Electron emission and the photoelectric effect
The door of a shopping mall opens by itself when a beam of light is cut; the sensor inside works by the photoelectric effect. Link to earlier classes : electrons in metals from Chapter 3 and the electromagnetic spectrum from Chapter 8.
Free electrons in a metal are held inside by the surface. The least energy needed to remove an electron is the work function φ₀ of the metal, usually quoted in electron-volt. Electrons can be freed by heating (thermionic emission), by a strong field (field emission) or by light (photoelectric emission). In the photoelectric effect, light of high enough frequency falling on a metal surface knocks electrons out of it.
Rule 1: Threshold frequency — Electrons are emitted only if the frequency of light is at least the threshold frequency f₀ = φ₀ ÷ h, equivalently if the wavelength is at most λ₀ = h c ÷ φ₀ (h = 6.63 × 10⁻³⁴ J s). Below f₀ no electron comes out, however bright the light.
- Easy : Sodium has φ₀ = 2.3 eV ⇒ f₀ = 2.3 × 1.6 × 10⁻¹⁹ ÷ 6.63 × 10⁻³⁴ = 5.6 × 10¹⁴ Hz.
- Medium : The threshold wavelength is 1240 ÷ 2.3 = 539 nm. Red light of 600 nm (2.07 eV) cannot eject electrons from sodium.
- Exam : Platinum has φ₀ = 5.65 eV, so λ₀ = 1240 ÷ 5.65 = 219 nm, in the ultraviolet.
Rule 2: Units — 1 eV = 1.6 × 10⁻¹⁹ J and E (eV) = 1240 ÷ λ (nm).
- Easy : 4.3 eV = 4.3 × 1.6 × 10⁻¹⁹ = 6.88 × 10⁻¹⁹ J.
- Medium : Metals with a low work function (caesium, sodium) respond to visible light; zinc and platinum need ultraviolet.
- Exam : The photoelectric effect was first observed by Hertz (1887) and studied by Hallwachs and Lenard.
f₀ = φ₀/h; no emission below f₀; 1 eV = 1.6 × 10⁻¹⁹ J.
Believing that a very bright light of low frequency can eject electrons.
Memory clue : "Frequency opens the door; brightness only brings more people."
Spot the difference
| Emission | Energy supplied by |
|---|---|
| Thermionic | heat |
| Field | strong electric field |
| Photoelectric | light |
Practice questions
- 1Define the work function of a metal.
- 2How many joules are there in 1 eV?
- 3Who first observed the photoelectric effect?
- 4Find the threshold frequency of sodium (φ₀ = 2.3 eV).
- 5Find its threshold wavelength.
- 6Can light of 600 nm eject electrons from sodium?
- 7Express a work function of 4.3 eV in joules.
- 8Find the threshold wavelength of platinum (φ₀ = 5.65 eV).
Show answers (try them yourself first)
- 1The minimum energy needed to remove an electron from the metal surface.
- 21.6 × 10⁻¹⁹ J.
- 3Heinrich Hertz (1887).
- 4f₀ ≈ 5.6 × 10¹⁴ Hz.
- 5λ₀ ≈ 539 nm.
- 6No; a photon of 600 nm has 2.07 eV, less than 2.3 eV.
- 76.88 × 10⁻¹⁹ J.
- 8λ₀ ≈ 219 nm (ultraviolet).
Experimental study of the photoelectric effect
Light meters in cameras measure brightness through the current produced by light on a photosensitive plate. Link to earlier classes : potential difference and current from Chapters 2 and 3.
In the experiment, light falls on a metal plate (emitter) in an evacuated tube, and the plate collector is kept at a chosen potential. The current measured is the photocurrent. The collector voltage which just stops the most energetic electrons is the stopping potential V₀, so eV₀ = K_max.
