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Background

Irving Kaplan, in Article 6.7: The Compton Effect of Nuclear Physics (2nd Ed.) explains the Compton effect as follows:

Compton (1923) was able to show that when a beam of monochromatic x-rays was scattered by a light element such as carbon, the scattered radiation consisted of two components, one of the same wavelength as that of the incident beam, the second of slightly longer wavelength.

The explanation for the observation of unshifted line is that

If the collision is such that the electron is not detached from the atom, but remains bound, the rest mass $m_0$ of the electron must be replaced by the mass of the atom, which is several thousand times greater. The calculated value of $\Delta \lambda$ then becomes much too small to be detected. A photon which collides with a bound electron, therefore, does not have its wavelength changed, and this accounts for the presence of the unshifted spectral line.

Now consider a carbon atom (vibrating in a bulk) moving towards X-ray photon (having frequency $\nu \neq(\delta E_n)/h$, where $\delta E_n$ refers to any possible excitation of electron) and photon is moving towards the carbon atom. photon strikes the electron of carbon atom. Their velocities are such that same energy is given back to photon which is given to carbon atom when photon collides and is thrown back at 180 degrees. The photon has same energy as before hence same frequency. (Everything is taken in a particular laboratory frame of reference.) The energy of atom and photon are not changed only their direction of motion are changed.


$%Note the visible separation between the background and the questions$ With this in mind, then

  1. Is the photon ejected after collision the same photon which was thrown in? If not, what happened to the previous photon? What is going on with the energy carried by the previous photon? Since energy cannot be absorbed, where does the energy go?

  2. When this photon collides (perpendicularly), is it at rest at the very moment of collision?

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  • $\begingroup$ Related: physics.stackexchange.com/q/35177/2451 $\endgroup$ – Qmechanic Dec 7 '13 at 21:29
  • $\begingroup$ @Qmechanic This question is talking of compton scattering. It is different from the other questions because they talk about reyleigh and thomson scattering in which light is absorbed and then re-emitted. $\endgroup$ – user31782 Dec 9 '13 at 5:24
  • $\begingroup$ Please consider adding comment before downvoting. $\endgroup$ – user31782 Dec 24 '13 at 20:04
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    $\begingroup$ This question is good in its present state. Note how far it has gone from its initial state. If you put in the work on clarifying it before posting, you avoid downvotes and you help your answerers in having something clear to address, and in not having to hit a moving target. $\endgroup$ – Emilio Pisanty Mar 14 '14 at 13:01
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    $\begingroup$ And just to piggyback on @EmilioPisanty, it would also prevent your posts from becoming community wikis from excessive editing so you can actually keep the rep you earn from it. 24 revisions is a lot and reflects more work should have been done to clarify it before posting. $\endgroup$ – tpg2114 Mar 14 '14 at 16:21
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In particle physics there exists elastic scattering for all interactions: change of direction but not of energies.

When a photon penetrates into a medium composed of particles whose sizes are much smaller than the wavelength of the incident photon, the scattering process, also known as Rayleigh scattering, is also elastic. In this scattering process, the energy (and therefore the wavelength) of the incident photon is conserved and only its direction is changed. In this case, the scattering intensity is proportional to the fourth power of the reciprocal wavelength of the incident photon.

How probable it is to be reflected by 180degrees would be a matter for a specific calculation, not for the weak in mathematical skills and numerical methods. It all depends on the boundary and the atoms/material which is the target for scattering. It is still a matter of active research to calculate the cross sections for such processes.

The nonrelativistic limit of Rayleigh scattering amplitude on 2s electrons of neutral and partially ionized atoms is obtained by making use of the Green Function method. The result takes into consideration the retardation, relativistic kinematics and screening effects. The spurious singularities introduced by the retardation in a nonrelativistic approach are cancelled by the relativistic kinematics. For neutral and partially ionized atoms, a screening model is considered with an effective charge obtained by fitting the Hartree-Fock charge distribution with pure Coulombian wave functions corresponding to a central potential of a nucleus with Z eff as the atomic number. The total cross section of the photoeffect on the 2s electrons is also calculated from the imaginary part of the forward scattering amplitude by means of the optical theorem.

