22.1 - Photon energy and momentum

22.1 - Photon energy and momentum

Describe photons as quanta of electromagnetic energy, calculate their energies in joules and electronvolts, and use their momentum to explain a mechanical transfer from light to a surface.

Quanta of electromagnetic radiation

Can light transfer momentum to a surface even though photons have no rest mass? It can. The photon model connects the energy of electromagnetic radiation with a momentum that can change when light is absorbed or reflected.

You already know that light can interfere and diffract. Those observations require its wave behaviour. When electromagnetic radiation exchanges energy with matter, it also shows a particulate aspect: energy is transferred in discrete quanta called photons.

Photon

A photon is a quantum of electromagnetic energy.

The photon is not a little object riding inside a separate electromagnetic wave. Photons are the quanta of the radiation itself. “Particle” here names the discrete energy-and-momentum behaviour, rather than a claim that every light experiment can be described as classical balls following visible paths.

For radiation of one frequency, each photon has the same energy. Increasing the number of photons increases the total energy without changing each photon's energy. Changing frequency changes the energy of an individual photon. There is no requirement that all electromagnetic radiation have one common photon energy.

At fixed beam area and frequency, a more intense beam delivers more photons each second. It does not mean that each photon is more energetic. This distinction will explain why changing light's colour and changing its intensity have different effects in the next lesson.

Model limit

Use the model appropriate to the observation: wave behaviour accounts for diffraction and interference, while discrete photon exchanges account for effects such as photoelectric emission. Neither picture licenses treating photons as tiny material balls in every situation.

The first step is to state clearly what the quantum belongs to.

Photon energy and the electronvolt

A higher-frequency photon transfers a larger quantum of energy. Planck's constant gives the proportionality between that energy and frequency. Combining it with the free-space wave relation connects energy to wavelength.

Photon energy

E=hf=hcλE=hf=\frac{hc}{\lambda}

EE is photon energy in J, hh is Planck constant in J s\mathrm{J\,s}, ff is frequency in Hz, cc is the speed of light in free space in m s−1\mathrm{m\,s^{-1}}, and λ\lambda is wavelength in free space in m. The wavelength form follows from c=fλc=f\lambda. Doubling frequency doubles photon energy; doubling wavelength halves it.

Not on the data sheet: learn it.

Photon energies are often inconveniently small in joules. The electronvolt, eV, is the energy transferred to an electron when it moves through an accelerating p.d. of magnitude one volt. Since energy transfer has magnitude charge multiplied by p.d.,

1 eV=(1.60×10−19 C)(1 V)=1.60×10−19 J1\,\mathrm{eV}=(1.60\times10^{-19}\,\mathrm{C})(1\,\mathrm{V}) =1.60\times10^{-19}\,\mathrm{J}

The elementary-charge value is supplied, but this conversion is not a separate formula on the general sheet. Multiply an energy in eV by 1.60×10−191.60\times10^{-19} to obtain J; divide joules by that factor to obtain eV. An electronvolt is an energy unit, not a potential difference and not the charge on an electron.

Worked example 1: a visible photon's energy

Question. Radiation has free-space wavelength 5.20×10−7 m5.20\times10^{-7}\,\mathrm{m}. Calculate its photon energy in J and eV. Use h=6.63×10−34 J sh=6.63\times10^{-34}\,\mathrm{J\,s}, c=3.00×108 m s−1c=3.00\times10^{8}\,\mathrm{m\,s^{-1}} and 1 eV=1.60×10−19 J1\,\mathrm{eV}=1.60\times10^{-19}\,\mathrm{J}.

Plan. Find energy in SI units, then change the energy unit.

E=hcλ=(6.63×10−34)(3.00×108)5.20×10−7 J=3.825×10−19 JE/eV=3.825×10−191.60×10−19=2.390625\begin{aligned} E&=\frac{hc}{\lambda} =\frac{(6.63\times10^{-34})(3.00\times10^{8})}{5.20\times10^{-7}}\,\mathrm{J} =3.825\times10^{-19}\,\mathrm{J}\\ E/\mathrm{eV}&=\frac{3.825\times10^{-19}}{1.60\times10^{-19}}=2.390625 \end{aligned}

Answer. 3.83×10−19 J3.83\times10^{-19}\,\mathrm{J}, or 2.39 eV2.39\,\mathrm{eV} (3 s.f.).

Check. These are the same energy in different units. An energy in eV is not multiplied by the elementary charge again when it is already expressed in joules.

A beam's total energy is the sum of its photon energies. If every photon has energy EE, a photon arrival rate N˙\dot N gives beam power P=N˙EP=\dot N E. Here N˙\dot N is a number per second, PP is in W and EE is in J. This counting relation follows from energy per second and is not printed separately on the data sheet.

Worked example 2: power and photon arrival rate

Question. A monochromatic beam has power 1.0 mW1.0\,\mathrm{mW} and photon energy 4.0×10−19 J4.0\times10^{-19}\,\mathrm{J}. Determine the number of photons passing a cross-section each second.

Plan. Divide joules per second by joules per photon.

N˙=PE=1.0×10−3 J s−14.0×10−19 J=2.5×1015 s−1\dot N=\frac{P}{E}=\frac{1.0\times10^{-3}\,\mathrm{J\,s^{-1}}}{4.0\times10^{-19}\,\mathrm{J}} =2.5\times10^{15}\,\mathrm{s^{-1}}

Check. A modest macroscopic power can involve an enormous photon rate because each photon carries a tiny energy.

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