Photoelectric Effect: Is the Zinc Specimen Faulty, or the Tube?
A student is given a two-electrode photocell whose cathode is zinc. The manufacturer states that the work function of the zinc cathode is 4.30 eV; the collector electrode is a different metal whose surface is tarnished and not identified. Available sources are a 50 mW 405 nm violet LED, a 5 mW 532 nm green laser, and a low-pressure mercury lamp fitted with a 254 nm filter that delivers only about 1 µW to the cell. The stopping potential is measured with a reversing potential divider and a sensitive microammeter. With the 254 nm source the student measures V_s = 0.30 V. With the 405 nm LED at full power the photocurrent is zero. The student concludes that the manufacturer has quoted the wrong work function for zinc. Use hc = 1240 eV nm.
1. Decide which of the three sources can eject electrons from the zinc cathode, supporting your answer with photon energies.
Read the worked answer
Photon energies from E = hc/λ = 1240/λ(nm) eV:
- 254 nm: E = 1240/254 = 4.88 eV
- 405 nm: E = 1240/405 = 3.06 eV
- 532 nm: E = 1240/532 = 2.33 eV
Emission requires E > φ = 4.30 eV. Only the 254 nm mercury line qualifies; the predicted maximum kinetic energy is 4.88 − 4.30 = 0.58 eV, i.e. a stopping potential of 0.58 V.
The violet LED gives 3.06 eV, well below threshold, so no electrons whatsoever - and this is not an intensity problem. Intensity sets the number of photons, not the energy of each one; the 50 mW LED delivers enormously more power than the microwatt mercury lamp yet ejects nothing, because every photon it emits is individually too weak. The green laser is even further below threshold.
This pair of observations is the cleanest evidence for the quantum picture: a higher-power, longer-wavelength beam produces no effect while a far weaker, shorter-wavelength beam does. Subsisting on 'brighter light eventually wins' is the classical expectation that the photon model overturns.
2. The measured stopping potential is 0.30 V rather than 0.58 V. Evaluate the student's conclusion that the work function is wrong.
Read the worked answer
The student has overlooked that the measured p.d. is between two electrodes, not relative to one.
Einstein's equation describes the energy of the electron relative to the vacuum level of the emitting surface. In a sealed cell, the zinc cathode and the collector are different metals with different work functions, and when they are joined externally their Fermi levels equalise, creating a contact potential difference in series with the applied voltage. The voltmeter therefore reads
eV_s = hf − φ_collector
not hf − φ_zinc.
Check the number: hf − eV_s = 4.88 − 0.30 = 4.58 eV. A collector with an effective work function of 4.58 eV (perfectly plausible for a tarnished metal) reproduces the reading exactly. The zinc is probably fine.
The student's reasoning is defective in a precise way: one equation with two unknowns (φ_zinc and φ_collector) cannot solve for either. A single-wavelength measurement can never distinguish a wrong cathode work function from a contact potential, so the conclusion is unsupported even if the zinc were at fault.
3. Describe how a set of measurements would separate the two work functions, and say what the gradient and intercept give.
Read the worked answer
Measure V_s for several monochromatic wavelengths with the same cell, using mercury lines at 254, 313, 365, 405 and 436 nm.
The working equation is V_s = (h/e)f − φ_collector/e, which is linear in f.
- Gradient = h/e, from which h follows. This is the standard determination and is unaffected by the contact potential, because the offset is constant across wavelengths.
- Intercept = −φ_collector/e, so the intercept gives the collector work function, not the cathode's.
To obtain the zinc work function, either:
- Fit a collector of zinc, so that φ_collector = φ_cathode and the intercept yields 4.30 eV; or
- Measure the contact potential independently (for example with a Kelvin probe) and add it back to the measured stopping potentials.
Diagnostic value: if the gradient returns the accepted h/e but the intercept is 4.58 eV, the cell asymmetry is confirmed and the manufacturer's figure stands. If the gradient is also wrong, the fault is elsewhere - a drifting supply, or photocurrent from the anode, or the filter leaking a second wavelength.