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Electrical Breakdown/Avalanche of Air Classical Mechanics Collision Theory

  • #1

Homework Statement



Electrical breakdown (e.g, lightning) is caused by an avalanche process. If a free electron gains enough energy between collisions to ionize the neutral molecules when they collide with them, then those two electrons will gain enough energy between collisions to ionize the next neutral molecules they collide with and the ionization of the gas grows exponentially. Conversely, if the electron doesn’t gain enough energy to ionize the molecule, it will lose energy in collisions with neutral molecules and will eventually recombine.

(a) Estimate how large the electric field must be for breakdown to occur. The electron needs to gain ∼ 200 eV of energy between collisions and the maximum cross section for ionization of air molecules is about 3 × 10−16 cm2.

(b) If the electric field is large enough for breakdown to occur, estimate how fast the current channel of free electrons forms.

Homework Equations



V = E d

d = mean free path

The Attempt at a Solution



V = 200 V since the electron gains 200 eV.
d is calculated by [tex] \frac{V}{N_{A} \sigma} [/tex], where [tex]N_{A}[/tex] is Avogadro's number and in STP with V = 22.4 x 10-3 m.3.

Don't know if this exactly right. I get d = 1240.3 nm and an electric field for breakdown to occur of [tex] 161.25 \times 10^{6} \frac{V}{m} [/tex] while I read that it should be around [tex] 3 \times 10^{6} \frac{V}{m}[/tex].

Also, for part (b) I don't even know what they are asking for.

Any help before Tuesday, January 18 at 1:00 Pacific Time (GMT -0800) would be greatly appreciated.
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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hi alimerzairan! :smile:

(have a theta: θ and a sigma: σ and try using the X2 icon just above the Reply box :wink:)

not really my field :redface:, but anyway …

you don't seem to have used the cross-section …

it occurs to me that if the electron misses that cross-section, it'll be able to travel further and pick up more energy before its next collision, so the average d will be greater, and E = V/d will be smaller …

does that work? :smile:

(btw, did you mean 1:00 am? that's an unusual deadline :confused:)
 
  • #3
Hi Tiny-Tim,

I do use the cross section for the mean free path, d.

I guess you are not in the United States since we are the only country that still uses a.m. and p.m. when the rest of the world works under 24-hour clock. Don't know why this country doesn't change, but what I mean by 1:00 is 13:00 Pacific time GMT (-0800).

Does anyone understand the second part of the problem. I am primarily stuck on that.
 

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