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Topic: Frequency of collusion  (Read 4472 times)

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Offline Sophia7X

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Frequency of collusion
« on: December 27, 2012, 06:44:16 PM »
Typical fluorescent lamp: 120 cm in length and 3.6 cm in diameter, Ar pressure of 3 Torr.
The electric field between the tube cathode and the anode continuously transfers energy to the electrons. The electrons redistribute the energy among themselves, quickly reaching the temperature on the order of 11,000 K. Similarly, neutral Ar atoms also quickly equilibrated thermally among themselves. However, because of a very large mass mismatch, collisions between electrons and Ar atoms are extremely inefficient in transferring the electrons' energy to Ar atoms. Hence, the argon temperature in a is much lower than the electrons' temperature.

What is the total frequency of electron-Ar collisions in the tube having volume of 4.9e-3 m^3 and concentration of free electrons = 5.0e17 m^-3, the mean collision time of an electron with Ar atoms is 7.9e-10 s?

I did [(4.9e-3)(5.0e17)]/(7.9e-10) = 3.1e24 s^-1 but the answer is 7.9e27.

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Offline Borek

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Re: Frequency of collusion
« Reply #1 on: December 27, 2012, 07:07:09 PM »
I wonder if you should not use data given to calculate average distance between Ar atoms and time it takes for an electron to travel between these atoms.
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Offline Sophia7X

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Re: Frequency of collusion
« Reply #2 on: December 27, 2012, 07:21:41 PM »
Solutions didn't make any mention of that. However, it said to calculate surface area first. But it doesn't show how the surface area was used in calculation.

(Surface area = 0.27 m^2).
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Offline curiouscat

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Re: Frequency of collusion
« Reply #3 on: December 27, 2012, 09:47:13 PM »
I did [(4.9e-3)(5.0e17)]/(7.9e-10) = 3.1e24 s^-1 but the answer is 7.9e27.

I got the same answer.

Offline curiouscat

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Re: Frequency of collusion
« Reply #4 on: December 27, 2012, 09:48:55 PM »
I wonder if you should not use data given to calculate average distance between Ar atoms and time it takes for an electron to travel between these atoms.

That would make perfect sense except isn't that exactly the same as the " mean collision time of an electron with Ar atoms" already given in the Question?

Offline Borek

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Re: Frequency of collusion
« Reply #5 on: December 28, 2012, 03:59:17 AM »
I wonder if you should not use data given to calculate average distance between Ar atoms and time it takes for an electron to travel between these atoms.

That would make perfect sense except isn't that exactly the same as the " mean collision time of an electron with Ar atoms" already given in the Question?

My thinking is "collision time" and "time between collisions" are two separate things.

I can be easily wrong.
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Offline curiouscat

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Re: Frequency of collusion
« Reply #6 on: December 28, 2012, 05:58:32 AM »

My thinking is "collision time" and "time between collisions" are two separate things.


Ah! Got what you mean. Could be right, especially since the other approach is failing in getting the expected answer.

Offline curiouscat

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Re: Frequency of collusion
« Reply #7 on: December 28, 2012, 06:01:55 AM »
Solutions didn't make any mention of that. However, it said to calculate surface area first. But it doesn't show how the surface area was used in calculation.

(Surface area = 0.27 m^2).

With the given tube dimensions I get neither the area nor volume of tube mentioned. What gives?

Offline Borek

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Re: Frequency of collusion
« Reply #8 on: December 28, 2012, 07:05:53 AM »
IMHO volume and area are off by the difference between diameter and radius.
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Offline Sophia7X

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Re: Frequency of collusion
« Reply #9 on: December 28, 2012, 01:23:16 PM »
Hmm, I think it's an error in the answer key. It says 2πrh = 0.27 m^2 but it
should be half that value since it stated that diameter = 3.6 cm.

That means their given volume should actually be 1.22e-3 m^3, which gives 7.7e23 s^-1 if you do (concentration of electrons*volume)/(mean collision time).

Perhaps the answer is wrong because the answer key explicitly states frequency = (1/t)(neV), t being the mean collision time and ne being concentration of free electrons.

Also, if you took the time between collisions into account, wouldnt that give a lower frequency? The solution is apparently 7.9e27, which is several orders higher than what I thought.

This is the iChO 2012 problem set, I'm wondering why there are so many errors.
In another problem, where you had to draw some structure but the answer key showed carbon having 2 double bonds and a single....
Entropy happens.

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