Rule 1: Findings of the experiment — (i) For a given frequency, the photocurrent is proportional to the intensity of light. (ii) The stopping potential, and hence the maximum kinetic energy, does not depend on the intensity. (iii) K_max rises linearly with the frequency. (iv) There is a threshold frequency. (v) The emission is almost instantaneous (less than 10⁻⁹ s), even for faint light.
- Easy : Doubling the intensity doubles the saturation current but leaves V₀ unchanged.
- Medium : V₀ = 1.5 V means K_max = 1.5 eV = 2.4 × 10⁻¹⁹ J.
- Exam : The fastest electrons then have v = √(2 × 2.4 × 10⁻¹⁹ ÷ 9.1 × 10⁻³¹) = 7.3 × 10⁵ m s⁻¹.
Rule 2: Failure of the wave theory — The wave theory predicts that (a) a brighter light should give electrons more energy, (b) any frequency should work if the light is bright enough, and (c) a faint light needs some time to build up the energy. All three disagree with the observations.
- Easy : Wave theory cannot explain a threshold frequency.
- Medium : It cannot explain why K_max does not depend on intensity.
- Exam : Einstein's photon picture explains all five findings.
In the figure : Photocurrent against collector voltage for two intensities (same frequency): different saturation currents, the same stopping potential.
I (saturation) ∝ intensity; V₀ depends on frequency only; eV₀ = K_max.
Thinking that brighter light gives faster electrons. It gives more electrons.
Memory clue : "Intensity: how many. Frequency: how fast."
Spot the difference
| Change | Effect on current | Effect on V₀ |
|---|---|---|
| Higher intensity | increases | none |
| Higher frequency | none (for same photon rate) | increases |
Practice questions
- 1Define the stopping potential.
- 2What is the effect of intensity on the maximum kinetic energy?
- 3How long is the time lag between illumination and emission?
- 4If V₀ = 1.5 V, find K_max in eV and in joule.
- 5Find the speed of the fastest electrons.
- 6What happens to V₀ if the frequency is increased?
- 7What happens to the photocurrent and V₀ if the intensity is doubled at the same frequency?
- 8Give two reasons why the wave theory fails.
Show answers (try them yourself first)
- 1The minimum retarding potential which stops the fastest photoelectrons (eV₀ = K_max).
- 2None.
- 3Less than 10⁻⁹ s (practically instantaneous).
- 41.5 eV = 2.4 × 10⁻¹⁹ J.
- 5v ≈ 7.3 × 10⁵ m s⁻¹.
- 6V₀ increases.
- 7The photocurrent doubles; V₀ is unchanged.
- 8It cannot explain the threshold frequency, nor the independence of K_max from intensity, nor the absence of a time lag.
Einstein's photoelectric equation
Solar panels turn light into current; each photon that is absorbed frees one electron. Link to earlier classes : the energy of a photon from Chapter 8 and conservation of energy.
Einstein proposed that light is made of particles of energy called photons, with energy E = hf. A photon gives all its energy to one electron: a part φ₀ frees the electron and the rest becomes its kinetic energy.
Rule 1: Einstein's equation — h f = φ₀ + K_max, that is K_max = h f − φ₀ and e V₀ = h f − φ₀. Photons with hf < φ₀ cannot eject electrons.
- Easy : Light of 400 nm (E = 1240 ÷ 400 = 3.10 eV) on sodium (φ₀ = 2.3 eV) gives K_max = 0.80 eV and V₀ = 0.80 V.
- Medium : Light of 600 nm has only 2.07 eV, less than 2.3 eV, so there is no emission at all.
- Exam : Since eV₀ = hf − φ₀, doubling the intensity does not change K_max, but doubles the number of photons and so the number of electrons.
Rule 2: The V₀–f graph — The graph of V₀ against f is a straight line with slope h ÷ e = 4.14 × 10⁻¹⁵ V s. It cuts the frequency axis at f₀ and does not depend on the intensity of light.
- Easy : Millikan measured the slope and confirmed the equation, finding h.
