3.2.1.3
A photon is a quantum of electromagnetic energy
Particles, Antiparticles & Photons — AQA A-Level Physics
Key Definition
Photon — A massless discrete packet (quantum) of electromagnetic energy.
$$\begin{aligned}
E &= hf \\
&= \frac{hc}{\lambda}
\end{aligned}$$
- Electromagnetic energyThe capacity to do work. Measured in joules (J). is not transferred continuously — it comes in discrete packets.
- Each photonA quantum (discrete packet) of electromagnetic radiation. Its energy is proportional to its frequency. carries a specific amount of energyThe capacity to do work. Measured in joules (J). determined by its frequencyThe number of complete oscillations passing a point per unit time. Measured in hertz (Hz)..
- Higher $frequencyThe number of complete oscillations passing a point per unit time. Measured in hertz (Hz). = higher photonA quantum (discrete packet) of electromagnetic radiation. Its energy is proportional to its frequency. energyThe capacity to do work. Measured in joules (J).. Energy$ is inversely proportional to wavelengthThe minimum distance between two points on a wave that are in phase (e.g. crest to crest). Measured in metres (m)..
$$E = hf$$
- $E$: energy of the photonA quantum (discrete packet) of electromagnetic radiation. Its energy is proportional to its frequency. (J)
- $h$: Planck's constant (6.63 × 10⁻³⁴ J s)
- $f$: frequencyThe number of complete oscillations passing a point per unit time. Measured in hertz (Hz). (Hz)
$$E = \frac{hc}{\lambda}$$
- $E$: energy of the photon (J)
- $h$: Planck's constant (J s)
- $c$: speed of light (3.00 × 10⁸ m s⁻¹)
- $λ$: wavelengthThe minimum distance between two points on a wave that are in phase (e.g. crest to crest). Measured in metres (m). (m)
Worked Example
Light of wavelengthThe minimum distance between two points on a wave that are in phase (e.g. crest to crest). Measured in metres (m). 490 nm and powerThe rate of energy transfer. Measured in watts (W). 3.6 mW is incident on a surface and completely absorbed. Calculate the number of photons hitting the surface in 2.0 s.
Show Solution
1
List known values
- $\lambda = 490 \times 10^{-9}$ m
- $P = 3.6 \times 10^{-3}$ W
- $t = 2.0$ s
2
Calculate the energy of one photon
$$E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34}) \times (3.0 \times 10^{8})}{490 \times 10^{-9}} = 4.06 \times 10^{-19} \text{ J}$$
3
Find photons per second
$$\frac{P}{E} = \frac{3.6 \times 10^{-3}}{4.06 \times 10^{-19}} = 8.9 \times 10^{15} \text{ s}^{-1}$$
4
Multiply by time
$$N = 8.9 \times 10^{15} \times 2.0 = 1.8 \times 10^{16}$$
Answer
$N = 1.8 \times 10^{16}$ photons
Examiner Tips and Tricks
- Learn the definition: 'a discrete quantity/packet/quantum of electromagnetic energy'.
- Planck's constant and the speed of light are on the data sheet, but memorising them speeds up calculation questions.