Concerning the electromagnetic capacity for an electron

[Kidphysics]

From what I understand, if a photon hits let's say an atom of Hydrogen its single electron will jump to a higher energy state until it releases it's own photon and returns to a lower state. My question is that what if a high energy gamma ray hits the Hydrogen atom? The electron will fly a long distance away from the atom ionizing it, but the photoelectric effect shows that there is some limit to what an electron can absorb as far as radiation is concerned. So my question is that indeed if a high energy gamma ray hits a Hydrogen atom, some of it will be absorbed by the electron ionizing it, but what about the rest of the energy, does it just zap the nucleus?

 

[Dickfore]

I have never heard of such a limit.

 

[Kidphysics]

I have never heard of such a limit.

So in theory one electron could absorb a high powered gamma ray and be extremely kinetically energized?

The reason I assume that there is a limit is from the photoelectric effect (taken from wiki)

1. For a given metal and frequency of incident radiation, the rate at which photoelectrons are ejected is directly proportional to the intensity of the incident light.

Electrons can absorb energy from photons when irradiated, but they usually follow an "all or nothing" principle. All of the energy from one photon must be absorbed and used to liberate one electron from atomic binding, or the energy is re-emitted. If the photon energy is absorbed, some of the energy liberates the electron from the atom, and the rest contributes to the electron's kinetic energy as a free particle.[citation needed]

 

[Dickfore]

So in theory one electron could absorb a high powered gamma ray and be extremely kinetically energized?

The reason I assume that there is a limit is from the photoelectric effect (taken from wiki)

1. For a given metal and frequency of incident radiation, the rate at which photoelectrons are ejected is directly proportional to the intensity of the incident light.

Electrons can absorb energy from photons when irradiated, but they usually follow an "all or nothing" principle. All of the energy from one photon must be absorbed and used to liberate one electron from atomic binding, or the energy is re-emitted. If the photon energy is absorbed, some of the energy liberates the electron from the atom, and the rest contributes to the electron's kinetic energy as a free particle.[citation needed]

How did you draw the conclusion that you had from the above quote? Specifically, the bolded part.

 

[Kidphysics]

How did you draw the conclusion that you had from the above quote? Specifically, the bolded part.

I assumed there was a limit because I interpreted the photoelectric effect incorrectly. Also I did not assume it was possibly for a single electron to be able to absorb such powerful energy but now I believe that if a gamma ray were to hit an electron it would absorb it totally and become an extremely kinetically energized free particle, yes?

 

[Dickfore]

It does not matter what you believe, but what actually happens. Also, when you say "extremely energetic", it means you compared the gamma-ray energy with another energy scale and it turned out to be much bigger. What is your reference energy scale and why would that energy be significant for the present discussion?

 

[Kidphysics]

It does not matter what you believe, but what actually happens. Also, when you say "extremely energetic", it means you compared the gamma-ray energy with another energy scale and it turned out to be much bigger. What is your reference energy scale and why would that energy be significant for the present discussion?

Not believe, it's what I deducted from the bold print, which seems to answer my question. So can you tell me what actually happens or more Socratic teaching for me? I have no reference energy scale other than my intuition.

 

[Dickfore]

It's a two-body process:

An incident photon is absorbed and we have an outgoing electron and an ionized atom flying away. However, this is an inelastic collision, since the kinetic energies of the outgoing particles are smaller than the incident photon energy. This energy difference is due to the binding energy of the electron. We can account for this difference in the relativistic expressions by assuming the rest mass of the outgoing ion to be:

$$M_{\mathrm{ion}} = M_{\mathrm{atom}} - m_{e} + \frac{B}{c^{2}}$$

Then, the laws of conservation of energy and momentum are:

$$\hbar \omega + M_{\mathrm{atom}} c^{2} = E_{e} + E_{\mathrm{ion}}$$

$$\hbar \mathbf{k} = \mathbf{p}_{e} + \mathbf{p}_{\mathrm{ion}}$$

Because of the high energies involved, we should use relativistic expressions:

$$E = \sqrt{(m c^{2})^{2} + (p c)^{2}}$$

Also, for the photon we have the relation $\omega = c k$.

 

[Kidphysics]

It's a two-body process:

An incident photon is absorbed and we have an outgoing electron and an ionized atom flying away. However, this is an inelastic collision, since the kinetic energies of the outgoing particles are smaller than the incident photon energy. This energy difference is due to the binding energy of the electron. We can account for this difference in the relativistic expressions by assuming the rest mass of the outgoing ion to be:

$$M_{\mathrm{ion}} = M_{\mathrm{atom}} - m_{e} + \frac{B}{c^{2}}$$

Then, the laws of conservation of energy and momentum are:

$$\hbar \omega + M_{\mathrm{atom}} c^{2} = E_{e} + E_{\mathrm{ion}}$$

$$\hbar \mathbf{k} = \mathbf{p}_{e} + \mathbf{p}_{\mathrm{ion}}$$

Because of the high energies involved, we should use relativistic expressions:

$$E = \sqrt{(m c^{2})^{2} + (p c)^{2}}$$

Also, for the photon we have the relation $\omega = c k$.

Thank you. Your responses have helped clear up a lot of things for me today.