Photons and the photoelectric effect

In an ordinary experiment, red light cannot release electrons from some metals however much its intensity is increased. Weaker light of higher frequency may produce a current. Why does the energy in one portion matter, as well as the total energy delivered?

When light is absorbed, it transfers energy in discrete portions. A photon is a quantum of electromagnetic radiation; complete absorption transfers its whole energy. Frequency ν, in hertz, sets the energy of one portion. This does not remove wave behaviour: light still interferes and diffracts. Do not picture a photon as a small, cut-off fragment of a wave. Quantum theory connects how light propagates with how it exchanges energy.

The relation is E=h\nu=hc/\lambda. E is in joules, and the Planck constant h≈6,63·10⁻³⁴ Дж·с sets the size of an energy quantum. Here c≈3·10⁸ м/с and λ is the vacuum wavelength in metres. Doubling frequency doubles photon energy; doubling λ halves it. The units J·s times s⁻¹ give J. Do not combine a wavelength in matter with vacuum c without accounting for the medium's refractive index.

Keep frequency fixed and increase the light flux. More photons cross the same area in the same time, but each still has energy hν. Increasing frequency instead changes the energy of each portion. These two experiments can deliver the same total energy with different photon numbers and energies. This distinction explains the photoelectric threshold: one portion may be insufficient for an electron even when the whole beam carries plenty of energy.

For electron energies, the electronvolt is convenient: approximately 1\,\text{эВ}=1{,}60\cdot10^{-19}\,\text{Дж}. It is the energy associated with an elementary charge crossing a potential difference of 1 V. By definition |e|≈1,60·10⁻¹⁹ Кл is positive; the electron charge is −|e|. An electronvolt is not a voltage: energy is charge times potential difference. The numerical connection for one electron does not make the units interchangeable.

A photon transfers momentum as well as energy: in vacuum p=E/c=h/\lambda, with p in kg·m/s. Doubling frequency doubles momentum; doubling λ halves it. A photon's rest mass m₀ is zero. There is no contradiction: the familiar p=mv applies to a slow massive object and is not a universal definition of momentum. Momentum transfer by light can exert pressure on an absorber.

We can formally express photon energy in mass units: m_\gamma=E/c^2. This is the mass equivalent of energy, in kilograms, not a nonzero rest mass. The equivalent grows with energy while the photon's invariant mass stays zero. A photon has no rest frame: we cannot come to rest beside it as we could beside a moving trolley. Without this qualification, the historical notation suggests a misleading picture of a massive light particle.

In the external photoelectric effect, light releases electrons from a material into the space outside it. An electron is bound to the material. Escaping through the surface needs a minimum energy contribution: the work function A. It is measured in joules or electronvolts and depends on the material and surface condition. Changing the surface can change the threshold even with the same light source. Here work means the energy needed for release, not the duration of illumination.

In the single-photon model, one electron absorbs one photon. The balance h\nu=A+E_{\text{к,max}} says that the work function must be supplied first; the remainder can become kinetic energy. The maximum applies to electrons with no additional losses; others can emerge with less energy. Increasing hν at fixed A raises this upper limit by the same amount. If hν<A, a negative kinetic-energy result means emission is impossible, not motion with negative energy.

Imagine many photons, each falling just short of A. In the ordinary single-photon process considered here, increasing their number does not combine many portions into one emission event. Intensity therefore cannot overcome the frequency threshold. At sufficient frequency, a larger photon flux can release more electrons and increase the photocurrent. Multiphoton processes under other conditions lie outside this model and do not contradict its stated conclusion.

At the emission threshold, the energy left for motion approaches zero: h\nu_0=A, so \nu_0=A/h. Threshold frequency, in hertz, increases with work function: stronger binding requires more energetic photons. For a fixed surface, ν must be at least ν₀ in the ideal threshold model. If A is in eV and h in J·s, first make the energy units consistent. Dividing numbers expressed in different energy units gives the wrong frequency.

Since λ=c/ν in vacuum, the lowest allowed frequency corresponds to the longest allowed wavelength: \lambda_0=c/\nu_0=hc/A. This threshold wavelength λ₀ is measured in metres. Doubling A halves it, so shorter waves are needed. The traditional name red limit means the long-wavelength boundary; it need not lie in visible red light. For λ>λ₀ a photon lacks enough energy even if the source is bright.

How can we measure the electrons' maximum energy? Apply a field that slows them on their way to the collector. At the stopping potential Uз, even the most energetic electrons fail to reach it. Uз is the smallest voltage magnitude that achieves this, in volts. With charge in coulombs, E_{\text{к,max}}=|e|U_\text{з} gives energy in joules. Doubling the maximum energy requires twice Uз. The circuit polarity provides retardation; the energy equation uses the positive magnitude.

To find the emission speed of a nonrelativistic electron, use E_{\text{к,max}}=mv_{\max}^2/2, where m≈9,11·10⁻³¹ кг. Thus v_max grows as the square root of energy: quadrupling energy doubles speed. First check that v≪c. For electron energies that bring speed close to c, the classical formula is insufficient. Neither the photoelectric energy balance nor the definition of energy requires treating this approximation as exact outside its range.

Photons of energy 3 eV fall on a metal with work function 2 eV. Find the electrons' maximum kinetic energy and the stopping potential.

The photons have energy 3 eV and the work function is 2 eV. Since 3 > 2, single-photon emission is energetically possible.

The maximum energy left for motion is Eк,max = E − A = 3 − 2 = 1 эВ.

An electron loses 1 eV of kinetic energy when crossing a retarding potential difference of 1 V. Therefore Uз = 1 В.

The result describes the most energetic electrons. Electrons with extra losses have less energy and stop reaching the collector at a lower voltage.

Eк,max = 1 эВ, Uз = 1 В. Of each photon's energy, 2 eV is needed for release, leaving at most 1 eV for electron motion.

Frequency sets the energy of one light quantum. At the same frequency, greater intensity means more quanta.

Work function sets the threshold. The difference hν−A is the maximum photoelectron energy, not a common energy for all electrons.

Photons have energy and momentum despite zero rest mass. E/c² is a separate mass equivalent of energy.

Next we will explore binding energy and the statistical transformations of atomic nuclei.