Photoelectric Work Function from Three LEDs with a Drifting Electrometer
A student uses a phototube with an unknown cathode and three LEDs whose datasheet wavelengths are 450 nm, 550 nm and 650 nm. For each, the student finds the stopping potential by adjusting a reversing supply until the photocurrent just vanishes, recording 1.15 V, 0.62 V and 0.15 V respectively. The photocurrent at 650 nm is only about 2% of the current at 450 nm, and the electrometer's zero drifts by roughly ±0.08 V between runs; the student did not re-zero between measurements. Computing hf − φ = eV_s for each point, the student obtains work functions of 1.61, 1.63 and 1.76 eV and reports the average, 1.67 eV, as the cathode's work function. A classmate argues that the 650 nm point should simply be thrown out and that the intensity dependence has not been tested at all.
1. Identify which data point is least trustworthy, quantify why, and describe the checks that would confirm or repair it.
Read the worked answer
Photon energies: 1240/450 = 2.76 eV, 1240/550 = 2.25 eV, 1240/650 = 1.91 eV. Subtracting the stopping potentials gives work functions of 1.61, 1.63 and 1.76 eV. The first two agree to 0.02 eV; the third is 0.13 eV away and is the weak point.
The reason is signal-to-noise. At 650 nm the electrons leave with only about 0.3 eV, the photocurrent is 2% of the 450 nm value, and the ±0.08 V zero drift is more than half the measured stopping potential. A 0.08 V offset shifts the inferred work function by 0.08 eV, enough to manufacture the apparent trend. Repairs: re-zero the electrometer before every run, repeat 650 nm several times and average, reverse the supply polarity to check for a symmetric offset, and vary the LED drive current. The stopping potential must be independent of intensity; if it moves, the reading is an artifact, not photophysics.
2. Using the reliable data, produce a defensible value of the work function and the threshold wavelength, and explain why averaging the three points is the wrong procedure.
Read the worked answer
Fit the relation V_s = (h/e)f − φ/e. The frequencies for 450 and 550 nm are 6.66 × 10¹⁴ and 5.45 × 10¹⁴ Hz, so the slope is (1.15 − 0.62)/(1.21 × 10¹⁴) = 4.38 × 10⁻¹⁵ V·s, giving h = 4.38 × 10⁻¹⁵ × 1.602 × 10⁻¹⁹ = 7.0 × 10⁻³⁴ J·s, about 6% above the accepted value and consistent with the stated uncertainties. The intercept yields φ ≈ 1.62 eV.
Including the 650 nm point in the same fit gives a slope of 4.9 × 10⁻¹⁵, a Planck constant 18% too large, which is the quantitative cost of the drifting point. Averaging three work functions is wrong because it weights the noisiest point equally with the cleanest; a fit weighted by each point's uncertainty is the correct procedure. With φ = 1.62 eV the threshold wavelength is 1240/1.62 = 765 nm.
3. Predict the outcome of two further tests, one at longer wavelength and one varying intensity, and explain how relativistic effects enter.
Read the worked answer
The 450 nm electrons carry 1.15 eV = 1.84 × 10⁻¹⁹ J, so v = √(2K/m) = √(2 × 1.84 × 10⁻¹⁹/9.11 × 10⁻³¹) ≈ 6.4 × 10⁵ m/s, about 0.2% of c. Classical mechanics is fully adequate; relativistic corrections matter only as photon energy approaches the 511 keV electron rest energy.
At 800 nm the photon energy is 1.55 eV, below the 1.62 eV work function, so no photoelectrons should appear at any intensity; any signal there is thermionic emission, leakage or stray light, and it is a good diagnostic for a dirty tube. At 700 nm (1.77 eV) expect a stopping potential near 0.15 V, small enough to be confused with drift, so it must be measured with the electrometer freshly zeroed. Doubling the intensity at 550 nm should double the saturation current while leaving V_s at 0.62 V. If V_s changes with intensity, the photon model is not at fault; the measurement is.