Photocell Response to Two LEDs
A student tests a photocell with a metal cathode of work function 2.20 eV. Two monochromatic LEDs are available: LED A at 450 nm and LED B at 620 nm. The student connects a variable reverse voltage and a microammeter, and finds that LED A produces a stopping potential of about 0.56 V, while LED B gives no current even at zero bias. The student claims LED B is broken. The instructor points out that the same photocell responds to LED A and that the dark current is negligible. Planck's constant is 6.63×10^-34 J s, the speed of light is 3.00×10^8 m/s, and 1 eV = 1.60×10^-19 J. The student must decide whether LED B can eject electrons, calculate the photon energies and the maximum kinetic energy for LED A, and explain what would change if the intensity of LED A were doubled while the reverse voltage stayed at 0.56 V.
1. Calculate the photon energy in eV for each LED. Use these values to decide whether LED B can eject electrons from the cathode.
Read the worked answer
Photon energy E = hc/λ.
For LED A, λ = 450 nm = 4.50×10^−7 m: E_A = (6.63×10^−34 × 3.00×10^8)/(4.50×10^−7) = 4.42×10^−19 J. Dividing by 1.60×10^−19 J/eV gives E_A = 2.76 eV.
For LED B, λ = 620 nm = 6.20×10^−7 m: E_B = (6.63×10^−34 × 3.00×10^8)/(6.20×10^−7) = 3.21×10^−19 J = 2.00 eV.
The work function is 2.20 eV. LED B's photons have less energy than the work function, so a single photon cannot free an electron. LED B is not necessarily broken; it simply cannot produce photoemission from this cathode.
2. Calculate the maximum kinetic energy and stopping potential expected for LED A, and compare with the measured 0.56 V.
Read the worked answer
For LED A, Einstein's photoelectric equation gives K_max = E_A − φ = 2.76 eV − 2.20 eV = 0.56 eV.
The stopping potential is the reverse voltage that just stops the most energetic electrons: eV_s = K_max, so V_s = 0.56 V.
This agrees with the measured 0.56 V, which supports the photon model and the quoted work function. The small difference, if any, would be within the LED linewidth and measurement uncertainty. LED B would give K_max = 2.00 − 2.20 = −0.20 eV, which is not physical; no electrons are emitted.
3. The student doubles the intensity of LED A and keeps the reverse voltage at 0.56 V. Predict what happens to the stopping potential and to the photocurrent, explaining the reasoning.
Read the worked answer
At exactly 0.56 V reverse bias, the most energetic electrons are just stopped, so the photocurrent is essentially zero. The stopping potential depends on the photon frequency (or wavelength) and the cathode work function, not on intensity.
Doubling the intensity doubles the number of photons per second, so it doubles the number of photoelectrons per second, but each electron still has the same maximum kinetic energy. Therefore the stopping potential remains 0.56 V. If the reverse voltage were lowered slightly below 0.56 V, the current would approximately double; at 0.56 V it stays near zero. This is the decisive distinction between the classical wave prediction and the photon explanation.