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Ion Beam collision calculations

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LAN_403

MArked One

Sun Oct 21 2012, 02:59PM

Registered Member #3688 Joined: Mon Feb 14 2011, 07:39PM
Location: Europe
Posts: 38

Greetings.

I was wondering if anyone could help me out with an interesting problem I've run into:

I'm trying to calculate the necessary distance for an ion beam travelling at 100-200 KeV with a current of say 60 amps (obviously pulsed) to collide with another beam travelling at the same velocity with the same current density, and dissipate most (90%) of its energy via collision.

To put this in clearer terms, I'm looking to calculate the time a beam must spend inside another beam for 90% of the particles to experience collision.

I'm thinking of mapping the particle beams to two 3d probability distributions, then simulating collision and thus estimating the necessary "time window" for their interaction.

This then raises the question of how the Actual distribution looks: is it even, with A constant density or are there "modes" of activity?

Also, how does one calculate the theoretical amount of ions in the beam? (the ions in questions are h+ or d+)

if Anyone has Any resources or programs they could point me to, I would greatly Appreciate it!

Here's a list of values that I'm Aware of (As of now). I will update this list when I get new data:

BEAM:

Voltage: 100-200 KV
Amperage: 60A
Pulse duration: 41-50 microseconds, may be shortened if the PSU is redesigned
60A is approximately 374,490,560,595,610,759,043 protons per second.

OTHER:

Vacuum level: submicron/1 micron (depends on outgassing)
Acceleration method: linear Accelerator
Pulse waveform: probably irregular or one-pulse sinusoidal.

Once Again, Any input will be Appreciated!

Kind Regards,
Ivan Kozlov

Sulaiman

Sun Oct 21 2012, 10:17PM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3141

I think that answer to the easy part of the question is
the electrical charge of a proton is 1.602176565(35)Ã—10âˆ’19 Coulomb
so 60A is approximately 374,490,560,595,610,759,043 protons per second.

How do you intend to produce such an ion current?

MArked One

Mon Oct 22 2012, 01:37AM

Registered Member #3688 Joined: Mon Feb 14 2011, 07:39PM
Location: Europe
Posts: 38

That's instantaneous pulse current. The design uses A vector inversion generator to create 500J pulses. The values for current were derived by division against the estimated pulse time.

We Assumed that the pulse current would equal beam current. Are we mistaken?

Thank you for the input: the first post will be updated with the new data

Ash Small

Mon Oct 22 2012, 07:12AM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

I believe space charge will be an issue here.

The protons will repel each other, so the beam will tend to spread out.

Focussing will be an issue here.

If you plan to use deuterium, then you are probably considering fusion, so, if you haven't done so already, it will be worth reading up on 'fusion cross-section' for additional information.

Dr. Slack

Mon Oct 22 2012, 07:21AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

It's all to do with cross section. There are two issues. A) what is the cross section of a particle? B) what is the effective cross section of the beam, given a particle density?

If you pretend to be a particle, and collide yourself with the other beam, you can count what happens particle by particle. You have a C/T chance of hitting the first one, where C is the cross section of the particle, and T is the total area of the beam. The chance of hitting the second particle is not independent, there is a small chance it might be hiding behind the first one, which reduces the total chance of hitting any particle from 2C/T to slightly less than that.

As you travel down the beam and have met T/C particles, when the chance of collision would be exactly 100% if they all spread out with no shadowing, the actual chance of hitting a particle is (e-1)/e, or about 63%. You can do this exponential decay thing for other numbers of particles met, ie length of beam.

Obviously the beam will not be uniform density, so you can either 1) integrate over the area or 2) just take the densest part, as the less desne parts will not be contributing significantly.

What's the collision cross section per particle? This varies with what you define as collision. The area for deflecting the particle by at least 1 degree will be relatively large. The area for dissipating 90% of the energy of the collision will be much less. If in fact, colliding two ions can dissipate energy? Won't they collide elastically? Unless they of opposite sign? Would they need a third particle to carry off excess energy?

Uspring

Mon Oct 22 2012, 11:28AM

Registered Member #3988 Joined: Thu Jul 07 2011, 03:25PM
Location:
Posts: 711

MarKed One wrote:

I'm trying to calculate the necessary distance for an ion beam travelling at 100-200 KeV with a current of say 60 amps (obviously pulsed) to collide with another beam travelling at the same velocity with the same current density, and dissipate most (90%) of its energy via collision.

Say you have a beam intensity of i (particles/s) an interaction length l, i.e. the length that the beams cross, a cross section s of your beam and v the speed of the particles. Then the time the particle spends in the interaction region is l/v. The number of particles in the interaction region is i*l/v. The density (per area) of the particles in the region is i*l/(v*s).
Assuming e.g. an interaction length l of 10cm and a beam cross section of 1mm^2, you get a particle density of about 6e14/cm^2.
To calculate probability of a collision between a proton of the other beam traversing the interaction region you need the cross section of proton-proton events. At 200keV you'll solely see Rutherford scattering. The cross sections are dependent on the energy and are extremely low. A very conservative much too large estimate could be gotten from the Bohr radius a, i.e. 5e-9cm. That would make the interaction probability to a^2 * density = 1.5e-2.

In short, I don't think it'll work.

Proud Mary

Mon Oct 22 2012, 12:36PM

Registered Member #543 Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992

There are a good number of research papers on collider ion beam modeling, so while I know nothing at all about the subject, I'm perfectly certain that something more than a bit of basic arithmetic is required.

Ash Small

Mon Oct 22 2012, 01:23PM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

Uspring wrote ...

At 200keV you'll solely see Rutherford scattering.

If, as the OP states in the first post, deuterium is used there will be some fusion.

fusion cross section for d+ is here:

If I've interpreted it correctly, the fusion cross section for two D+ ions, each at 200keV (total 400keV) is ~6.5^-2 Barns.

Uspring

Mon Oct 22 2012, 04:13PM

Registered Member #3988 Joined: Thu Jul 07 2011, 03:25PM
Location:
Posts: 711

Proud Mary wrote:

There are a good number of research papers on collider ion beam modeling, so while I know nothing at all about the subject, I'm perfectly certain that something more than a bit of basic arithmetic is required.

I haven't said anything about having 1mm wide 60A ion beam along 10cm length being feasible. That will definitely be a challenge. Aside from this, what is wrong with the calculation?

@Ash Small:
Thank you for the cross section info. The value of 6.5e-26 cm^2 together with the beam density guess I made would result in a fusion probability of 4e-11. So you're right, it will happen, but not so often.

hboy007

Wed Oct 24 2012, 10:14AM

Registered Member #1667 Joined: Sat Aug 30 2008, 09:57PM
Location:
Posts: 374

Interesting topic, unfortunately I can't come up with a suitable expression right now.
Two things came to my mind:
* integrate the coulomb scattering cross section over a small forward solid angle to approximate the fraction of particles that are still "on course" after collisions in a differential length segment, then find a solution to the differential equation that takes both beams into account - they mutually attenuate one another.

* do a simulation with GEANT

this is the no-brainer but the framework is mature enough to produce interesting results for your research.

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