Photoelectric and Compton Effects

[Red_CCF]

I was posting on another thread and it got me wondering: how come a photon is absorbed in the photoelectric effect but it is only partially absorbed in the Compton Effect? How come the electron in the Compton Effect does not absorb the photon entirely like the photoelectric effect? If photon is a particle as well how come it does not bounce off of a metal plate?

Thanks for any help that you can provide

 

[kanato]

In the Compton effect the photon cannot be completely absorbed because momentum and energy must be conserved. In the photoelectric effect, you have electrons in a metal which form a set of states with a continuous band of energies, so the system is much more flexible as to whether or not a photon can be absorbed. Not all the photons will be absorbed in the photoelectric effect, many will be scattered or reflected.

 

[Red_CCF]

In the Compton effect the photon cannot be completely absorbed because momentum and energy must be conserved.

Why would the collision be inelastic if the photon is completely absorbed?

 

[kanato]

Work out the change in energy and momentum if the photon is completely absorbed:

Initial energy is energy of photon plus rest energy of electron (p is momentum of photon):

E_i = pc + mc^2

Final energy is entirely the electron (p is now momentum of the electron):

E_f = \sqrt{(pc)^2 + (mc^2)^2}

Set them equal and square and you get:

(pc)^2 + 2 pmc^3 + (mc^2)^2 = (pc)^2 + (mc^2)^2

Cancel terms until you get p = 0. This tells you that energy and momentum can only both be conserved if the photon has zero momentum, and hence zero energy, so there is no physically meaningful result.

 

[jtbell]

Also note that in the photoelectric effect, the "object" that the photon "collides" with is a multi-particle system, which makes for many ways to distribute the photon's energy and momentum so that they are still conserved in the end.

 

[Red_CCF]

Work out the change in energy and momentum if the photon is completely absorbed:

Initial energy is energy of photon plus rest energy of electron (p is momentum of photon):

E_i = pc + mc^2

Final energy is entirely the electron (p is now momentum of the electron):

E_f = \sqrt{(pc)^2 + (mc^2)^2}

Set them equal and square and you get:

(pc)^2 + 2 pmc^3 + (mc^2)^2 = (pc)^2 + (mc^2)^2

Cancel terms until you get p = 0. This tells you that energy and momentum can only both be conserved if the photon has zero momentum, and hence zero energy, so there is no physically meaningful result.

Thanks so much for clearing that up

 

[vin300]

Why would the collision be inelastic if the photon is completely absorbed?

Inelatic collision refers to depletion of photon energy