Photoelectric effect: extracting h and the work function from stopping potentials
A student uses a vacuum photocell with an unknown metal cathode. Light from LEDs of wavelength 400 nm and 500 nm is shone on the cathode, and the stopping potential is measured with a variable reverse voltage. For 400 nm the stopping potential is 0.90 V; for 500 nm it is 0.28 V. When the intensity of the 500 nm LED is doubled, the photocurrent doubles but the stopping potential remains 0.28 V. The student's partner says this proves light behaves as a wave, because brighter light gives more current. The student also finds that a 650 nm LED produces no photocurrent even at maximum intensity. The photocell has a small reverse leakage current, and the voltmeter has a zero error of ±0.02 V.
1. Use the two stopping potentials to determine Planck's constant and the work function of the cathode. State the equation and show the substitution.
Read the worked answer
Einstein's photoelectric equation: eV_s = hc/λ − φ, so φ = hc/λ − eV_s.
Using the 400 nm data: photon energy = hc/λ = 1240 eV nm / 400 nm = 3.10 eV. Then φ = 3.10 − 0.90 = 2.20 eV. Using 500 nm: 1240/500 = 2.48 eV, so φ = 2.48 − 0.28 = 2.20 eV. The two values agree.
For h, subtract the two equations: e(V₁ − V₂) = hc(1/λ₁ − 1/λ₂). Thus h = e(V₁ − V₂)/[c(1/λ₁ − 1/λ₂)].
ΔV = 0.90 − 0.28 = 0.62 V. 1/λ₁ − 1/λ₂ = 1/(400 × 10⁻⁹) − 1/(500 × 10⁻⁹) = 2.50 × 10⁶ − 2.00 × 10⁶ = 0.50 × 10⁶ m⁻¹. Denominator = 3.00 × 10⁸ × 0.50 × 10⁶ = 1.50 × 10¹⁴ m s⁻¹ m⁻¹.
h = (1.60 × 10⁻¹⁹ × 0.62)/(1.50 × 10¹⁴) = 6.61 × 10⁻³⁴ J s. This is close to the accepted value, within the precision of the data. The work function is 2.20 eV, or about 3.5 × 10⁻¹⁹ J.
2. Explain why doubling the intensity at 500 nm doubles the current but leaves the stopping potential unchanged. Evaluate the partner's claim that this proves the wave model.
Read the worked answer
In the photon model, intensity is proportional to the number of photons arriving per second (for fixed wavelength). Doubling the intensity doubles the number of photons, so twice as many photoelectrons are emitted per second and the photocurrent doubles, provided the cell is not saturated by space charge. However, each photon still has energy hc/λ = 2.48 eV, so the maximum kinetic energy of an emitted electron and the stopping potential are unchanged. That is exactly what is observed.
The partner's conclusion is the opposite of what the data support. A classical wave model would predict that a more intense wave carries more energy, so electrons could gain more energy and the stopping potential would increase. The observation that V_s depends on frequency but not intensity is strong evidence for quantised photon absorption, not for a continuous wave energy transfer. The wave model can explain the current increase, but it fails on the stopping potential.
3. The 650 nm LED produces no current. Calculate the photon energy and explain whether this is consistent with the work function. Then predict what would happen if a 350 nm LED with half the intensity of the 400 nm LED were used, and discuss the effect of the voltmeter zero error on your value of φ.
Read the worked answer
At 650 nm, photon energy = 1240/650 = 1.91 eV. This is below the work function of 2.20 eV, so a single photon cannot free an electron; no photocurrent is expected, consistent with observation.
At 350 nm, photon energy = 1240/350 = 3.54 eV. The maximum kinetic energy would be 3.54 − 2.20 = 1.34 eV, so the stopping potential would be 1.34 V. Half the intensity means roughly half the photon arrival rate, so the photocurrent would be about half that for the 400 nm LED at the same original intensity, but the stopping potential would still be 1.34 V because it depends on photon energy, not intensity.
A zero error of ±0.02 V shifts both stopping potentials by the same amount. The difference ΔV is unchanged, so the calculated h is unaffected. The work function, however, is found from φ = hc/λ − eV_s, so it carries an uncertainty of about ±0.02 eV from the zero error, plus any uncertainty in h and λ.