Energy and momentum of a photon
| English | Chinese | Pinyin |
|---|---|---|
| photon | 光子 | guāng zi |
| quantum | 量子 | liàng zǐ |
| electronvolt | 电子伏特 | diàn zi fú tè |
| momentum | 动量 | dòngliàng |
| rest mass | 静止质量 | jìng zhǐ zhì liàng |
Light comes in packets
- Light is not a smooth stream — it arrives in tiny packets called photons 光子.
- Each photon is a quantum 量子 of electromagnetic energy, travelling at the speed of light.
- A solar cell, in effect, counts them.
Photon energy
- A photon's energy is $E = hf = \dfrac{hc}{\lambda}$, where $h = 6.63 \times 10^{-34}\ \text{J}\cdot\text{s}$.
- Higher frequency (shorter wavelength) → more energy. A γ-ray photon dwarfs a radio photon.
The electromagnetic spectrum: as frequency increases (wavelength decreases), each photon carries more energy. A gamma-ray photon has billions of times more energy than a radio photon.
A photon is a packet of energy, E = hf
Energy of a photon
E = h·f
Photon energy is proportional to frequency — the gradient is Planck's constant h.
The energy of a photon is:
$E = hf = \dfrac{hc}{\lambda}$ — proportional to frequency.
Higher-frequency photons carry more energy.
$E = hf$, so energy rises with frequency — a γ-ray photon carries far more than a radio one.
The electronvolt 电子伏特
- On the atomic scale we use the electronvolt: $1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$.
- It is the energy an electron gains across a $1\ \text{V}$ p.d. (A visible photon is a few eV.)
A charged zinc plate loses its charge — the gold leaf falls — when ultraviolet light shines on it
A photon has energy $3.2 \times 10^{-19}\ \text{J}$. What is this in eV? ($1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$)
$\dfrac{3.2 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.0\ \text{eV}$.
Photon momentum 动量
- A photon also carries momentum: $p = \dfrac{E}{c} = \dfrac{h}{\lambda}$.
- It has zero rest mass 静止质量, yet a real momentum — which is why light can push on a surface.
The maximum kinetic energy of photoelectrons rises linearly with frequency, reaching zero at the threshold frequency $f_0$
A photon's momentum is:
$p = \dfrac{E}{c} = \dfrac{h}{\lambda}$.
A photon has zero rest mass but still carries momentum.
Its momentum is $\dfrac{E}{c}$ — non-zero even though its rest mass is zero. This gives radiation pressure.
You've got it
- photon energy $E = hf = \dfrac{hc}{\lambda}$ (higher $f$ → more energy)
- $1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$
- photon momentum $p = \dfrac{h}{\lambda}$ (zero rest mass, non-zero momentum)