# Which particle aspect is required to explain photoelectric or Compton effect?

What do we mean when we say that it requires the particle nature of radiation i.e., photons, to explain photoelectric or Compton effect?

I don't understand which particle nature is used to explain these experiments. We cannot say that the momentum and energy are distinctive properties of a particle. Waves, too, have these properties.

So the question is which particle aspect is relevant for explaining the above-mentioned experiments? How do I make it understandable to a High school student?

What do we mean when we say that it requires the particle nature of radiation i.e., photons, to explain photoelectric or Compton effect?

Both the photon and the electron are point particles in the elementary particle table.

We cannot say that the momentum and energy are distinctive properties of a particle. Waves, too, have these properties.

Classically a particle is described not only by energy and momentum but also by a specific center of mass point at (x,y,z) which allows to determine a unique trajectory given the initial conditions. If it is a point particle more so. Impact points are also clearly known given the trajectory. A classical particle hits a wall at a specific (y,z) point.

Neither of these two processes ( photoelectric and Compton) is good for displaying the particle nature of the elementary particles, because their detection depends on the probability distribution measurable by the accumulation of many events.

The photoelectric effect demonstrates that photons exist as individual energy packets and not a continuum classical wave carrying energy. The same with Compton scattering, where the photon gives off part of its energy to the electron. But both cannot be used as a clear demonstration of a classical particle behavior, because both depend on probability distributions for the trajectories, there is no uniqueness of trajectory for the same boundary conditions.

IMO the best teaching demonstration of the dual nature of elementary particles is given by the double slit experiment.

The particle nature is the footprint of each individual electron , which is a unique x,y point on the screen, as expected by a particle trajectory. Its position seems random , until the wave nature is displayed in the accumulation. It is a probability distribution which is describable as the complex conjugate square of the wavefunction describing "electron scattering on two slits"

A better demonstration of a particle nature is this bubble chamber picture of a pi mu e decay

The pion from the main interaction decays to

Then the muon decays :

The track until the decay point acts like a classical trajectory , at that momentum and the same magnetic field it will have the same circular track. At the interaction point the probabilistic nature enters, which is tied up with the wave nature of the elementary particle: there is a probability distribution for the distributions of how the mu+ and nu_mu share the momentum energy, and the same for the sharing of the e+ nu_e and anti-nu_mu in the second decay, and it is probability distributions that display a wave nature.

## Instataneous response

Classical electromagnetic radiation impinging on a metal plate would require a quite long time to accumulate enough energy to kick out electrons. Assume a power of $P = 1 \, \text{W}$ on an area of $A = 1 \, \text{m}^2$ (corresponding to $\sim 5 \cdot 10^{18}$ surface atoms). How long will it take an atom to gather the $e \Phi \sim 5 \, \text{eV} \approx 8 \cdot 10^{-19} \, \text{J}$ required for ionizazion? If the power is distributed among all surface atoms equally, this would take $t = \frac{8 \cdot 10^{-19} \, \text{J}}{5 \cdot 10^{18} \cdot 1 \, \text{W}} = 4 \, \text{s}$ until an atom has enough energy to release one electron. But what you observe is that as soon as you switch on the light the electron emission starts. This alone doesn't necessarily require an explanation with quantized light, as one might come up with the idea of the light intensity being not equally distributed over the metal plate or with some mechanisms in the metal that tend to concentrate the energy at single atoms (and by this violating thermodynamics).

## Energy of the released electrons

The other point is that a classical electromagnetic field does not dictate the energy of the emitted electrons. Especially in the case of "classical" mechanisms that redistribute the energy there should be a large spread in the energies. But what you observe when you illuminate a metal plate with monochromatic light of photon energy $h \nu$ is that the kinetic energy of the electrons is very sharply peaked at $E_{\text{el}} = h \nu - e \Phi$ (photon energy minus binding energy of the metal).

Moreover if you increase the light flux this does not alter the energy of the electrons – it's only their number that increases. Again there is no classical explanation for that. The quantum mechanical explanation of light quanta (photons) that interact one-on-one with charge quanta (electrons) solves both – the problem with the instantaneous response and the one about the energy-flux-dependence.

Further discussion can be found for example in this answer to a similar question, maybe going a bit too far for explaining it to a high school student.

Compton observed that the wavelength of the scattered X-rays was always greater than the wavelength of the incoming X-rays. This shift in wavelength (frequency) - Compton shift cannot be explained using classical electromagnetic wave theory.

In the low-energy regime, the scattering of light by a free charged particle is called Thomson scattering. In the classical picture, the charged particle oscillates in the electric field of the incoming light and emits radiation with the same frequency, but with an intensity depending on the intensity of the incoming light.

In Compton scattering, the scattered light has a greater wavelength, which means that it has a lower energy. This shift depends on the scattering angle and not on the intensity of the incoming light.

Regarding the photoelectric effect, if you treat the light as an electromagnetic wave, you would expect that the photoelectrons will be ejected with some time-lag because the electromagnetic wave transfers energy continuously. But that is not what happens.

Also, from a classical perspective, the intensity of light should influence the kinetic energy of the photoelectrons since a higher intensity means a higher electric field. But what happens is that the intensity of light influences the number of photoelectrons, not their kinetic energy. Another argument is that the photoelectric effect happens only if the incoming light has enough energy. In the classical picture, the photoelectric effect should occur independent of the frequency given that the light is intense enough. But the photoelectric effect happens only if the frequency of the light exceeds a threshold frequency.