Einstein's photoelectric equation and the photon picture
Light comes in photons of energy E = hν. One photon frees one electron, so hν = φ₀ + K_max: the extra energy becomes kinetic energy.
Why the wave theory fails
| Question | Wave theory predicts | Experiment shows |
|---|---|---|
| Does K_max depend on intensity? | Yes: brighter light should give electrons more energy | No: K_max does not depend on intensity |
| Is there a threshold frequency? | No: light of any frequency should work if it is bright enough | Yes: no emission below ν₀ |
| Is there a time lag? | Yes: with dim light, an electron needs a long time to collect enough energy | No: emission is instantaneous |
Einstein's photoelectric equation (1905)
- 1Einstein took light to be made of energy quanta (photons), each of energy E = hν.
- 2In the photoelectric effect, one photon is absorbed by one electron, which takes all of its energy at once.
- 3Part of this energy, the work function φ₀, is used to pull the electron out of the metal.
- 4★The rest becomes kinetic energy. The fastest electrons have K_max = hν − φ₀.
- 5Using the stopping potential: eV₀ = hν − φ₀.
- 6Emission needs K_max ≥ 0, so hν ≥ φ₀. The threshold frequency is ν₀ = φ₀/h, and so K_max = h(ν − ν₀).
★ key step
The photon picture
- Light is a stream of energy packets called photons.
- Each photon has energy E = hν = hc/λ and momentum p = h/λ, and moves at the speed of light c.
- All photons of one frequency have the same energy. Brighter light only means more photons each second.
- Photons have no charge, so electric and magnetic fields do not deflect them.
- In the photoelectric effect, one photon gives all its energy to one electron, at once.
How the equation explains each law
- K_max depends only on ν and φ₀: each electron absorbs one photon of energy hν, so brighter light does not raise K_max.
- Threshold frequency: if hν < φ₀, one photon cannot free an electron, however many photons arrive.
- Photocurrent ∝ intensity: intensity means the number of photons per second. More photons free more electrons.
- No time lag: the electron gets the photon's energy in one packet, not slowly over time.
- V₀ against ν is a straight line: V₀ = (h/e)ν − φ₀/e, with slope h/e for every metal. Millikan used this slope to measure h and confirmed the equation.
Worked example: light of 400 nm on a metal
Light of wavelength 400 nm falls on a metal of work function 2.0 eV. Find (a) the energy of each photon in eV, (b) the maximum kinetic energy of the photoelectrons, (c) the stopping potential, (d) the maximum speed of the electrons. (h = 6.63 × 10⁻³⁴ J s, c = 3 × 10⁸ m/s, e = 1.6 × 10⁻¹⁹ C, m = 9.1 × 10⁻³¹ kg)
- (a) E = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸) ÷ (4 × 10⁻⁷) J = 4.97 × 10⁻¹⁹ J.
- In eV: 4.97 × 10⁻¹⁹ ÷ 1.6 × 10⁻¹⁹ = 3.11 eV.
- (b) K_max = hν − φ₀ = 3.11 eV − 2.0 eV = 1.11 eV = 1.11 × 1.6 × 10⁻¹⁹ J = 1.78 × 10⁻¹⁹ J.
- (c) eV₀ = K_max, so V₀ = 1.11 V.
- (d) ½mv² = K_max, so v = √(2 × 1.78 × 10⁻¹⁹ ÷ 9.1 × 10⁻³¹) = √(3.91 × 10¹¹) ≈ 6.3 × 10⁵ m/s.
(a) 3.11 eV; (b) 1.11 eV (1.78 × 10⁻¹⁹ J); (c) 1.11 V; (d) about 6.3 × 10⁵ m/s
Worked example: finding h and φ₀ from data
For a metal, the stopping potential is 0.4 V for light of frequency 6 × 10¹⁴ Hz and 1.2 V for 8 × 10¹⁴ Hz. Find (a) the value of Planck's constant these readings give, (b) the threshold frequency, (c) the work function in eV. (e = 1.6 × 10⁻¹⁹ C)
- Slope = ΔV₀/Δν = (1.2 − 0.4) V ÷ (2 × 10¹⁴ Hz) = 4 × 10⁻¹⁵ V s.
- (a) h = e × slope = 1.6 × 10⁻¹⁹ C × 4 × 10⁻¹⁵ V s = 6.4 × 10⁻³⁴ J s. This is an experimental estimate from two readings; the accepted value is 6.63 × 10⁻³⁴ J s.
- (b) V₀ = slope × (ν − ν₀): 0.4 = 4 × 10⁻¹⁵ × (6 × 10¹⁴ − ν₀).
- 6 × 10¹⁴ − ν₀ = 0.4 ÷ (4 × 10⁻¹⁵) = 1 × 10¹⁴, so ν₀ = 5 × 10¹⁴ Hz.
- (c) φ₀ = hν₀ = 6.4 × 10⁻³⁴ × 5 × 10¹⁴ J = 3.2 × 10⁻¹⁹ J = 2.0 eV.
(a) about 6.4 × 10⁻³⁴ J s, an experimental estimate (accepted value 6.63 × 10⁻³⁴ J s); (b) 5 × 10¹⁴ Hz; (c) 2.0 eV
The camera in your mobile phone and the solar panels on Indian rooftops both work on the one-photon, one-electron idea. Einstein's Nobel Prize (1921) was for this law.
3 marks: “Write Einstein's photoelectric equation. Use it to explain why there is a threshold frequency and why the stopping potential does not depend on intensity.” hν = φ₀ + K_max. If hν < φ₀, no electron can escape, so ν₀ = φ₀/h. Intensity changes only the number of photons, not hν, so K_max and V₀ stay the same.
3 marks: “Light of frequency 1.2 × 10¹⁵ Hz falls on zinc (φ₀ = 3.4 eV). Find the maximum kinetic energy of the photoelectrons and the stopping potential. (h = 6.63 × 10⁻³⁴ J s, 1 eV = 1.6 × 10⁻¹⁹ J)” hν = 6.63 × 10⁻³⁴ × 1.2 × 10¹⁵ = 7.96 × 10⁻¹⁹ J = 4.97 eV. K_max = 4.97 − 3.4 = 1.57 eV ≈ 2.5 × 10⁻¹⁹ J. V₀ = 1.57 V.
1 marks: “What is the slope of the graph of stopping potential against frequency?” h/e, the same for every metal.
Photon pays the fee, electron keeps the change: hν = φ₀ + K_max.