Photocell Practical: Faulty Cell or Faulty Diagnosis?
In a physics practical a student tests a vacuum photocell whose cathode is specified as a caesium-based material of work function 2.20 eV. She uses four LEDs, at 470 nm, 530 nm, 590 nm and 620 nm, each behind a diffuser, and measures the stopping potential with a freshly calibrated digital voltmeter. She obtains 0.40 V and 0.12 V for the first two, and essentially zero for the last two, although a small current of about 0.3 microampere still flows at 620 nm even at the largest reverse voltage she can apply. The lab assistant says the cell must be faulty, because setting the 470 nm LED brighter gives a much larger current but the same stopping potential. The student must decide whether the cell is faulty, deduce the work function of the cathode actually in use, and account for the residual current. (hc = 1240 eV nm.)
1. Compute photon energies, predict the stopping potentials for a 2.20 eV cathode, and compare with her readings to decide whether the cell is faulty.
Read the worked answer
Photon energies, E = 1240/lambda eV:
- 470 nm -> 2.64 eV
- 530 nm -> 2.34 eV
- 590 nm -> 2.10 eV
- 620 nm -> 2.00 eV
Predicted stopping potentials for phi = 2.20 eV, using eV0 = E - phi:
- 470 nm: 2.64 - 2.20 = 0.44 V
- 530 nm: 2.34 - 2.20 = 0.14 V
- 590 nm: 2.10 < 2.20 -> no emission, V0 = 0
- 620 nm: 2.00 < 2.20 -> no emission, V0 = 0
Comparison with observation. Measured 0.40 V and 0.12 V, against 0.44 V and 0.14 V predicted. Both are low by about 0.02-0.04 V - a small, consistent offset, exactly what a contact potential difference between the two metals in the circuit, or a slight zero error, produces. The two absorbing LEDs show no emission, as predicted. The qualitative pattern (emission above a threshold wavelength, none below) is correct and the discrepancy is uniform, so there is no evidence of a fault in the cell.
Deduced work function. From each emitting LED, phi = E - eV0: 470 nm: 2.64 - 0.40 = 2.24 eV 530 nm: 2.34 - 0.12 = 2.22 eV Mean about 2.23 eV, within 0.03 eV of the specified 2.20 eV. The cell is behaving as a 2.2 eV cathode should; the assistant's conclusion is wrong.
2. The assistant argues that a brighter LED giving more current but the same stopping potential proves a fault. Evaluate this argument and state what observation would actually indicate a faulty cell.
Read the worked answer
The argument is backwards. Intensity is a measure of photons per second, not of photon energy. Raising the intensity raises the number of photoelectrons ejected per second, which raises the saturation current - and that is exactly what she saw. But every electron still leaves with the same maximum kinetic energy, because that depends only on the energy of a single photon (E = hc/lambda) and on the work function: eV0 = hc/lambda - phi. So the stopping potential is independent of intensity. Bright light gives more current, not faster electrons.
This is a confirmation, not a fault. The intensity test is the standard way to demonstrate that light arrives in quanta; a cell that passed it is behaving correctly. If the stopping potential had risen when the LED was made brighter, that would signal a problem - for example a voltage drop across a high resistance in the measuring circuit, or a cell where space-charge or leakage makes the measured V0 depend on photocurrent.
What would actually indicate a fault.
- A stopping potential that varies with intensity at fixed wavelength.
- Emission at 590 or 620 nm (below the threshold), which would mean the cathode has a lower work function than specified.
- A plot of V0 against 1/lambda whose slope differs appreciably from hc/e = 1240 eV nm.
- A current that persists at zero illumination (dark current) - which must be checked here, see part (c). The right next step is not to condemn the cell but to run a control with all LEDs off, or the cathode covered.
3. Account for the 0.3 microampere that still flows at 620 nm even at the maximum reverse voltage, and explain how the student should test her explanation.
Read the worked answer
Photon energy check. 1240/620 = 2.00 eV, which is below the deduced work function of about 2.23 eV. No photoelectron can be emitted from the cathode at this wavelength, so the current cannot be photoelectric emission from the cathode - however large the reverse voltage, there are simply no energetic enough photons.
Candidate explanations.
- Dark/leakage current. Even with no light, a small current flows: thermionic emission from a low-work-function caesium cathode is tiny but not strictly zero at room temperature, and there is always some leakage across the insulator between anode and cathode. A fraction of a microampere is a normal dark current for such a cell.
- Stray light. Room lighting, or light leaking past the diffuser, contains wavelengths shorter than 560 nm and can eject electrons even though the 620 nm LED cannot. This is a real photocurrent, just not caused by the LED.
- Photoelectric emission from the anode. The reverse voltage accelerates electrons towards the anode; if any stray short-wavelength light strikes the anode, it can emit electrons which then travel to the cathode, giving a current of the 'wrong' sign.
How to test. Switch the LED off (or slide a card over the window) and read the current again. If it stays at ~0.3 microampere, it is dark current or stray light, and the 620 nm result is fully explained without any fault. If it falls to zero, then stray light during the LED exposure is the culprit - repeat with the apparatus in a dark box. Either way, the observation is an experimental artefact, not evidence of a defective photocell.