[Physics] Compton effect and Photoelectric effect


Why in Compton effect an electron (which is very lightly bounded) can't absorb whole energy of incident photon but in photoelectric effect the electron absorb whole energy of the incident photon?
In Compton effect the electron is not totally free, if it absorb whole energy then it does not violet the basic postulate of relativity. But in this case we consider the collision of incident photon with the free electron (though it is not completely free) is elastic. Why it is not consider here some energy (though small) loss to free the electron from the metal? Please answer my question analytically.

Best Answer

The two processes are different.

In the Compton effect the energy of the incident photon (~ many keV or Mev) is very much larger than the binding energy of the electron in the atom (~ eV) and so a target electron bound in an atom can be considered as essentially free.
In the Compton effect the target electron is ejected from the atom and the process can be treated using "billiard ball" dynamics.
The equation for the change in wavelength derived by use of the conservation of momentum and energy shows that there is a finite maximum amount of energy that can be transferred from the photon to an electron in this process.

$\lambda' - \lambda = \dfrac {h}{m_{\rm e} c}(1 - \cos \theta)$ where $\lambda' - \lambda$ is the change in wavelength of the photon and $\theta$ is the scattering angle whose maximum value is $180^\circ$.

The scattered photon having lost some of its energy can then undergo further collisions with electrons in the material.

The photons responsible for the photoelectric effect have energies of order of a few electron-volts ($450\; \rm nm\approx 2.8 \;eV$) and for metals it is the conduction (free) electrons which are ejected. A free electron is given enough energy by the photon it has absorbed to overcome a potential barrier (work function energy) to escape with the rest of the energy of the photon from the surface of the metal.

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