- Medium : If V₀ rises by 1.656 V when f rises by 4.0 × 10¹⁴ Hz, h = 1.6 × 10⁻¹⁹ × 1.656 ÷ 4.0 × 10¹⁴ = 6.6 × 10⁻³⁴ J s.
- Exam : Different metals give parallel lines with different intercepts on the frequency axis.
In the figure : V₀ against f for two metals: parallel straight lines of slope h ÷ e starting from their threshold frequencies.
hf = φ₀ + K_max; eV₀ = hf − φ₀; slope of V₀–f = h/e.
Using the work function in joules with hf in eV. Keep one unit.
Memory clue : "Photon energy = ticket price (φ₀) + leftover kinetic energy."
Spot the difference
| Quantity | Depends on |
|---|---|
| K_max | frequency and metal (not intensity) |
| Photocurrent | intensity |
| Slope of V₀–f | only h/e (same for all metals) |
Practice questions
- 1Write Einstein's photoelectric equation.
- 2What is a photon?
- 3What condition connects hf₀ and φ₀?
- 4Light of 400 nm falls on sodium (φ₀ = 2.3 eV). Find K_max.
- 5Find the stopping potential in that case.
- 6Will light of 600 nm cause emission from sodium?
- 7V₀ rises by 1.656 V when f rises by 4.0 × 10¹⁴ Hz. Find h.
- 8What is the slope of the V₀–f graph?
Show answers (try them yourself first)
- 1hf = φ₀ + K_max.
- 2A quantum of light with energy E = hf.
- 3hf₀ = φ₀.
- 4K_max = 3.10 − 2.3 = 0.80 eV.
- 5V₀ = 0.80 V.
- 6No; hf = 2.07 eV < φ₀.
- 7h = eΔV ÷ Δf ≈ 6.6 × 10⁻³⁴ J s.
- 8h ÷ e = 4.14 × 10⁻¹⁵ V s.
Particle nature of light: energy and momentum of a photon
A laser beam, even a very weak one, is detected as individual clicks by a sensitive detector, each click being one photon. Link to earlier classes : the electromagnetic waves of Chapter 8.
A photon behaves like a particle with energy E = hf, momentum p = h ÷ λ and zero rest mass. It travels at the speed c. The intensity of a light beam is the number of photons crossing unit area per second multiplied by the energy of each.
Rule 1: Photon energy and momentum — E = h f = h c ÷ λ and p = E ÷ c = h ÷ λ.
- Easy : λ = 500 nm ⇒ E = 2.48 eV.
- Medium : The momentum is p = 6.63 × 10⁻³⁴ ÷ 5 × 10⁻⁷ = 1.33 × 10⁻²⁷ kg m s⁻¹.
- Exam : The energy of one photon of 633 nm is 6.63 × 10⁻³⁴ × 3 × 10⁸ ÷ 6.33 × 10⁻⁷ = 3.14 × 10⁻¹⁹ J.
Rule 2: Number of photons — If the power is P, the photons emitted per second is n = P ÷ (h f) = P λ ÷ (h c).
- Easy : A 1 mW laser of 633 nm emits n = 10⁻³ ÷ 3.14 × 10⁻¹⁹ = 3.2 × 10¹⁵ photons per second.
- Medium : Doubling the intensity at fixed frequency doubles n.
- Exam : The energy of a photon is fixed by the frequency; a brighter beam has more photons, not more energetic ones.
E = hf = hc/λ; p = h/λ; photons have no rest mass.
Taking brighter light to have bigger photons.
Memory clue : "Colour sets the photon's energy; brightness sets their number."
Spot the difference
| Property | Photon |
|---|---|
| Rest mass | zero |
| Charge | zero |
| Speed | c in vacuum |
Practice questions
- 1Write the energy of a photon in terms of frequency.
- 2Write the momentum of a photon in terms of wavelength.
- 3What is the rest mass of a photon?
- 4Find the energy of a 500 nm photon in eV.
- 5Find its momentum.
- 6Find the photons emitted per second by a 1 mW laser of 633 nm.
