E.2 · IB Physics
The photoelectric effect
Light frees electrons from a metal only above a threshold frequency. Below it, no amount of intensity helps, which is evidence that light arrives as quanta of energy hf.
Predict first
Start from the default settings, where electrons are emitted. If you raise the Intensity (photon rate) slider from 5 to 10, what happens to KEmax?
KEmax = hf − φ depends only on the frequency and the metal. Intensity changes the number of photons per second, so the electron rate doubles but KEmax does not change.
Check it in the model: Move the Intensity (photon rate) slider from 5 to 10 and compare KEmax and Electron rate in the readout.
photons emitted electrons KEmax line
KEmax = hf − φ ; f₀ = φ ÷ h
eVs = KEmax
h = 6.63 × 10⁻³⁴ J s, e = 1.60 × 10⁻¹⁹ C
rate = 2I × 10¹⁵ photons s⁻¹, one electron per photon above f₀
Try this
- Set f = 4.0 ×10¹⁴ Hz with φ = 2.0 eV and raise the intensity from 1 to 10. Does any electron appear? Now move f above f₀.
- Choose caesium, then zinc. Compare f₀ and KEmax at the same frequency, and find the lowest frequency that frees electrons from zinc.
- At fixed f above f₀, double the intensity. Which readouts change (rate, photocurrent) and which do not (KEmax, stopping voltage)?
Common error: Raising the intensity does not give the electrons more energy: it sends more photons, so it changes the number of electrons per second, while KEmax = hf − φ depends only on frequency and metal.