# Photoelectric effect cant occur with a free electron

1. Sep 16, 2008

### EngageEngage

1. The problem statement, all variables and given/known data
The question asks me to prove that the photoelectric effect cannot occur with a free electron. ie. one not bound to an atom. A hint is also provided: Consider the reference frame in which the total momentum of the electron and incident photon are zero.

2. Relevant equations

3. The attempt at a solution
I've been thinking about how to prove this for like a week, but i cannot figure out how to do so. I know that in the case of a free electron, we will have a work function potential of 0V. As for the hint, I cannot really figure out what it is trying to get me to do. In the zero momentum frame, we are moving at a velocity that gives the electron a momentum equal to the momentum of the photon. This direction is the same direction that the photon moves in. The electron momentum will be the same, since p=E/c, which is a constant after an energy is predetermined in all frames (right?). After the collision (or absorption in this case), in this frame, the electron should stay stationary, since this is the zero momentum frame. If I were to move back to the lab frame, I would see the electron moving in the same direction as the photon was moving.

This actually brings me to more confusion. Why don't we just get Compton scattering at this point? Why should the photon be absorbed by the electron anyways?

If anyone could please push me into the right direction I would appreciate it greatly!

2. Sep 16, 2008

### Vanadium 50

Staff Emeritus
What's the mass of the electron after the collision in the center of mass system? What's the total energy? Does this make sense?

3. Sep 16, 2008

### granpa

is it possible that a lone electron would indeed oscillate in the presence of a light wave but would always emit exactly the same amount of light as it absorbs (sort of like free electrons in a metal reflect light that shines on it)?

4. Sep 16, 2008

### EngageEngage

Before in CM:
$$E_{CM,before}=E_{\gamma}+E_{e}$$

$$E_{e}=\sqrt{(pc)^{2}+(mc^{2})^2}$$

but, the momentum of the electron is the same as of the photon

$$E_{e}=\sqrt{E_{\gamma}^{2}+(mc^{2})^2}$$

$$E_{CM,before}=E_{\gamma}+\sqrt{E_{\gamma}^{2}+(mc^{2})^2}$$

After:

$$E_{CM,after}=mc^{2}$$

How could I include the excitation of the electron into the above expression, which will account for the absorption of the photon? Is this where this breaks down?

Thank you for all the help!

5. Sep 17, 2008

### Vanadium 50

Staff Emeritus
What excitation of the electron? (And that's a hint)

6. Sep 17, 2008

### EngageEngage

So you're saying that there is no excitation. Why is that so? Didn't the electron absorb the photon (or at least some of its energy)?

So are you saying that now I can write:

$$=E_{\gamma}+\sqrt{E_{\gamma}^{2}+(mc^ {2})^2}=mc_{2}$$

by conservation of energy in the CM frame?

7. Sep 17, 2008

### EngageEngage

if I can write that then I clearly have a contradiction because the energy before is significantly larger than the energy after. am i moving in the right direction? once again thank you for helping me out.

8. Mar 23, 2009

### sp416

I actually have been assigned this same question and am still a bit confused.

Is the idea to say that if Pelectron=Pphoton, then in the case of the photoelectric effect occurring we assume that the electron absorbs the photon thereby increasing it's total energy.

Would we then do a total energy balance where Et before absorption is Et=sqrt((hv)2+(mec2)2)

and after the absorption, the total energy would be just...Et=mec^2 since photons have no mass so the electron after absorbing the photon has just rest energy?

Is this a valid interpretation of the problem? I'm very curious about what's really going on here...I feel a little dull to be honest.

Last edited: Mar 23, 2009
9. Nov 17, 2010

### videomaniac

...

The first term under the radical for the energy is incorrect. Since this is the total relativistic energy of the electron, you must use the kinetic energy of the electron, pc, after the collision. If you work out the simultaneous conservation of linear momentum and total energy you'll indeed get a contradiction, that the rest mass of the electron is zero. It isn't necessary to go to a C.M. frame for the analysis, the lab frame will work fine. Was this out of the problem set in Chpt. 3 of Tipler's "Modern Physics"?

10. May 20, 2011

### chucklingchuc

conservation of momentum:
P_gamma + P_electron = P'_gamma + P'_electron (eq. 1)
Since P_electron = 0 (assumed), P'_gamma = 0 (it is absorbed) eq. 1 becomes:
P_gamma = P'_electron (eq. 2)

conservation of energy:
E_gamma + E_electron = E'_gamma + E'_electron (eq. 3)
Since E'_gamma = 0 (it is absorbed), eq. 3 becomes:
E_gamma + E_electron = E'_electron (eq. 4)

Using the following identities along with eq.2 and 4, you end up with m_electron = 0 which is obviously false.

E_gamma = h*v
P_gamma = h*v/c
E_electron = KE_electron+m_electron*c^2
P_electron^2*c^2 = E_electron^2-m_electron^2*c^4

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