photon energy and momentum

Interference, diffraction and polarization show that light has wave properties, yet energy exchange with matter occurs in discrete portions.

Einstein proposed that light propagates and is absorbed in quanta with energy $E = h\nu = hc/\lambda$, where $h$ is Planck's constant, $\nu$ is frequency and $\lambda$ is wavelength.

Special relativity gives the energy–momentum relation $E^2 = (pc)^2 + (m_0 c^2)^2$.

For a photon $m_0 = 0$ (a photon cannot be at rest), so $E = pc$, hence the momentum is $p = E/c = h/\lambda$.

The momentum direction coincides with the direction of light propagation, which appears in radiation pressure and in the Compton effect.

Thus the photon is a massless particle with nonzero momentum, and its energy and momentum are related through the speed of light $c$.

Find the energy (in joules and electron-volts) and momentum of a photon with wavelength $\lambda = 500$ nm.

A laser pointer with power $P=1.0$ mW emits light of wavelength $\lambda=650$ nm. (a) How many photons are emitted per second? (b) What force does the light exert on a black absorbing target if the beam strikes it completely?

An electron and a positron are at rest and annihilate, producing two identical photons that fly apart in opposite directions. The electron rest energy is $m_ec^2=0.511$ MeV. Find the energy, momentum and wavelength of each photon.

Self-study plan with a simulation (safe, no real radiation sources).

1. Open the PhET simulation 'Photoelectric Effect'.
2. Choose a metal (for example, sodium) and for wavelengths 400, 450, 500, 550 and 600 nm record the stopping voltage $V_s$.
3. For each wavelength calculate the frequency $\nu=c/\lambda$, the photon energy $E=h\nu$ and the momentum $p=h/\lambda=E/c$; fill in a table.
4. Plot $eV_s$ versus $\nu$. The slope should be close to Planck's constant $h$, and the intercept with the $\nu$ axis should be close to the red threshold of the photoelectric effect.
5. Double the intensity and verify that $V_s$ and the photon energy do not change while the photocurrent increases. Explain this using the number of photons per second $N=P/E$.
6. Write a 3–4 sentence conclusion linking photon energy and momentum.

Show that photon energy and momentum increase with frequency (and decrease with wavelength) and are related by $E=pc$, and provide benchmarks along the electromagnetic spectrum.

A two-panel diagram.

(a) A graph of $E$ versus $\nu$: a straight line with slope $h$; points for radio waves, visible light, X-rays and gamma rays are marked on it.

(b) A sketch of a photon beam: for each photon the momentum vector $\vec p$ points along the beam, and the vector length is proportional to $1/\lambda$; next to it the relation $E=pc=h\nu$ is written.

A smartphone screen shows a green area of wavelength $\lambda = 550\ \text{nm}$ and gives off about $P = 0.10\ \text{W}$ of green light. Each photon carries $E = hc/\lambda = \frac{6.63\times10^{-34} \times 3.00\times10^{8}}{5.50\times10^{-7}} = 3.61\times10^{-19}\ \text{J} = 2.25\ \text{eV},$ so the screen emits $N = P/E \approx 2.8\times10^{17}$ photons every second.

This is the photon picture: $E = hf = hc/\lambda$ with $h = 6.63\times10^{-34}\ \text{J·s}$, and each photon also carries momentum $p = E/c = h/\lambda = 1.20\times10^{-27}\ \text{kg·m/s}$. Multiplying by the number of photons per second gives the total momentum carried away each second, $3.3\times10^{-10}\ \text{kg·m/s}$, i.e. a light-pressure force of order $3\times10^{-10}\ \text{N}$ — which is why a $0.20\ \text{kg}$ phone does not visibly recoil.

A microwave oven's magnetron runs at $f = 2.45\ \text{GHz}$, so $\lambda = c/f = 12.2\ \text{cm}$. One photon has $E = hf = 6.63\times10^{-34} \times 2.45\times10^{9} = 1.62\times10^{-24}\ \text{J} = 1.01\times10^{-5}\ \text{eV}.$ An $800\ \text{W}$ oven therefore delivers $N = P/E \approx 4.9\times10^{26}$ photons every second, and each of them carries $p = E/c = 5.4\times10^{-33}\ \text{kg·m/s}$.

The concept supplies both relations: $E = hf$ turns a measured frequency into the energy of a single quantum, and $p = E/c$ shows that even a massless quantum has momentum. The value $1.01\times10^{-5}\ \text{eV}$ explains why such photons make water molecules rotate and heat food, yet cannot ionise atoms, whose ionisation energies are of order $1$–$10\ \text{eV}$.

A golf laser rangefinder fires a $905\ \text{nm}$ infrared pulse of energy $3.0\times10^{-7}\ \text{J}$ lasting $10\ \text{ns}$ at a flag $150\ \text{m}$ away and times the echo. One photon of that pulse has $E = hc/\lambda = 2.20\times10^{-19}\ \text{J} = 1.37\ \text{eV}$, so the pulse contains $3.0\times10^{-7}/2.20\times10^{-19} \approx 1.4\times10^{12}$ photons.

Each photon carries momentum $p = h/\lambda = 7.3\times10^{-28}\ \text{kg·m/s}$; a pulse reflected straight back changes the momentum of every photon by $2p$, so over the $10\ \text{ns}$ pulse the light pushes the flag with about $2.0\times10^{-7}\ \text{N}$. The same $E = hc/\lambda$ fixes the photon energy the detector must register, which is why near-infrared light is chosen for the low-noise sensor.