A photon is the quantum of electromagnetic radiation. Unlike an ordinary particle with rest mass, a photon has zero
rest mass but still carries both energy and momentum. That is one of the most important ideas in modern physics,
because it connects wave properties such as wavelength and frequency with particle-like quantities such as energy and
momentum.
The basic photon energy relation is
Photon energy formulas.
\[
\begin{aligned}
E &= hf \\
&= \frac{hc}{\lambda}
\end{aligned}
\]
Here, \(E\) is the photon energy, \(h\) is Planck’s constant, \(f\) is the frequency, \(c\) is the speed of light,
and \(\lambda\) is the wavelength. Since light in vacuum always satisfies
\[
c = f\lambda,
\]
you can move easily between frequency and wavelength. A shorter wavelength means a higher frequency, and therefore a
larger photon energy.
Photon momentum
Even though photons have no rest mass, they still carry momentum. The momentum of a photon is given by
Photon momentum formulas.
\[
\begin{aligned}
p &= \frac{h}{\lambda} \\
&= \frac{E}{c}
\end{aligned}
\]
This relation is important in radiation pressure, scattering, and quantum mechanics. It explains why light can exert
force on matter and why photons participate in momentum conservation even without rest mass.
Useful eV–nm shortcut
In many practical problems, especially in optics and atomic physics, energy is expressed in electronvolts and
wavelength in nanometers. Combining constants gives a very convenient approximation:
Convenient conversion rule.
\[
\begin{aligned}
E(\mathrm{eV}) \approx \frac{1240}{\lambda(\mathrm{nm})}
\end{aligned}
\]
This shortcut is extremely useful for visible and ultraviolet photons. It comes directly from the exact relation
\(E = hc/\lambda\), after inserting constants and converting units.
Sample example: \(\lambda = 500\ \mathrm{nm}\)
For a photon with wavelength \(500\ \mathrm{nm}\), first convert the wavelength to meters:
\[
\lambda = 500\times 10^{-9}\ \mathrm{m} = 5.00\times10^{-7}\ \mathrm{m}.
\]
The frequency is then
\[
\begin{aligned}
f &= \frac{c}{\lambda} \\
&= \frac{3.00\times10^8}{5.00\times10^{-7}} \\
&\approx 6.00\times10^{14}\ \mathrm{Hz}.
\end{aligned}
\]
The energy is
\[
\begin{aligned}
E &= \frac{hc}{\lambda} \\
&\approx 3.97\times10^{-19}\ \mathrm{J}.
\end{aligned}
\]
Converting to electronvolts,
\[
E \approx 2.48\ \mathrm{eV}.
\]
The momentum is
\[
\begin{aligned}
p &= \frac{h}{\lambda} \\
&\approx 1.33\times10^{-27}\ \mathrm{kg\cdot m/s}.
\end{aligned}
\]
These values correspond to a visible-green photon. They show clearly that visible light carries a small but nonzero
momentum and an energy of only a few electronvolts.
Physical interpretation
As wavelength decreases, both energy and momentum increase. This is why X-rays and gamma rays are much more energetic
than visible light, while radio photons are much less energetic. The same relation explains why high-frequency light
is more effective in phenomena such as the photoelectric effect.
At a more advanced university level, the same ideas connect to relativistic particle physics. For a photon, the
relativistic energy-momentum relation reduces to
\[
E = pc,
\]
which is perfectly consistent with \(p = h/\lambda\) and \(E = hf\).
| Quantity |
Relation |
Meaning |
| Wave-speed relation |
\(c=f\lambda\) |
Links frequency and wavelength in vacuum |
| Photon energy |
\(E=hf=hc/\lambda\) |
Energy rises as wavelength decreases |
| Photon momentum |
\(p=h/\lambda=E/c\) |
Photons carry momentum despite zero rest mass |
| Convenient optics rule |
\(E(\mathrm{eV})\approx1240/\lambda(\mathrm{nm})\) |
Fast wavelength-to-energy conversion |
| Relativistic photon relation |
\(E=pc\) |
Special case of the energy-momentum relation for massless particles |