Photoelectric effect: stopping voltage that will not behave
A student illuminates a zinc cathode (work function W = 4.3 eV) in a vacuum photocell with monochromatic ultraviolet light. She measures the stopping voltage U_s needed to reduce the photocurrent to zero, using light of two wavelengths: λ₁ = 250 nm gives U_s = 0.66 V, and λ₂ = 280 nm gives U_s = 0.13 V. She expects a linear relation between U_s and 1/λ. Her graph is a good straight line, but the slope corresponds to a Planck constant about 15 % larger than the accepted value, and the intercept gives a work function of about 3.5 eV instead of 4.3 eV. She suspects the apparatus is broken.
1. Write the photoelectric equation in the form used for the graph and identify what the slope and intercept should be. Then assess whether a 15 % error in h is plausible from a real apparatus.
Read the worked answer
Einstein's equation: eU_s = hc/λ − W, so
U_s = (hc/e)(1/λ) − W/e.
A plot of U_s against 1/λ should be a straight line with
- slope = hc/e,
- intercept (extrapolated to 1/λ = 0) = −W/e.
Expected slope: hc/e = (6.63×10⁻³⁴ × 3.00×10⁸)/1.60×10⁻¹⁹ = 1.24×10⁻⁶ V·m = 1.24 V·µm. Expected intercept: −4.3 V.
Check the data: for λ₁ = 250 nm, hc/λ = 1240/250 = 4.96 eV, so U_s = 4.96 − 4.3 = 0.66 V ✓. For λ₂ = 280 nm, 1240/280 = 4.43 eV, U_s = 0.13 V ✓. The two data points are exactly consistent with the accepted h and W!
So the '15 % error' cannot come from these two points - it must come from how the student fitted or plotted them (e.g. using wavelength in nm without converting, or mislabelling the axis). The apparatus is probably fine; the analysis is suspect.
2. Given that the two data points are internally consistent with the accepted constants, explain how a student could still obtain a slope 15 % too large and an intercept of −3.5 eV.
Read the worked answer
A single systematic scaling error in the horizontal variable reproduces both symptoms. If the student plots U_s against 1/λ but uses λ in nanometres and then mislabels or rescales the axis (e.g. plots against 10⁹/λ but treats it as 10⁶/λ, or mixes units between points), the apparent slope changes by the unit factor while the intercept - read where 1/λ → 0 - is dragged to a wrong value.
More concretely: with only two points, any error in one point's wavelength assignment tilts the line. Suppose λ₂ is recorded as 265 nm instead of 280 nm (misread scale). Then the second point moves left, the line steepens, and the extrapolated intercept shifts upward toward −3.5 V. A 15 % slope error corresponds to about a 15 % error in the horizontal separation of the two points - well within the range of a misread monochromator scale.
Diagnosis: with only two points the fit is not over-determined, so a bad point cannot be detected. The fix is more wavelengths and a proper least-squares fit.
3. Design an improved measurement plan that would let the student test the apparatus rather than guess, including how to detect a systematic error.
Read the worked answer
Plan:
- Take ≥ 6 wavelengths spanning 240–320 nm (where U_s is comfortably measurable) and record U_s for each, approaching the stopping point from both directions to avoid backlash.
- Plot U_s vs 1/λ with λ in metres; fit a straight line by least squares. Extract slope = hc/e and intercept = −W/e with uncertainties.
- Check linearity: residuals should scatter randomly; a systematic curvature signals a problem (e.g. contact potential, stray light, or a non-ideal cathode).
- Independent check of h: the slope alone gives h; compare with the accepted value. A genuine 15 % discrepancy would be far outside typical photoelectric uncertainties (~1–2 %), so it would point to a real fault (e.g. a contact potential difference between cathode and anode adding a constant offset to U_s - this shifts the intercept, not the slope).
Key diagnostic: a constant offset (contact potential) shifts the intercept while leaving the slope correct; a scale error (wrong wavelength calibration) changes the slope. Distinguishing these two is exactly the judgement the experiment tests.