Short on time? This page is the recap. For the apparatus diagram, the graph and the worked examples, use the full study guide.
Recap
- The photoelectric effect is the emission of electrons when electromagnetic radiation falls on a photoactive material.
- The work function φ is the minimum energy needed to remove an electron from the material. The exam gives you its value.
- Emission needs a minimum frequency, the threshold frequency f₀. At or above f₀, electrons are emitted even in dim light.
- K_max depends on frequency, not on the number of photons. This is evidence that light comes in photons.
- In the stopping-potential experiment, the collector is made negative relative to the emitter and the potential difference is raised until the current is zero.
Key relationships
| Relationship | Symbols and units | Use it to |
|---|---|---|
| E = hf = hc/λ | h = 6.63 × 10⁻³⁴ J·s = 4.14 × 10⁻¹⁵ eV·s; hc = 1240 eV·nm | find photon energy |
| K_max = hf − φ | all energies in the same unit (J or eV) | find K_max or φ |
| f₀ = φ/h, λ₀ = hc/φ | Hz, m (or nm) | find the threshold |
| eV_s = K_max | V_s in V; K_max in eV equals V_s in V | link stopping potential to K_max |
| 1 eV = 1.60 × 10⁻¹⁹ J | convert units |
| On a graph of K_max against f | It gives |
|---|---|
| Slope | h (the same for every material) |
| Horizontal intercept | f₀ |
| Vertical intercept (extended) | −φ |
For V_s against f, the slope is h/e and the vertical intercept is −φ/e.
Mistakes to avoid
- Brighter light, faster electrons. No: brighter light gives more electrons per second at the same K_max.
- Saving up energy. One electron cannot combine energy from two photons. Below f₀ nothing is emitted.
- Treating K_max ∝ f. The line does not go through the origin.
- Wavelength the wrong way round. Emission needs λ shorter than λ₀.
- Mixing eV and J in one equation.
- Reading the dashed extension below f₀ as real electrons.
Quick self-check
- Light of 300 nm falls on a metal with φ = 3.10 eV. What is K_max? (1240 ÷ 300 = 4.13 eV, so K_max = 1.03 eV)
- The intensity of that light is tripled. What happens to K_max and to the current? (K_max unchanged; current about three times larger)
- A metal has φ = 2.48 eV. What is its threshold wavelength? (1240 ÷ 2.48 = 500 nm)
Next: practice questions.