22.2 Photoelectric effect
- Syllabus
- 9702–2028–2029
- Topic
- 22.2
- Level
- A2
Learning objectives
22.2.1Photoelectrons may be emitted from a metal surface when it is illuminated by• understand that photoelectrons may be emitted from a metal surface when it is illuminated by electromagnetic radiation22.2.2The terms threshold frequency and threshold wavelength• understand and use the terms threshold frequency and threshold wavelength22.2.3Photoelectric emission in terms of photon energy and work function energy• explain photoelectric emission in terms of photon energy and work function energy22.2.4Hf = Φ + 1/2 mvmax 2• recall and use hf = Φ + 1/2 mvmax 222.2.5Why the maximum kinetic energy of photoelectrons is independent of• explain why the maximum kinetic energy of photoelectrons is independent of intensity, whereas the photoelectric current is proportional to intensity
Photoelectrons can be emitted when light of sufficient frequency illuminates a metal
The photoelectric effect is emission of electrons from a metal when incident photons have enough energy to overcome the surface work function.
Increasing intensity above threshold increases the number of emitted electrons, while frequency controls each photon’s energy and maximum electron speed.
Dim ultraviolet light can eject electrons while bright red light may not if red photon energy is below threshold.
Photoemission is not caused by accumulating energy over time from sub-threshold photons in the simple model.
Threshold frequency is the minimum frequency for photoemission, with an equivalent threshold wavelength
Threshold frequency f₀ is the minimum photon frequency that ejects electrons; threshold wavelength λ₀ is the maximum wavelength, related by c=f₀λ₀.
Below f₀ or above λ₀ no electrons are emitted regardless of intensity in the ideal model.
A metal with threshold frequency 6.0×10¹⁴ Hz has threshold wavelength about 500 nm.
Threshold frequency is not the frequency of emitted electrons, and higher intensity cannot compensate for lower photon frequency.
Photoelectric emission compares photon energy with the metal work function
A photon ejects an electron if hf≥Φ, where Φ is the work function; any excess becomes the electron’s kinetic energy.
Use Φ as the minimum energy needed to escape the surface and connect it to threshold frequency via Φ=hf₀.
Increasing frequency above threshold increases maximum photoelectron kinetic energy by the excess photon energy.
Work function is not the same as binding energy of every electron in the bulk and does not depend on beam intensity in this model.
Use hf=Φ+Kmax for speed, stopping potential and graphs
hf=Φ+Kmax=Φ+½mevmax2=Φ+eVs
One photon supplies hf. The maximum-energy electrons spend Φ escaping and keep the remainder Kmax; a stopping potential Vs removes that maximum kinetic energy because Kmax=eVs.
For f=1.10×10¹⁵ Hz and Φ=5.80×10⁻¹⁹ J: Kmax=hf−Φ=1.49×10⁻¹⁹ J, so vmax=√(2Kmax/me)=5.72×10⁵ m s⁻¹ and Vs=Kmax/e=0.933 V.
Kmax=hf−ΦKmaxagainstf:gradienth;f−axisinterceptf0=Φ/h;extrapolatedK−axisintercept−Φ
Different metals give parallel Kmax–f lines because h is universal. A larger work function shifts the threshold to a larger frequency and the extrapolated intercept to a more negative value.
Use maximum—not average—kinetic energy. If Φ is in eV, either keep the whole energy calculation in eV or convert consistently; do not write Kmax=Vs without the charge factor when using joules.
Maximum photoelectron kinetic energy depends on frequency, not light intensity
For photoelectrons, the maximum kinetic energy is set by the energy of one incident photon: K_max=hf−Φ.
At a fixed frequency above threshold, raising intensity sends more photons per second, so the photocurrent can rise, but each photon still supplies the same energy and K_max is unchanged.
Doubling the intensity of ultraviolet light doubles the possible emission rate in an ideal experiment, not the stopping potential.
Intensity is not a substitute for frequency: sub-threshold photons do not eject electrons simply because the beam is brighter.