The photoelectric effect: determination of Planck's constant
Aim: to illuminate a vacuum photocell with the spectral lines of a mercury lamp selected by interference filters, record current–voltage curves, determine the stopping potential for each frequency, and obtain h and the work function from Einstein's equation.
Hg lamp, filter wheel, photocell & electrometer
Current–voltage characteristics
Positive U retards the photoelectrons. Each colour is one filter. The zoomed region near the cut-off is where V₀ is decided.
Observation table
Graph: stopping potential V₀ against frequency ν
Results
Theory
A photon of energy hν frees an electron from the cathode with at most
$$ E_{max} = h\nu - \phi_c .$$A retarding voltage stops the fastest electrons when \(eV_0 = E_{max} + (\phi_c - \phi_a)\). The contact potential between the cathode and anode cancels \(\phi_c\), which leaves
$$ V_0 = \frac{h}{e}\,\nu - \frac{\phi_a}{e} .$$(Millikan, 1916.) The slope of V₀ against ν gives h/e. The intercept gives the anode work function, not the cathode's.
Finding V₀ properly
Photoelectrons leave with a spread of energies. Near the cut-off, the current falls off as \(I \propto (E_{max} - eV)^2\), so the curve meets the axis tangentially. Reading the zero-current voltage therefore systematically underestimates V₀, by more for higher frequencies, which biases h low. In addition, the anode emits a small reverse current. The rigorous method subtracts the reverse-current plateau and extrapolates \(\sqrt{I - I_{plateau}}\) linearly to zero. Both methods are computed here, so you can see the difference.
Procedure
- Select a filter and sweep U from −1 V (accelerating) to +2.5 V (retarding). Note the saturation current and the reverse-current plateau.
- Repeat for all five lines: 365, 405, 436, 546 and 578 nm.
- Check that changing the intensity changes the saturation current but not V₀. This is the key observation that classical wave theory cannot explain.
- Plot V₀ against ν and fit: h = e × slope.
Viva questions
Why is V₀ independent of intensity?
Each electron absorbs a single photon. Intensity changes the number of photons, not the energy of each one.
Why does the yellow line give a negative V₀ here?
hν(578 nm) = 2.14 eV is less than the anode work function, so the photoelectrons need an accelerating field to reach the anode.
What is the threshold frequency?
ν₀ = φ_c/h. Below it, no electrons are emitted however bright the light.