Thanks for trying to answer me Mr Chi but my problem remain,I will make myshelf clear.
If the wavelength of the X-ray photon before the collision with the electron is ë then after the collision the wavelength of the X-ray becomes bigger because the X-ray photon looses a part of its momentum because of the interaction with the electron.
Let’s call this new bigger wavelength ë’. When we are using X-rays the Compton effect is noticeable because the difference ë’-ë is measurable, but when we are using visible photons the ë’ is equal to ë and we can not notice the Compton effect.
And my question is the following:
The wavelength does not change because the ë of the visible photon is 100000000 times bigger than the diameter of the electron and so the photon actually can not see the electron? or because the changing of the momentum during the collision is so small that the difference between ë’ and ë can not be measured in our labs?
What is realy happening I am very curious,aren't you?
The wavelength shift in the Compton effect depends only on the mass of the particle with which the effect takes place. The maximal shift occurs when the photon bounces back at 180 degrees. For The Compton effect with a photon and an electron' the change in wavelength is about 0.02 Angstroms. So when your photon is in the visible light, you just cannot see this difference.
You are obviously talking about the equation of Compton:
ë´- ë= L(1- cosö).
Where L is equal to L=h/mc=0.02426 Angstroms (for the electron) so the shift ë´- ë of the photon is between 0 < ë´- ë < 0.04852 Angstroms
(because 0< 1-cosö < 2).
We know that the modern spectrometers has resolving power of 0.01 Angstroms so we can measure shifts of the wavelength which are bigger than 0.01 Angstroms. This means that:
ë´- ë > 0,01 Á and ëc(1- cosö) > 0.01 A if we do the calculations we will get that we can measure shifts of the wavelength when the change of the angle is bigger than 54 degrees.
If my thoughts are correct then the Compton effect is not noticeable when the change of the direction of the visible photon during the collision with the electron is smaller than 54 degrees??? Is that correct????????????????
Can we say that the Compton effect with a visible photon and an electron it is noticeable only when ö>54 degrees???????????????
The frequency of a monochromatic source is never completely monochromatic (Due to Heisenberg's uncertainty principle), it has a finite spectral width (linewidth). The natural linewidth (Linewidth due to HUP), is usually around 8 MHz, which corresponds to a wavelength difference of around 0.03 angstroms in the visible, which is roughly the same wavelength change induced by the compton shift.
Hence it is quite impossible to spectrally differentiate between the Compton shifted radiation and the original source.
I think that this not so “simple” as the collision between an X-ray photon and an electron. This is because here appears the photoelectric effect.
The free electrons of the metal can absorb only photons that have energy bigger than the extraction work of the metal.
So if the X-ray has energy equal or bigger than the extraction work then there is no X-ray rebounding because it is absorb by the electron. But if the X-ray’s energy is less than the extraction work we can notice the Compton effect.
"The only thing I know is that
I don't know anything"
The Compton effect occurs primarily in the absorption of high X-ray energy and low atomic numbers. The effect takes place when high X-ray energy photons collide with an electron. Both particles may be deflected at an angle to the direction of the path of the incident X-ray. The incident photon having delivered some of its energy to the electron emerges with a longer wavelength. These deflections, accompanied by a charge of wavelength are known as Compton scattering.
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