slope h that meets the axis at ν₀; different metals give parallel lines.
AP tip: To decide whether electrons are ejected, compare the energy of one photon with the work function (threshold energy), never the total energy or the intensity of the light. Watch the units (wavelength in m, eV converted to J), and check the size of your answer: a visible photon carries about 10⁻¹⁹ J.