Examples of photon in the following topics:

 It does so through a process of optical amplification based on the stimulated emission of photons.
 When such an electron decays without external influence, it emits a photon; this process is called "spontaneous emission. " The phase associated with the emitted photon is random.
 Therefore, after the atom decays, we have two identical outgoing photons.
 In stimulated emission process, a photon (with a frequency equal to the atomic transition) encounters an excited atom, and a new photon identical to the incoming photon is produced.
 The result is an atom in the ground state with two outgoing photons.

 The probability of pair production in photonmatter interactions increases with increasing photon energy, and also increases with atomic number ($Z$) of the nucleus approximately as $Z^2$.
 The photon must have enough energy to create the mass of an electron plus a positron.
 The electron and positron can annihilate and produce two 0.511 MeV gamma photons.
 A photon decays into an electronpositron pair.
 Describe process of pair production as the result of photon interaction with nucleus

 A photon is an elementary particle, the quantum of light.
 Photons are emitted in many natural processes.
 During all these processes, photons will carry energy and momentum.
 Energy of photon: From the studies of photoelectric effects, energy of a photon is directly proportional to its frequency with the Planck constant being the proportionality factor.
 In the case of a photon with zero rest mass, we get $E = pc$.

 It results in a decrease in energy (increase in wavelength) of the photon (which may be an Xray or gamma ray photon), called the Compton Effect.
 Part of the energy of the photon is transferred to the scattering electron.
 The Compton Effect is the name given to the scattering of a photon by an electron.
 Energy and momentum are conserved, resulting in a reduction of both for the scattered photon.
 Studying this effect, Compton verified that photons have momentum.

 In Compton scattering the photon always loses energy to an electron initially at rest.
 Inverse Compton scattering corresponds to the situation where the photon gains energy from the electron because the electron is in motion.
 where $\theta$ is the angle that the photon makes with the $x$axis in the lab frame.
 where $\Theta$ is the angle between the incident and scattered photon in the restframe of the electron.
 If we assume that the photon distribution is isotropic, the angle $\langle \cos\theta \rangle = 0$.

 Let's say that you have a blackbody spectrum of temperature $T$ of photons passing through a region of hot plasma ($T_e$).You can assume that $T \ll T_e \ll m c^2/k$.
 What is the brightness temperature of the photons in the RayleighJeans limit after passing through the plasma in terms of the Compton $y$parameter?
 Let's suppose that the blackbody photons are from the cosmic microwave background.What is the difference in the brightness temperature of the photons that pass through the cluster and those that don't (including the sign)?
 What is the inverse Compton emission from a single electron passing through a gas of photons field in terms of the energy density of the photons and the Lorentz factor of the electron?
 What is the total inverse Compton emission from the region if you assume that the synchrotron emission provides the seed photons for the inverse Compton emission?

 Compton explained the Xray frequency shift during the Xray/electron scattering by attributing particlelike momentum to "photons".
 If the scattered photon still has enough energy left, the Compton scattering process may be repeated.
 Photons with an energy of this order of magnitude are in the xray range of the electromagnetic radiation spectrum.
 Therefore, you can say that Compton effects (with electrons) occur with xray photons.
 A photon of wavelength $\lambda$ comes in from the left, collides with a target at rest, and a new photon of wavelength $\lambda '$ emerges at an angle $\theta$.

 In this case, the number of photons cannot change.
 Also let's assume that the number of photons is small so the number of photons of a particular energy is
 Because the photons and electrons are in equilibrium the average change in the energy of a photon must be zero
 so $\alpha = 4$ and we find that the fractional change in the photon's energy per scattering is
 Essentially the Compton $y$parameter tracks how the energy of a photon changes as it passes through a cloud of hot electrons.Specifically, the energy of a photon will be $E=e^y E_i$ after passing through a cloud of nonrelativistic electrons with $kT \gg E

 Couldn't you think about scattering as the absorption and reemission of a photon and include the process in the absorption coefficients and source functions?
 The formalism that we have developed so far doesn't allow there to be a correlation between the properties of an absorbed photon and the emitted photon.
 On the other hand, the initial direction and energy of a scattered photon are generally highly correlated with the photon's final momentum.
 We can first look at a process in which the photon is scattered into a random direction without a change in energy.

 Let's consider an isotropic distribution of photons and derive the total power emitted by an electron passing through.
 Let $v dE$ be the density of photons having energy in the range $dE$.
 The number of photons in a box over the energy range is a Lorentz invariant
 Let's assume that the change in the energy of the photon in the rest frame of the electron is negligible compared to the change in the energy of the photon in the lab frame, i.e.
 The rate of decrease of the total initial photon energy is