# Migration Path of Electrons in Ionized Air (Gamma Radiation)

1. Jul 16, 2013

### sapratz

I would love some clarification on a gamma ray process. This is what I understand so far:

Electrons are accelerated at 19MeV at a cathode which is releasing gamma ray photons with an energy of 1.9MeV. Is it possible to generalize the emission number of photons (roughly) to be equal to the 19MeV/1.9MeV (per electron fired)?

Assuming that is okay, we can move on to the range of the gamma photons in air. If I have calculated the decreasing exponential curve describing the number of photons present at each point in the gamma ray beam, can I take the derivative of this function spatially to be able to calculate the number of collisions at any given point?

Assuming that's alright, I have the number of collisions happening at any given point and I have calculated that the interactions are mostly compton (from high energy photons). I used the generalized for for maximum energy transfer:

Emax = 1.9MeV(2*(3.718/8.436)) = 1.67MeV transferred to each electron, with the remainder producing a photon at a frequency of 5.56e19Hz.

(We are only looking at the first wave of interactions, not the successive photon interactions)

so we have the energy of an ionized electron and the number of electrons that are being ejected at any given point from the beam.

My REAL question (assuming all that other stuff is correct) is can we calculate the charges moving at each point to essentially develop a current profile at any given point along the beam?

And also, just conceptually, does anyone have any idea as to the path the electrons will take after they are released? how does the system balance the potential difference developed? Does it arc to ground, or does it loop back to the start of the beam?

Thanks,
Scott

2. Jul 16, 2013

### Staff: Mentor

Why should it emit photons with exactly 1.9 MeV?
You want 10 1.9MeV-photons per incoming electron? I think it would be a great achievement if you get something close to 1.

Sure.

This is the maximal energy transfer, the actual energy transfer can be lower.
In addition, you get pair production.

It should be possible to calculate the estimated rate of electrons everywhere.

They ionize some other atoms, emit some bremsstrahlung and so on - the typical processes for fast electrons.
You create a lot of free charges in ionizing processes. I guess those are sufficient to balance the currents.

3. Jul 16, 2013

### sapratz

I don't know how to navigate the the forum very well, so I'll just copy paste, but thanks for the response.

Why should it emit photons with exactly 1.9 MeV?

-The particle guys gave us that number, it's the average energy per photon released from the accelerator. Why do you think this number is odd?

You want 10 1.9MeV-photons per incoming electron? I think it would be a great achievement if you get something close to 1.

-Where does the conservation of energy come into play here? I assumed that since the electrons were colliding with an energy and were experiencing breaking radiation until absorption they were going to emit photons of equal energy. That's where I got the number.

This is the maximal energy transfer, the actual energy transfer can be lower.
In addition, you get pair production.

- how much lower is the energy do you think? does the percentage transferred depend on the amount of energy contained in the photon? i was trying to determine what the angle of ejection was to calculate the amount of energy transferred, but I couldn't find anything on it. I didn't know if there was a distribution of probabilities based on the element/compound type but I was having trouble.

They ionize some other atoms, emit some bremsstrahlung and so on - the typical processes for fast electrons.

- so there is no "path" that the electrons will flow? we are essentially trying to determine the magnetic field generated by the resulting interactions from the gamma interactions. Do the electrons tend to correct the potential difference by being absorbed back at the earlier parts of the gamma radiator? basically do the ions arc back to the positive ions created?

thanks

4. Jul 17, 2013

### Staff: Mentor

You can quote things by putting [noparse]

Last edited: Jul 17, 2013
5. Jul 17, 2013

### ZapperZ

Staff Emeritus
I am still unsure where the gamma photons are coming from. It appears that this is entirely from the electrons that was accelerated to 19 MeV? Over what distance are they accelerated to give that 1.9 MeV gamma? I've accelerated electrons to 8 MeV over a distance of about 10 cm, and we detect no gamma. In fact, what we detect is "beam loading" in the RF cavity (i.e. it "sucked" in EM field).

Now, if the 19 MeV electrons were to crash onto, say, the stainless steel pipe or some high-Z material, then I definitely know those generate lots of gamma via Bremsstrahlung (I used to deal with accelerator safety and shielding issues a lot!).

Zz.

6. Jul 18, 2013

### sapratz

Yes, the 19MeV electrons are colliding with a very High-Z material and that is how the 1.9MeV photons are being produced

7. Jul 18, 2013

### ZapperZ

Staff Emeritus
Then you should have given this vital info, because it also means that you have left out a very important piece of information: the thickness of this material.

In such a Bremsstrahlung radiation, most of the radiation emitted is along the direction of the path of the electrons, i.e. the forward direction. How much of these gamma photons that got through is, obviously, a direct function of the thickness of the material due to the attenuation effects inside the material.

So as you have described the problem, there is at least one major piece of info that is missing.

Zz.

8. Jul 18, 2013

### sapratz

120mil tantalum :)

so the average energy of the gamma photons released is 1.9MeV and the number that is actually released is dependent upon the thickness and density of the plate material? That makes sense I guess. Any ideas on some formulas for this calculation?

Also, do you think there will be current traveling on the plate due to electron interactions?