QUESTION: Does the intensity of scattered photons increase in compton's effect , when compared to the photon at the source of light? If so, why ? Doesn't this violate the law of energy conservation?
ANSWER: First thing you need to know:
In physics, "intensity" refers to the amount of energy going through an area in a given amount of time. So we could say, "this light has an intensity of 1 joule per square meter per second."
If all photons in a beam of light have the same frequency (or energy, or wavelength), then intensity refers to the number of photons in the beam. No decrease in the number of photons, no decrease in the intensity.
If the number of photons in a beam remains the same, then a change in energy in any of these photons would mean a decrease in intensity.
So I'm not sure if you really meant "intensity," or are confusing the "intensity" of the incoming photons with their "energy."
But there is no increase in either the intensity or the energy of the incoming photons.
Compton scattering is not that much different than a collision between a moving billiard ball and a stationary billiard ball.
The x-ray photon comes in, strikes an electron (which is ejected), and the photon bounces off at an angle. The photon coming off at an angle has a lower energy, just like a billiard ball would lose a little of its kinetic energy. The energy lost by the photon is sent into the ejected electron. Thus, no violation of energy conservation.
This is all I would prefer to discuss about this topic.
---------- FOLLOW-UP ----------
QUESTION: Hello and thanks for the swift reply!
You said that a change in the photons energy would DECREASE the energy, but doesn't that depend on the frequency , ie if the frequency becomes lower as the number of photons remain the same , the intensity will decrease and vice versa.?
Thank you so much once again !
And by what I meant as a shift or increase in the intensity of a scattered photon in compton's scattering, is related to the picture in the link below:
> if the frequency becomes lower as the number of photons remain the same,
> the intensity will decrease and vice versa?
That would be correct.
The gif you showed was, I conclude, using the word "intensity" in a SLIGHTLY different way than I would prefer it be used. The word USUALLY means "energy intensity," but it can be used to mean "numerical intensity" as well. The graphs in that gif simply show that the absolute number of photons has two peaks, that the peak is at a higher wavelength (ie, lower frequency), and that the wavelength shift depends on the angle from the incident beam. Use of photons makes the latter fact easy to explain, but classical electrodynamics make it impossible to do so.