Einstein's photoelectric equation
Each photon gives all its energy hν to one electron: K_max = hν − φ₀. So the stopping potential rises with frequency, with slope h/e.
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
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.
How to draw the V₀ against ν graph
- Put frequency ν on the horizontal axis and stopping potential V₀ on the vertical axis.
- Draw a straight line starting on the ν axis at ν₀ and rising to the right. Label ν₀.
- Extend the line backwards as a dashed line to cut the V₀ axis below the origin at −φ₀/e.
- Write slope = h/e beside the line.
- For a metal with a larger work function, draw a parallel line starting at a larger ν₀.
Worked example: using Einstein's equation
The threshold frequency of a metal is 5 × 10¹⁴ Hz. Light of frequency 8 × 10¹⁴ Hz falls on it. Find (a) the work function in eV, (b) the maximum kinetic energy of the photoelectrons, (c) the stopping potential. (h = 6.63 × 10⁻³⁴ J s, e = 1.6 × 10⁻¹⁹ C)
- (a) φ₀ = hν₀ = 6.63 × 10⁻³⁴ J s × 5 × 10¹⁴ Hz = 3.315 × 10⁻¹⁹ J.
- In eV: 3.315 × 10⁻¹⁹ ÷ 1.6 × 10⁻¹⁹ ≈ 2.07 eV.
- (b) K_max = h(ν − ν₀) = 6.63 × 10⁻³⁴ × 3 × 10¹⁴ J = 1.989 × 10⁻¹⁹ J ≈ 1.99 × 10⁻¹⁹ J.
- In eV: 1.989 × 10⁻¹⁹ ÷ 1.6 × 10⁻¹⁹ ≈ 1.24 eV.
- (c) eV₀ = K_max, so V₀ ≈ 1.24 V.
(a) about 2.07 eV; (b) about 1.99 × 10⁻¹⁹ J, or 1.24 eV; (c) about 1.24 V
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
Einstein received the Nobel Prize (1921) for this law, not for relativity. The same one-photon, one-electron idea is at work in the camera sensor of every mobile phone.
3 marks: “Write Einstein's photoelectric equation. Use it to explain (a) why there is a threshold frequency, (b) why K_max does not depend on intensity, (c) why emission is instantaneous.” K_max = hν − φ₀. (a) If hν < φ₀, no emission, so ν₀ = φ₀/h. (b) Each electron absorbs one photon; intensity only changes the number of photons. (c) Energy is given in one packet, so no time lag.
2 marks: “Light of frequency 1.5 times the threshold frequency falls on a metal. If the frequency is halved and the intensity doubled, will photoelectrons be emitted?” New frequency = 0.75ν₀, which is below ν₀. So no photoelectrons are emitted, whatever the intensity.
5 marks: “Draw the graph of stopping potential against frequency for a metal. Explain how Planck's constant and the work function can be found from it.” Straight line starting at ν₀ on the ν axis; V₀ = (h/e)ν − φ₀/e; slope = h/e, so h = e × slope; φ₀ = hν₀, or e × the magnitude of the intercept on the V₀ axis.
The photon pays the exit fee φ₀; the change is K_max. So eV₀ = hν − φ₀.