Gamma Ray Ionization Path to Overcome Potential

In summary: Are you saying that even though the photons have well above the energy required to ionize the electrons, that they won't ionize them?Naturally, the photons would not ionize the electrons if they have above the energy required.
  • #1
sapratz
27
0
If a gamma ray is being fired and is in turn colliding with electrons and ionizing the electrons in the gamma ray path, where do the electrons tend to? In what direction is the current most likely to flow for a high powered gamma ray ionization?
 
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  • #2
You mean what happens when free electrons collide with gamma ray photons?

Their momentum changes depending on the momentum of the gamma ray photons. The direction of the electrons (and thus the current), depends on their original momentum.
If they stood still, they will move in the direction of the gamma rays.But I'm not sure, I can half remember something that electrons need to transfer momentum to something else/heavy (a nucleus), but I can't quite remember what the situation was for this.
 
  • #3
Dreak said:
You mean what happens when free electrons collide with gamma ray photons?

Thanks for the response,

I am not asking about anything like compton scattering and all that, I am referring to the process of the electrons that have been ionized after the collision with the gamma ray photons. Where do the ionized particles rejoin the positive ions? where do they tend to travel to relieve the developed potential? do they tend to move in all directions randomly, or do the loop back into the beam to the positive particles?

Thanks
 
  • #4
I am not asking about anything like compton scattering and all that, I am referring to the process of the electrons that have been ionized after the collision with the gamma ray photons.
That happens via Compton scattering.
Where do the ionized particles rejoin the positive ions?
They do not.

do they tend to move in all directions randomly
Well, not against the direction of the incoming photons.An unrelated question: Which gamma ray intensity do you plan to have? Are you sure that those electrons are relevant at all? If this could be possible, what about a vacuum?
 
  • #5
mfb said:
An unrelated question: Which gamma ray intensity do you plan to have? Are you sure that those electrons are relevant at all? If this could be possible, what about a vacuum?


This photons have an individual energy of 2MeV. I just want to know the path of the electrons so I can deduce some information about the electric field produced. This is all happening in air. Where are the electrons that are being released via compton scattering going?
 
  • #6
sapratz said:
This photons have an individual energy of 2MeV.
An average energy of 2 MeV (according to your other thread).

I just want to know the path of the electrons so I can deduce some information about the electric field produced.
Okay, but first I am curious if that is relevant at all. If your produced electromagnetic fields are a factor 1 million below anything you could observe in your setup, it is pointless to calculate this in detail.

Where are the electrons that are being released via compton scattering going?
http://en.wikipedia.org/wiki/Klein–Nishina_formula
 
  • #7
sapratz said:
If a gamma ray is being fired and is in turn colliding with electrons and ionizing the electrons in the gamma ray path, where do the electrons tend to? In what direction is the current most likely to flow for a high powered gamma ray ionization?

Wait, back off a minute. How do you "ionize electrons"??!

Zz.
 
  • #8
sapratz said:
This photons have an individual energy of 2MeV. I just want to know the path of the electrons so I can deduce some information about the electric field produced. This is all happening in air. Where are the electrons that are being released via compton scattering going?

In case of low intensity ray (if you don't mean nuke-powered gazer), ionization level will be very little, and majority of photons will be absorbed by neutral atoms of the air. Thus, with very good approximation you may assume that ions and electrons impulses will be evenly distributed in all directions.

For careless estimation energy of impulse, where scattering on free electrons become significant:
Let's say, it takes 1km of air to absorb half of gamma-photons,
Amount of molecules per m2:
Volume(1000m3)/MolarDensity(0.0224m3/mol)*AvogardoConstant(6*10^23)=3*10^28
Energy density of impulse= 3*10^28*2Mev/m^2= 10^16 J/m^2
Due to secondary ionization it may be reduced by few orders, but still is quite big
 
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  • #9
Well, ionizing half the molecules would form a really dense and extremely hot plasma, probably destroying the equipment.

If we have 1kW of gamma rays, we get losses of 1W/m. With a narrow beam, this is sufficient to induce thermal convection, and to get a small current (at ~1MeV/electron, 1μW with a length of 1m).
 
  • #10
mfb said:
Well, ionizing half the molecules would form a really dense and extremely hot plasma, probably destroying the equipment.

Are you saying that even though the photons have well above the energy required to ionize the electrons, that they won't ionize them? naturally I am discussing rough number and concepts and the difference between expected value (1.9MeV) and the distribution is null. So let's keep this simple and just assume the values are 1.9MeV. I don't see how there is energy dissipation in this beam if there is no ionization, obviously there are tons of interactions happening here. The order of the generated EM field is obviously very low but that's why were are running these tests, we have expectation of the field generated from the machine, and also those EM fields are happening at a different time so they aren't really that big of an issue anyways. There are fields being produced that we are trying to understand here. I was looking for a way to be able to verify whether or not the unknown fields were from the electron interactions with the gamma energy photons.
 
