QCE Physics - Unit 4 - Quantum theory
Black-Body Radiation and the Photoelectric Effect | QCE Physics
Learn QCE Physics black-body radiation, Wien's law, photons, threshold frequency, work function and photoelectric effect data.
Updated 2026-06-15 - 4 min read
QCAA official coverage - Physics 2025 v1.3
Exact syllabus points covered
- Explain the concept of black-body radiation and the significance of the evidence it provides.
- Describe the photoelectric effect in terms of the photon.
- Describe the concepts of threshold frequency and work function.
- Solve problems involving blackbody radiation and the photoelectric effect using $\lambda_{max} = \frac{b}{T}$, $E = hf = \frac{hc}{\lambda}$, $E_k = hf - W$ and $W = hf_0$.
- Interpret data related to the photoelectric effect.
Black-body radiation and the photoelectric effect are two major reasons classical physics had to change. Classical wave ideas could not fully explain the observed data, so physicists developed the photon model: light transfers energy in discrete packets.
Original Sylligence diagram for physics blackbody photoelectric evidence.
Black-body radiation
A black body is an ideal object that absorbs all radiation incident on it and emits radiation with a spectrum that depends on temperature. As temperature increases, the emitted spectrum becomes more intense and the peak wavelength becomes shorter.
Wien's displacement law connects temperature and peak wavelength:
$ \lambda_{max}=\frac{b}{T} $
where $b$ is Wien's displacement constant.
This evidence mattered because classical physics predicted the wrong behaviour at short wavelengths. The observed spectrum made sense when energy exchange was treated as quantised rather than continuous.
Photon energy
In the photon model, light energy comes in packets. Each photon has energy:
$ E=hf $
Since $c=f\lambda$ for light in a vacuum, this can also be written:
$ E=\frac{hc}{\lambda} $
Higher frequency means higher photon energy. Shorter wavelength also means higher photon energy. Intensity affects the number of photons arriving per second, but frequency affects the energy of each photon.
Photoelectric effect
The photoelectric effect is the emission of electrons from a material when light of sufficiently high frequency strikes its surface. The minimum energy required to remove an electron from the material is the work function:
$ W=hf_0 $
where $f_0$ is the threshold frequency.
If the incoming photon has enough energy, the maximum kinetic energy of the emitted electron is:
$ E_k=hf-W $
If $hf<W$, no electrons are emitted, no matter how bright the light is. Increasing intensity below threshold increases the number of low-energy photons, but none of those photons individually has enough energy to release an electron.
Interpreting photoelectric data
QCE questions may use graphs of maximum kinetic energy against frequency. The graph has a straight-line form:
$ E_k=hf-W $
The gradient is Planck's constant $h$. The horizontal intercept is the threshold frequency $f_0$. The vertical intercept is $-W$ if energy is plotted in joules.
If stopping voltage is used, connect voltage to energy using:
$ E_k=qV_s $
For an electron, the charge magnitude is $e=1.60\times10^{-19}\ \mathrm{C}$. Be careful with electronvolts and joules. An electronvolt is an energy unit, not a charge unit:
$ 1\ \mathrm{eV}=1.60\times10^{-19}\ \mathrm{J} $
Worked example
Linking the evidence to the model
Black-body radiation suggests energy exchange is quantised. The photoelectric effect shows light behaves as photons in energy transfer. The double slit experiment still supports wave behaviour. The point is not that light is only a particle or only a wave; it is that different experiments reveal different aspects of quantum behaviour.