The Electron Velocity Needed To Ionize Oxygen Molecule

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SUMMARY

The discussion centers on the electron velocity required to ionize oxygen molecules, specifically noting that the ionization energy for O2 is 13.6 eV. Participants highlight that this energy can be achieved by accelerating electrons through a potential difference of 13.6 volts. To convert this energy into velocity, one must apply the kinetic energy formula (1/2 mv²) using the mass of an electron (9.1 x 10^-31 kg). Additionally, the conversation touches on designing a circular particle accelerator for air molecules and the complexities involved in calculating the resulting current and magnetic field.

PREREQUISITES
  • Understanding of ionization energy and its measurement in electron volts (eV).
  • Familiarity with kinetic energy calculations and the formula 1/2 mv².
  • Knowledge of particle accelerator design principles, specifically cyclotrons.
  • Basic concepts of electromagnetic fields and their relationship to charged particle movement.
NEXT STEPS
  • Research the NIST database for ionization energies of various gases.
  • Learn about the design and operation of cyclotrons and other particle accelerators.
  • Study the principles of elastic collision theory in particle physics.
  • Explore the relationship between current, magnetic fields, and particle motion in accelerators.
USEFUL FOR

Physicists, engineers, and researchers involved in particle physics, accelerator design, and ionization processes will benefit from this discussion.

Mohamed Nedal
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Hi ,

If I used electrons to ionize Oxygen gas by accelerating the electrons and then collide with Oxygen molecules , What is the electron velocity needed to do this ionization ?
 
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Hi,
Instead of saying it as velocity/speed, use energetic electrons... For oxygen you can find from NIST data base..Hope it helps,
Cheers, Rajini.
 
ya the site is so cool , but I didn't understood well how to get what I need ...

I have some of Oxygen gas and some electrons , I want to calculate the energy that I should give to electrons to be able to ionize the Oxygen gas .

Can you tell me a simple method to do this ?

Have a nice day ...
 
Hi,
have look at this NIST site
http://physics.nist.gov/PhysRefData/IonEnergy/tblNew.html
There you can find for O the ionization energy is 13.6 eV.
I really have no clue of calculation of these energies..
May be some information - http://dl.clackamas.cc.or.us/ch104-07/ionization_energy.htm
-Rajini.
 
Last edited by a moderator:
Do you really need to calculate the velocity for some reason?
(Because we don't usually bother with it, it's more convenient and practical to use the energy figure directly)

O2 needs to be hit by an electron with energy 13.6 eV to ionise.

That's just the energy of an electron that's been accelerated from zero through a potential difference of 13.6 volts. (not much is it?). To actually do that, you just use a vacuum tube with 13.6 volts on the anode - easy.

If you want to convert that to velocity, you need to multiply by e which is 1.6 x 10-19. That gives you the kinetic energy in Joules.
Then use the mass of an electron (9.1 x 10-31) to get the velocity.
(use 1/2 mv2)
 
ya i want to use the elastic collision theory to know the velocity or energy of Oxygen molecules after ionization with high-speed electrons ...

With knowing the ionization energy of one Oxygen molecule that equal 13.6 eV , So if I have ( n ) molecules that I want to ionize them , Should I multiply 13.6 eV with ( n ) to calculate the needed energy to ionize ( n ) molecules ?
 
Each electron will ionize one molecule (or maybe a few if it's very energetic).
Why would you need the total energy, if you want to know the speed of one oxygen molecule after collision? Or this is not what you want to find out?
 
Mohamed Nedal said:
ya i want to use the elastic collision theory to know the velocity or energy of Oxygen molecules after ionization with high-speed electrons ...

With knowing the ionization energy of one Oxygen molecule that equal 13.6 eV , So if I have ( n ) molecules that I want to ionize them , Should I multiply 13.6 eV with ( n ) to calculate the needed energy to ionize ( n ) molecules ?

Hitting oxygen atoms (or molecules) with electrons will not transfer any appreciable momentum to the oxygen. Oxygen is more than 50,000 times more massive than an electron.
It would be like an ant trying to move an elephant.

Ionisation is simply the removal of ONE electron from the oxygen molecule by collision and requires very little energy.
Somewhat similar to the ant knocking off one of the elephant's fleas.
 
Last edited:
Wonderful analogy :)

I want to design a circular particle accelerator for air molecules , So with knowing the required input energy to make ionization and the No. of particles inside the chamber , I can get the ion density or its No. inside this volume ...

And then accelerate them with electromagnetic fields , So they became as a trail of charged particles rotates and make a Current and then generate Magnetic Field ...

I want to calculate the value of this current and the magnetic field generated from the movement of these particles ... Can you help me ?
 
  • #10
Mohamed Nedal said:
Wonderful analogy :)

I want to design a circular particle accelerator for air molecules , So with knowing the required input energy to make ionization and the No. of particles inside the chamber , I can get the ion density or its No. inside this volume ...

And then accelerate them with electromagnetic fields , So they became as a trail of charged particles rotates and make a Current and then generate Magnetic Field ...

I want to calculate the value of this current and the magnetic field generated from the movement of these particles ... Can you help me ?

You are describing a cyclotron. http://en.wikipedia.org/wiki/Cyclotron

The current in the circulating beam is very weak. The resulting magnetic field is very much weaker than the fields that you need to make the ions move in the first place. A typical beam current in a small accelerator would be a few milli-amps at most - usually far less.
- typically a million times less.
 
  • #11
ya I know , I have another two questions please ...

1 - Is there some formula to get this weak current and magnetic field ?


2 - The particle's cyclotron resonance frequency which is : f = q*B/2*Pi*m

Is it the the frequency of the supplied current also ?
 
  • #12
Mohamed Nedal said:
ya I know , I have another two questions please ...

1 - Is there some formula to get this weak current and magnetic field ?


2 - The particle's cyclotron resonance frequency which is : f = q*B/2*Pi*m

Is it the the frequency of the supplied current also ?

1/ No, not really. It's a very complicated machine with lots of wasted energy. Most of the beam is lost because it's very difficult to focus and control.
I used to operate one when I was working at a University. My job was to sit and adjust the controls to keep the beam steady - it was very difficult. If I lost control, the beam could escape the right path and destroy parts of the machine.

2/ It's the frequency of the power supply to the main electrodes, yes.
 
  • #13
Yes its very hard thing , but I'm sure that there is some formula to get these values but I cannot find it ...

So if we want to calculate these values Theoretically , How to make this ?
 
  • #14
You can't. The beam current simply depends on how much current you 'inject' from the ion source less the amount you lose to inefficiencies in the ion source. You might get a microamp of beam out of a source that's taking a hundred amps of anode current.

It's just the same as if you have a circular wire with a current flowing to produce a magnetic field. How much current depends on how much you put in - nothing else.
 
  • #15
Unfortunately , It is ideal inefficient process ... I'll try to find some solution .

Thank you
 

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