Photoelectric effect
When light of suitable frequency falls on a metal surface, electrons are emitted. These are called photoelectrons. The effect was observed by Hertz and studied by Hallwachs and Lenard.
Key terms
- Work function (φ0): the minimum energy needed to free an electron from the metal surface.
- Threshold frequency (ν0): the minimum frequency of light that can eject electrons, ν0 = φ0/h. The threshold wavelength is λ0 = hc/φ0.
- Stopping potential (V0): the reverse potential that just stops the fastest photoelectrons, so that eV0 = Kmax.
Experimental observations
- Photocurrent is proportional to the intensity of light, for a fixed frequency above ν0.
- Kmax and V0 do not depend on intensity.
- Kmax increases linearly with the frequency of light.
- Below ν0 there is no emission, however intense the light.
- Emission is instantaneous, with a delay of less than about 10−9 s.
The wave theory of light predicted that more intensity should give more energy per electron and that a time lag should exist at low intensity. Neither is observed, which is why a new picture was needed.
Einstein’s photoelectric equation
Einstein proposed that light travels in packets of energy called photons, each carrying E = hν and momentum p = h/λ. One photon gives all its energy to one electron. Part of it frees the electron and the rest becomes kinetic energy.
Graphs you must know
V0 against ν: a straight line. The intercept on the frequency axis is ν0, and the slope h/e is the same for every metal.
Other graphs
- Kmax vs ν: a straight line with slope h and intercept −φ0 on the energy axis.
- Photocurrent vs intensity: a straight line through the origin.
- Photocurrent vs anode voltage: the current rises to a saturation value, which is higher for greater intensity. It falls to zero at −V0, which is the same for every intensity but larger for higher frequency.
- Different metals give parallel lines in V0 vs ν, since the slope is always h/e.
Wave nature of matter: de Broglie
If light can behave like a particle, de Broglie asked, can particles behave like waves? He proposed that every moving particle has an associated wavelength:
Useful forms
- From kinetic energy K: λ = h / √(2mK)
- Charge q accelerated through potential V: λ = h / √(2mqV)
- Electron accelerated through V volts: λ = 12.27 / √V Å (about 1.227 / √V nm)
The wavelength of everyday objects is far too small to detect because h is so tiny. A cricket ball moving at 40 m/s has λ of roughly 10−34 m. For electrons it is comparable to atomic spacing, so diffraction can be observed.
Davisson–Germer experiment
Davisson and Germer fired a beam of electrons at a nickel crystal and measured the scattered intensity at different angles. A sharp peak appeared at 54 V and a scattering angle of 50°. The wavelength found from Bragg’s law agreed with the de Broglie prediction.
- Bragg’s law: 2d sinθ = nλ
- Measured wavelength: about 1.65 Å
- de Broglie wavelength at 54 V: 12.27/√54 ≈ 1.67 Å
This was the first direct experimental proof of the wave nature of electrons. G. P. Thomson independently observed electron diffraction through thin metal foils.
Formula sheet
| Quantity | Formula |
|---|---|
| Photon energy | E = hν = hc/λ = 1240 / λ(nm) eV |
| Photon momentum | p = h/λ = E/c |
| Work function / threshold | φ0 = hν0 = hc/λ0 |
| Einstein’s equation | hν = φ0 + Kmax |
| Stopping potential | eV0 = Kmax = h(ν − ν0) |
| de Broglie wavelength | λ = h/p = h/√(2mK) |
| Electron through V volts | λ = 12.27/√V Å |
| Bragg’s law | 2d sinθ = nλ |
Worked numericals
1. Light of wavelength 400 nm falls on a metal of work function 2.3 eV. Find Kmax and the stopping potential.
Photon energy = 1240/400 = 3.1 eV. Kmax = 3.1 − 2.3 = 0.8 eV, so V0 = 0.8 V.
2. Find the energy of a photon of wavelength 6200 Å.
λ = 620 nm, so E = 1240/620 = 2.0 eV.
3. Find the de Broglie wavelength of an electron accelerated through 100 V.
λ = 12.27/√100 = 1.227 Å.
4. A metal has a threshold wavelength of 620 nm. Will light of wavelength 500 nm cause emission?
Emission needs λ ≤ λ0. Since 500 nm < 620 nm, the photon energy (2.48 eV) exceeds the work function (2.0 eV), so yes, with Kmax = 0.48 eV.
Quick questions
Why does increasing the intensity not increase Kmax?
Each electron absorbs one photon. Greater intensity means more photons per second, so more electrons are emitted, but the energy of each photon, hν, is unchanged.
Why is there no photoemission below the threshold frequency?
A single photon of lower frequency has less energy than the work function, and an electron cannot collect energy from several photons, so no electron escapes however intense the light.
What is the slope of the V0 vs ν graph?
The slope is h/e, a universal constant that is the same for every metal. Different metals give parallel lines with different intercepts.
Why can’t we see the wave nature of a moving cricket ball?
Its de Broglie wavelength, about 10−34 m, is far smaller than any size we can measure, so no diffraction or interference is observable.
What did the Davisson–Germer experiment prove?
It showed that electrons diffract from a crystal like waves, and that the measured wavelength matches h/p, confirming de Broglie’s hypothesis.
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