# Energy transfer from photon to an electron

1. May 16, 2015

### Jimmy Moriaty

Please tell me,why can't a photon transfer it's energy completely to a free electron?

2. May 16, 2015

### Orodruin

Staff Emeritus
This would necessarily violate the conservation of momentum.

3. May 16, 2015

### Drakkith

Staff Emeritus
So what happens when the electron absorbs the photon?

4. May 16, 2015

### Jimmy Moriaty

But in the photo-electric effect it's happeing.why?

5. May 16, 2015

### Orodruin

Staff Emeritus
Momentum is being transferred to something else and therefore it is conserved overall.

6. May 16, 2015

### Jimmy Moriaty

Then,why can't electron travel the same direction which the photon have been travelled?
(after absorbing the energy of photon)

Then, why can't electron travel the same direction which the photon have been travelled?(after absorbing the energy of photon)

7. May 16, 2015

### Drakkith

Staff Emeritus
Don't ask me, that's what I want to know!

8. May 16, 2015

### Orodruin

Staff Emeritus
It can, as long as there is something else involved to ensure momentum conservation. If the electron is free, it simply cannot absorb a photon due to momentum conservation.

9. May 16, 2015

### Staff: Mentor

Try it.... The electron changes its speed from 0 to $v$ as it absorbs the photon. The kinetic energy of the (non-relativistic, for simplicity) electron increases from zero to $mv^2/2$ and the momentum from 0 to $mv$. If the photon is to be completely absorbed and energy and momentum are to be conserved, then these values must be equal to the energy $p_{\gamma}c$ and the momentum $p_{\gamma}$ of the initial photon. A bit of algebra will pretty quickly convince you that this isn't possible - if the increase in the electron's energy is equal to the photon's total energy then the momentum won't balance, and vice versa.

10. May 16, 2015

### Jimmy Moriaty

Yeah!
You are right.There is a difference between those two velocities.
I've chose the wavelength of 0.071nm x-rays.
When I consider the E=Pc & P=mv I've got v=1.025*10^7 ms^-1
But when i chose E=Pc & E=1/2 mv^2 it's v=7.84*10^7 ms^-1
But what's the reason for this?

Last edited: May 16, 2015
11. May 16, 2015

### Staff: Mentor

You're assuming that there is a way for a photon to transfer all its energy and momentum to a single particle while respecting the laws of mechanics. The math exercise is showing that making this assumption inevitably leads to a contradiction.

12. May 16, 2015

### Staff: Mentor

The reason is that the mass of the electron is fixed. If the electron could become more massive then it would be possible to conserve both the energy and the momentum of the collision. This is why a free atom can absorb a photon but not a free electron.

13. May 16, 2015

### Drakkith

Staff Emeritus
So what happens? Does the free electron interact with the photon at all?

14. May 16, 2015

### Staff: Mentor

Yes. It scatters. So an electron and a photon come in, and an electron and a photon come out.

15. May 17, 2015

### Drakkith

Staff Emeritus
Ah, okay. Thanks, Dale.

16. May 17, 2015

### blue_leaf77

That's quite relativistic values. Just for an advice for a more accurate verification is to use relativistic formula, upon which you should find that $\hbar \omega + m_ec^2 = \sqrt{(\hbar \omega)^2 + (m_ec^2)^2}$ and you see the discrepancy very clearly. As Dale has said, this equation can be satisfied if $m_e$ on the RHS is allowed to increase.

Last edited: May 17, 2015
17. May 17, 2015

### ZapperZ

Staff Emeritus
This really needs to be clarified and corrected.

In the photoelectric effect, and in photoionization, the interaction is NOT with "free electrons". The electrons are either coupled to the lattice solid (as in the photoelectric effect) or bound in an atom. So your question about "free electron" and then asking about photoelectric effect is really comparing apples with oranges. It is not the same thing.

When photons hit either a metal or an atom, it is the whole solid or the whole atom that interacts. The energy bands and the energy levels that the electrons get promoted to are not present in free electrons. That is why the solid or the atom are the ones that are in the excited state. A consequence of this is that one or more electrons change state to a higher energy state. So while it is tempting to focus on just the electrons, since they are the ones that were changed, it is the entire solid or the entire atom that actually is responsible to allow for the energy absorption from the photons.

Zz.

18. May 18, 2015

### Jimmy Moriaty

Thank you everyone!

19. Jan 25, 2016

### drl

If a photon can give a packet of energy to an electron, overcoming its work function and freeing it up, can a low energy photon recover that same packet and cause the electron to revert to its bound state? I am assuming that an electron and an atom exist in particle form in their bound state and in a wave form in their free state.

Last edited: Jan 25, 2016
20. Jan 25, 2016

### Drakkith

Staff Emeritus
Not sure. That almost sounds like stimulated emission, but that process doesn't involve light of different wavelengths.

That would be incorrect. There aren't two separate states that objects transition between. They always behave according to their fundamental properties, some of which are 'wave-like' and some of which are 'particle-like'.