4.10.4Electrons, Waves and Photons
Work Function and Threshold Frequency
5 min read 6 parts Lesson 16 of 62 Unit quiz
Section 1 of 6Definitions
Work function
- The work function Φ is the minimum energy needed to remove an electron from the surface of a metal.
Threshold frequency
- The threshold frequency f₀ is the lowest frequency of radiation that can release a photoelectron from the metal.
Threshold wavelength
- The threshold wavelength λ₀ is the longest wavelength that can release a photoelectron,
- because wavelength is inversely related to frequency.
Metal dependent
- Each metal has its own work function, so its threshold frequency differs from that of other metals.
Section 2 of 6Relationships
Relationships
Link between them
- At the threshold the photon energy only just equals the work function, so hf₀ = Φ.
Frequency form
- Rearranging gives f₀ = Φ / h, which is the minimum frequency needed.
Wavelength form
- Since f = c/λ, the threshold wavelength is λ₀ = hc / Φ.
Higher work function
- A larger Φ means a higher threshold frequency and a shorter threshold wavelength.
| Condition | Result |
|---|---|
| f < f₀ (λ > λ₀) | No photoelectrons |
| f = f₀ (λ = λ₀) | Electrons just escape with no kinetic energy |
| f > f₀ (λ < λ₀) | Electrons escape with kinetic energy |
Section 3 of 6Deciding Whether Emission Occurs
1
Find the work function of the metal, converting from eV to joules if required.
2
Calculate the threshold frequency using f₀ = Φ / h, or the threshold wavelength using λ₀ = hc / Φ.
3
Compare the frequency of the incident radiation with f₀ (or its wavelength with λ₀).
4
If the frequency is lower, state that the photon energy is too small and no electron is emitted.
5
If higher, calculate the kinetic energy from hf − Φ.
Section 4 of 6Interpreting Graphs
Same gradient
- Graphs of KEmax against f for different metals are parallel lines, since each has the gradient h.
Different intercepts
- A metal with a larger work function has an x-intercept at a higher frequency.
Using the intercept
- Reading f₀ from the graph lets you calculate Φ using Φ = hf₀.
Using the y-intercept
- The vertical intercept gives −Φ directly, in joules or electronvolts depending on the axis.
Section 5 of 6Worked Example
Question
- The work function of a metal is 4.2 eV.
- Calculate its threshold frequency and threshold wavelength.
- (h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m/s, e = 1.60 × 10⁻¹⁹ C)
- Convert Φ to joules4.2 × 1.60 × 10⁻¹⁹ = 6.72 × 10⁻¹⁹ J.
- Threshold frequency f₀ = Φ / h = 6.72 × 10⁻¹⁹ / 6.63 × 10⁻³⁴ = 1.01 × 10¹⁵ Hz.
- Threshold wavelength λ₀ = c / f₀ = 3.00 × 10⁸ / 1.01 × 10¹⁵ = 2.96 × 10⁻⁷ m.
Answer f₀ = 1.0 × 10¹⁵ Hz and λ₀ = 296 nm, so only ultraviolet of shorter wavelength causes emission.
Section 6 of 6Exam Checklist & Tips
4.10.4 – Key Points for the Exam
- Work function Φ Minimum energy needed to remove an electron from the surface of a metal.
- Threshold frequency f₀ Minimum frequency of radiation that causes photoemission from a metal.
- Threshold wavelength λ₀ Maximum wavelength of radiation that causes photoemission from a metal.
- Electronvolt The energy gained by an electron accelerated through 1 V, equal to 1.60 × 10⁻¹⁹ J.
- Photon energy The energy carried by one photon, hf.
Common Mistakes
- Mixing up threshold frequency and wavelengthemission needs frequency above f₀ but wavelength below λ₀.
- Forgetting to convert the work function from eV to joules before dividing by h.
- Saying a very intense beam can compensate for frequency below threshold; it cannot.
- Stating that the work function depends on the incident radiation; it is a property of the metal.