# Mean Free Path/Atomic Diameter

1. Jun 25, 2010

### NeverGiveUp

1. The problem statement, all variables and given/known data
The pressure inside a tank of neon is 150 . The temperature is 25C.
how many atomic diameters does a neon atom move between collisions?

2. Relevant equations

I used the number density N/V=p/KbT and the mean free path equation, where pressure is in pasquals, and tried both that the atomic radius is 0.5 x 10^-10 m and then searched and found that the atomic radius of neon is 3.8 x 10-11 m and am wondering if this is my problem...not understanding the radius/diameter of neon.

3. The attempt at a solution oops see above, the answer is 61 in the back of the book, but I have not gotten that answer at all. Any help would be super appreciated!!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jun 27, 2010

### Redbelly98

Staff Emeritus
Welcome to Physics Forums.

Could you post your calculations, even if they don't give the same answer as in the back of the book? Also, what units go with the 150 pressure value given in the problem statement?

3. Jun 28, 2010

### NeverGiveUp

Sure, sorry about that. It's 150 atm, so 1519000 pascals, and then 298 K.

I first determined N/V=P/KbT.....the number density is ....3.69x10^27 m-3.

Then using the mean free path equation 1/(4* sqrt2*pi*3.69x10^27m-3*r^2) where r is the radius (from Knight's engineering physics book, but saw on the net that it often in diameter.....I'm new at some of this physics so a bit naive about it) where in the book, monatomic gases are 0.5 x 10^-10 m and so I wonder if this is where I'm messing up ...I did try a value of neon's radius of 3.8 x 10^-11 but still didn't come up with the final correct answer???

I get the mean free path to be with knight's radius value as 6.0 x 10^9 m....then divided that by the diameter (2r) to get (okay this is embarrassing because I'm reworking this in my calculator as I type this up and guess what....) I get 60.9 or 61. This is the correct answer. I seriously worked an hour on this simple intro problem and could not get that answer.

Well, thank you for making me redo this!

4. Jun 28, 2010

### Redbelly98

Staff Emeritus
Cool, glad it worked out.