Edit ,Replying to comment and additions to question

Is the photon ejected after collision is the same photon which was thrown in if not what happened to the previous photon and what is going on with the energy involved with previous photon in the above mentioned situation as energy cannot be absorbed.

You have to realize that elementary particles are indistinguishable except for their quantum numbers. There is no naming for individual electrons or photons so it does not make sense to talk of what happened to the "previous photon". Generically you can say that an elastically scattered photon is the "same" photon with the same energy E=h*nu . An inelastically scattered photon changes by giving up part of its energy to the crystal or the atom or the electron (kicking it out of its bound state) . A totally absorbed photon, if it has kicked an electron to a higher energy level, disappears from the picture its energy absorbed and its quantum numbers balanced in the reaction

When this photon collides (perpendicularly) at that very moment of collision is it at rest?

A photon can never be at rest, it is always traveling with velocity c in all inertial frames one can cook up. It is either absorbed or scattered . The probability of this happening is predicted by QED calculations, depending on the boundary conditions of the problem.

EDIT2 replying to comment:

the way a ball bounces back from a totally elastic wall, even classically it is not at rest" would you show me graph of speed of the ball vs time. and if i ask a gentle question What is the speed of ball(and photon) at the very moment of collision(t=t0). As i understand v=dx/dt at (t=t0) it is discontinues how can you say the speed of photon is c at t=t0

Let us not confused the vector velocity, which does change, with speed, which is a scalar and for the photon has the same value always. It is discontinuous at exact t=o classically though. In the classical case of the ball one would use delta functions to describe what happens at the impact on the vector going to infinitesimal times and space locations. At that point one would reach the quantum mechanical domain and with some very complex equations one would be able to describe t=0, with a delta(t) due to the Heisenberg Uncertainty principle. The speed of the ball will be the same after that delta( t), ( from energy conservation of the total elastic scattering) the velocity in the opposite direction.

Quantum mechanically there are Feynman diagrams that describe the interaction of photons with matter,where at the point of "contact" with the fields of the atoms the photon is "absorbed" to a virtual electron and then re emitted. Virtual means the electron is off mass shell, its " rest mass" different from zero for the electron. Once the photon is scattered it is on mass shell again, with rest mass=0 and speed c. Have a look at this chapter to understand the complexity of the matter. And here is a link with calculations and Feynman diagrams for elastic (Thomson and Ra)yleigh scattering.

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  • $\begingroup$ Can we have elastic scattering in _compton Effect_as i conceived in my question. Please let me know whether my question is technically correct or is worthless. $\endgroup$ – user31782 Dec 7 '13 at 15:15
  • $\begingroup$ Your question as it is is correct but you will need a serious course in physics to be able to grasp the complexity of scattering at the quantum level. For example if you had studied special relativity you would know that the photon is never at rest because it has a mass of zero and it travels with velocity c . $\endgroup$ – anna v Dec 7 '13 at 16:16
  • $\begingroup$ I just want to know. is the probability of such event is $0$. $\endgroup$ – user31782 Dec 7 '13 at 16:31
  • $\begingroup$ If the event you are talking about is for the photon to be at rest, the probability is zero by construction of the theory, which theory has been validated with innumerable experiments. $\endgroup$ – anna v Dec 7 '13 at 16:33
  • $\begingroup$ No the probability to reflect at $180^0$ in compton effect $\endgroup$ – user31782 Dec 7 '13 at 16:36
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Figured it would be interesting to look at some X-ray scattering cross-sections for carbon. A wide variety of scattering processes are involved. I suggest the OP have a look at the different scattering mechanisms involved.

Fig. 3-1. "Total photon cross section $\sigma_{\hbox{tot}}$ in carbon, as a function of energy, showing the contributions of different processes: $\tau$, atomic photo-effect (electron ejection, photon absorption); ,$\sigma_{\hbox{coh}}$ coherent scat­tering (Rayleigh scattering—atom neither ionized nor excited); $\sigma_{\hbox{incho}}$ incoherent scattering (Compton scattering off an electron);$\kappa_n$ pair production, nuclear field;$\kappa_n$ pair production, electron field; $\sigma_{\hbox{ph}}$ photonuclear absorption (nuclear absorption, usually followed by emission of a neutron or other particle)" From this Lawrence Berkeley document.

enter image description here

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