- 7Find the energy in joule of one photon of 633 nm.
- 8How does the number of photons change if the intensity is doubled at the same frequency?
Show answers (try them yourself first)
- 1E = hf.
- 2p = h ÷ λ.
- 3Zero.
- 4E = 2.48 eV.
- 5p ≈ 1.33 × 10⁻²⁷ kg m s⁻¹.
- 6n ≈ 3.2 × 10¹⁵ per second.
- 7E ≈ 3.14 × 10⁻¹⁹ J.
- 8It doubles.
Wave nature of matter and the de Broglie wavelength
An electron microscope can see far finer detail than a light microscope because electrons can have a very short wavelength. Link to earlier classes : the wave behaviour of light from Chapter 10.
Louis de Broglie suggested that if light, a wave, can behave like particles, then particles like electrons can behave like waves. A particle of momentum p has the de Broglie wavelength λ = h ÷ p. The wave nature of electrons was confirmed by the Davisson–Germer experiment, in which electrons were diffracted by a nickel crystal.
Rule 1: de Broglie wavelength — λ = h ÷ p = h ÷ (m v). For a charged particle accelerated from rest through a potential V, λ = h ÷ √(2 m q V). For an electron, λ = 1.227 ÷ √V nm (V in volt).
- Easy : An electron at 2 × 10⁶ m s⁻¹ has λ = 6.63 × 10⁻³⁴ ÷ (9.1 × 10⁻³¹ × 2 × 10⁶) = 0.364 nm.
- Medium : An electron accelerated through 150 V has λ = 1.227 ÷ √150 = 0.100 nm.
- Exam : The Davisson–Germer electrons at 54 V have λ = 1.227 ÷ √54 = 0.167 nm, which matched the wavelength obtained from the crystal's diffraction pattern.
Rule 2: Why we do not notice matter waves — λ becomes extremely small for large masses. For equal kinetic energy, a lighter particle has the longer wavelength: λ ∝ 1 ÷ √m.
- Easy : A 0.15 kg ball at 20 m s⁻¹ has λ = 6.63 × 10⁻³⁴ ÷ 3.0 = 2.2 × 10⁻³⁴ m, far too small to detect.
- Medium : For the same kinetic energy, λ(electron) ÷ λ(proton) = √(1.67 × 10⁻²⁷ ÷ 9.1 × 10⁻³¹) = 43.
- Exam : The electron microscope uses wavelengths of about 0.1 nm, much shorter than visible light (400 to 700 nm), so its resolution is far higher.
λ = h/p; electron: λ = 1.227/√V nm; heavier or faster means shorter λ.
Applying λ = 1.227/√V nm to a proton or an alpha particle. It is only for electrons.
Memory clue : "Lambda is h over p: more momentum, smaller wave."
Spot the difference
| Particle | λ for the same V |
|---|---|
| Electron | longest |
| Proton | shorter (÷ 43) |
| Alpha particle | shortest |
Practice questions
- 1Write the de Broglie relation.
- 2Which experiment confirmed the wave nature of electrons?
- 3Why do we not see the wave nature of a cricket ball?
- 4Find λ for an electron moving at 2 × 10⁶ m s⁻¹.
- 5Find λ for an electron accelerated through 150 V.
- 6Find λ for a 0.15 kg ball moving at 20 m s⁻¹.
- 7Find the ratio of the wavelengths of an electron and a proton of equal kinetic energy.
- 8Find λ for electrons accelerated through 54 V.
Show answers (try them yourself first)
- 1λ = h ÷ p = h ÷ mv.
- 2The Davisson–Germer experiment (electron diffraction by a nickel crystal).
- 3Its wavelength is extremely small because its momentum is large.
- 4λ ≈ 0.364 nm.
- 5λ ≈ 0.100 nm.
- 6λ ≈ 2.2 × 10⁻³⁴ m.
- 7λ_e ÷ λ_p = √(m_p ÷ m_e) ≈ 43.
- 8λ ≈ 0.167 nm.