Compton Effect Explanation Can someone brief me about Compton effect and why does this happen?
I searched everywhere read a CERN article too but couldn't understand it.
 A: When a high energetic photon (like the gamma or X ray photon) hit a charged particle like an electron, due to inelastic collision, the photon loses some energy and the electron get scattered. The energy lost by the photon will be equal to the energy gained by the scattered electron. This process of inelastic scattering of electron by a photon is called Compton scattering and the phenomenon is called Compton effect. This experiment ensures the particle nature of radiation like the photoelectric effect.  

Since the energy of the incident photon is reduced it's wavelength should increase (and frequency should decrease as per the relation:  
$$E=h\nu=\frac{hc}{\lambda}$$   
(This is why the yellow photon turned into a red photon in the animation).  
Hence the wavelength of scattered photon will be greater than that of the incident photon. This process differs from photo electric effect on the fact that in photoelectric effect, a photon is completely absorbed by the electron. The absorbed energy appears as the work function + the kinetic energy of the electron (in the case of metals).   
The photon usually work in the principle of "all or nothing" where the complete photon energy is either absorbed or not. i.e., a photon is not partially absorbed. This is used in explaining photoelectric effect. But, here due to inelastic scattering, the photon could transfer some of it's momentum ($\displaystyle{pc=\frac{E}{c}=\frac{h}{\lambda}}$) to the electron.  
The X-rays are highly energetic and have a binding energy far greater than that of an atomic electron (in the range of keV). So once incident on an atomic electron, the electron becomes free. Compton's experiment proved that light can behave as a stream of particle-like objects (quanta of energy), whose energy is proportional to the light wave's frequency. If the photon is of low but sufficient energy corresponding to visible light and soft X-rays, it can eject an electron from its host atom entirely (a process known as the photoelectric effect), instead of undergoing Compton scattering.   
Consider an electron at rest. An X-ray photon is coming from the left and will incident on the electron as shown.  
 
The electron gain some energy by transfer of momentum as expected in particle collisions. So the photon loses some energy and the electron gains some. Let $\lambda$ be the wavelength of the incident photon and $\lambda^\prime$ be that of the scattered photon. The original energy of the photon will be now equal to the sum of energy gained by the electron and the energy of the scattered photon as required by the conservation of energy.  Here $\theta$ represents the scattering angle of photon. Compton also included the possibility that the interaction of photon with electron would sometimes accelerate the electron to speeds sufficiently close to the velocity of light and would require the application of Einstein's special relativity theory to properly describe its energy and momentum. The basic principle used in the derivation of Compton scattering is the conservation of energy and momentum.  
Hence  
$$E_\gamma+E_e=E_{\gamma^\prime}+E_{e^\prime}\longrightarrow{conservation\space of\space energy}$$   
where the left hand side indicate the energy of photon and electron before collision an right hand side indicate the energy of photon and electron after collision. (The prime indicates that the parameter is associated with scattering).  
Also  
$$\vec{p_{\gamma}}+\vec{p_{e}}=\vec{p_{\gamma^\prime}}+\vec{p_{e^\prime}}$$
Since the initial momentum of electron at rest is zero, we write 
$$\vec{p_{\gamma}}=\vec{p_{\gamma^\prime}}+\vec{p_{e^\prime}}\longrightarrow{conservation\space of\space momentum}$$  
Now considering the relativistic effects, 
$$E_e=m_e c^2 \space \space \space (m_e-rest\space mass\space of\space electron)$$
$$E_{e^\prime}=\sqrt{(p_{e^\prime}c)^2+(m_e c^2)^2}$$  
Referring to the conservation of energy equation  
$$\frac{hc}{\lambda}+m_e c^2=\frac{hc}{\lambda^\prime}+\sqrt{(p_{e^\prime}c)^2+(m_e c^2)^2}$$  
Rearranging both sides and squaring  
$$(p_{e^\prime}c)^2= (\frac{hc}{\lambda}+m_e c^2-\frac{hc}{\lambda^\prime})^2-m_e^2 c^4$$   
or  
$$(p_{e^\prime}c)^2=(\frac{hc}{\lambda})^2+(\frac{hc}{\lambda^\prime})^2+(\frac{1}{\lambda}-\frac{1}{\lambda^\prime})2hcm_e c^2-\frac{2h^2c^2}{\lambda \lambda^\prime}\longrightarrow(1)$$  
From this expression we can find the magnitude of the scattered photon. It is to be seen that the momentum gained by the scattered electron will be greater than the momentum lost by the photon This is a consequence of relativistic effect since even though the initial momentum of the electron is zero, it have a rest energy).  
Now, from the conservation of momentum equation, we write  
$$\vec{p_{e^\prime}}=\vec{p_\gamma}-\vec{p_{\gamma^\prime}}$$  
By invoking the scalar product  
$$ p_{e^\prime}^2=\vec{p_{e^\prime}}\cdot \vec{p_{e^\prime}}=(\vec{p_\gamma}-\vec{p_{\gamma^\prime}})\cdot (\vec{p_\gamma}-\vec{p_{\gamma^\prime}})$$  
or by rule of cosines  
$$ p_{e^\prime}^2=p_\gamma^2+p_{\gamma^\prime}^2-2p_\gamma p_{\gamma^\prime} \cos\theta$$  
Multiplying both sides by $c^2$, we have  
$$ p_{e^\prime}^2 c^2=p_\gamma^2 c^2+p_{\gamma^\prime}^2 c^2-2c^2 p_\gamma p_{\gamma^\prime} \cos\theta$$
For a photon $E=pc=hc/\lambda$. So, the first two terms on the right hand side of the above equation represent the square of the energies of the incident and scattered photons respectively. Hence we write  
$$ p_{e^\prime}^2 c^2=(\frac{hc}{\lambda})^2+(\frac{hc}{\lambda^\prime})^2-\frac{2h^2c^2\cos\theta}{\lambda \lambda^\prime}\longrightarrow(2)$$  
Comparing both equations (1) and (2) we have  
$$(\frac{1}{\lambda}-\frac{1}{\lambda^\prime})2hcm_e c^2-\frac{2h^2c^2}{\lambda \lambda^\prime}=-\frac{2h^2c^2\cos\theta}{\lambda \lambda^\prime}$$  or  
$$(\lambda^\prime-\lambda)m_e c-h=-h\cos\theta\Rightarrow (\lambda^\prime-\lambda)m_e c={h}(1-\cos\theta)$$
or  
$${\color{red}{ \Delta\lambda=(\lambda^\prime-\lambda)=\frac{h}{m_ec} (1-\cos\theta)}}$$  
which gives the shift in the wavelength of the scattered and incident photon, called Compton Shift 
It is clear that when $\theta=0^0$, there will be no shift in wavelength, which means that if the incident photon travel undeviated, then there is no change in energy of the photon. This means there will be no electron on it's path. When $\theta=180^0$, then the incident photon is reflected back and the shift in wavelength will be maximum, means the photon and the shift corresponds to the maximum energy that the electron can gain. 
A: Compton scattering occurs when a photon impacts an atom, and is reflected with less energy, and at an angle. The larger the angle, the larger the share of the photons energy is absorbed. This proves that photons have a particle nature and posses quantized energy.
Imagine a golf ball hitting a pole and getting deflected. If the pole is more massive, there will be a sharper deflection.  That's the Compton effect.
