Photoelectric effect: LED threshold and stopping voltage
In a photoelectric-effect demonstration, students shine LEDs onto a clean metal surface with work function 2.20 eV. The LEDs have wavelengths 470 nm, 530 nm, and 640 nm. The 470 nm LED produces a stopping voltage of 0.44 V; the 530 nm LED produces 0.14 V; the 640 nm LED produces no current even at high intensity. A student says the 640 nm LED must be faulty because a brighter light should eject more electrons. Another says the 640 nm photons simply do not have enough energy. The students also compare a high-intensity 470 nm LED with a low-intensity 470 nm LED and find that the stopping voltage is the same but the current is different. Use hc = 1240 eV nm and treat the metal as having a single work function.
1. Calculate the photon energy for each LED and the maximum kinetic energy of emitted electrons for the 470 nm and 530 nm LEDs. Use these values to judge the two students’ claims about the 640 nm LED.
Read the worked answer
Use E_photon = hc/λ with hc = 1240 eV nm:
- 470 nm:
E = 1240/470 = 2.64 eV - 530 nm:
E = 1240/530 = 2.34 eV - 640 nm:
E = 1240/640 = 1.94 eV
For emission, E_photon > φ = 2.20 eV. The 470 nm and 530 nm photons qualify; the 640 nm photon does not. The maximum kinetic energies are
- 470 nm:
K_max = 2.64 − 2.20 = 0.44 eV - 530 nm:
K_max = 2.34 − 2.20 = 0.14 eV
These match the measured stopping voltages of 0.44 V and 0.14 V. The 640 nm photon has 1.94 eV, below the 2.20 eV work function, so no single photon can eject an electron. Increasing intensity increases the number of photons, not their individual energy, so it cannot overcome the threshold. The second student is correct; the first student’s “brighter light” argument fails because it confuses photon number with photon energy.
2. The 640 nm LED is replaced by a 400 nm LED of much lower intensity. Predict whether electrons are emitted and whether the stopping voltage changes compared with the 470 nm LED. Justify your answer.
Read the worked answer
For 400 nm,
E_photon = 1240/400 = 3.10 eV
This is above the 2.20 eV work function, so electrons are emitted even if the intensity is low. Each individual photon has enough energy; low intensity only means fewer photons per second, hence a smaller current.
The maximum kinetic energy is
K_max = 3.10 − 2.20 = 0.90 eV
so the stopping voltage is 0.90 V. This is higher than the 0.44 V for the 470 nm LED. The stopping voltage depends on photon energy and work function, not on intensity. Therefore changing from 470 nm to 400 nm changes the stopping voltage, while changing the intensity of a given wavelength does not.
3. Explain why the stopping voltage is the same for high- and low-intensity 470 nm light, but the photocurrent is different.
Read the worked answer
The stopping voltage measures the maximum kinetic energy of the emitted electrons:
eV_s = K_max = hc/λ − φ
For a fixed wavelength and metal, hc/λ − φ is fixed, so every emitted electron has the same maximum kinetic energy. Intensity does not change the energy of each photon; it changes the number of photons arriving per second. Therefore the stopping voltage is unchanged.
A higher intensity means more photons per second, so more electrons are ejected per second, giving a larger saturation current. The low-intensity beam still ejects electrons with the same maximum energy, but there are fewer of them. This is why the current changes while the stopping voltage does not.