  • #11
Graniar said:
In case of low intensity ray (if you don't mean nuke-powered gazer), ionization level will be very little, and majority of photons will be absorbed by neutral atoms of the air.

What do you consider a "low intensity ray"?
 
  • #12
sapratz said:
Are you saying that even though the photons have well above the energy required to ionize the electrons, that they won't ionize them?
I didn't say that, where did you get that impression?

naturally I am discussing rough number and concepts
Okay. The most important rough number is the gamma ray intensity. Can you give us some estimate for that?

So let's keep this simple and just assume the values are 1.9MeV.
Okay.

The order of the generated EM field is obviously very low but that's why were are running these tests
Which tests?
 
  • #13
ZapperZ said:
Wait, back off a minute. How do you "ionize electrons"??!

Zz.

How about - remove them from atoms and molecules, by fair means or foul?

But there are various possibilities.

A photon might undergo a Compton scattering off an electron that is effectively free, weakly bound to the outskirts of an atom.

In this case, the speed of the nucleus would be unaffected, because the nucleus would be a mere spectator that suddenly discovers its electron missing.

The angular distribution of the electron would then be given by Compton scattering - it would be impossible for the electron to travel towards the origin of the photon (because the photon is losing momentum and energy, not gaining).

On the other hand, imagine that a photon excites electron and nucleus as an atom.

In this case, the momentum of the electron would be balanced by the momentum of the recoil of the nucleus. The momentum of the photon would be irrelevant, and the electron could travel in any direction including towards the origin of photon.

So which of these processes would happen?
Or would both processes happen, but in every case just one or the other?
Or would a process of intermediate nature happen, where the nucleus/cation does receive some momentum, but the momentum/direction of the photon still matters?

Now, in every case, the electron ends up being much faster than the cation - whether the electron has equal momentum but higher energy, or whether the cation does not get momentum at all.

Am I right in guessing that the electron also travels further than the cation, before it is slowed down and captured into an anion?

Now, say imagine that the gamma rays are confined into a narrow ray.
Whether they undergo Compton scattering (whereby the electrons are emitted preferentially towards the ray, but can be emitted some angle to the side as well) or atomic ionization (wherby electrons are emitted in all directions), the electrons can leave the beam and many do, but cations due their small or no speed remain in or near the beam at first.

If the beam is weak then the effects of the beam are negligible. The cations stay where they are created in the beam, until they slowly diffuse away by Brownian motion. The anions also stay where they were created by stopping electrons, until they slowly diffuse and happen to meet the cations.

The beam thus creates a positively charged beam (of cations left behind by ionization) and a negatively charged shell (of anions from the stopped electrons). And also an electric field between the beam and the shell - inside the shell. Beam and shell together are neutral, so no electrostatic field outside the shell - in case of isotropic atomic ionization.

But if the ionization is at least partly from Compton scattering, then the electrons preferentially travel in the photon travel direction.

The negative shell is then displaced from the positive beam interior along the beam axis.

And this could create electrostatic field observable outside the shell.

Now, if the beam is strong, then for one the electrostatic fields would affect the ion behaviour appreciably. The cations would not only undergo free thermal diffusion out of the beam, but would also be propelled by electrostatic field to actively drift outside the beam. And the anions in the shell would actively drift towards the beam. Also the heating by the beam would cause the air and ions in the beam to expand, and then start to travel buoyantly in the direction that is up.

So... any comments on how the electrons would mainly be enitted, or what the electrostatic fields would look like? What would be the size of the shell?
 

1. What is gamma ray ionization?

Gamma ray ionization is a process in which gamma rays, a form of high-energy electromagnetic radiation, interact with atoms or molecules in a material, causing them to lose or gain electrons and become ionized.

2. How does gamma ray ionization overcome potential?

Gamma ray ionization can overcome potential barriers, such as the binding energy of an electron to an atom, by providing enough energy to remove the electron from the atom and creating an ion. This allows for the formation of new chemical bonds and reactions to occur.

3. What are some potential applications of gamma ray ionization?

Gamma ray ionization has many potential applications in various fields, including radiation therapy for cancer treatment, sterilization of medical equipment, and food preservation. It is also used in industrial processes, such as polymerization and cross-linking of plastics.

4. Is gamma ray ionization safe?

Gamma ray ionization can be dangerous in high doses, but in controlled and regulated settings, it can be used safely for various purposes. However, proper safety precautions should always be taken when working with gamma rays.

5. How is gamma ray ionization different from other types of ionization?

Gamma ray ionization differs from other types of ionization, such as alpha and beta particle ionization, in that it does not involve the emission of particles. Instead, gamma rays are electromagnetic radiation that can pass through materials and cause ionization by interacting with atoms or molecules along their path.

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