# Mean free path in 2d?

Let us say we have a collection of $n$ stationary balls per square meter each of diameter $d$ and we roll an identical ball between them. Then what is the free mean path of this ball?

My calculations would indicate that it is: $$\lambda=\frac{1}{2dn}$$ However in the book I am working through (namely Thermodynamics, Kinetic theory and statistical thermodynamics, 3rd ed by Sears and Salinger problem 10-7*) they seem to have used from there numerical answers: $$\lambda =\frac{1}{dn}$$ Is my form correct, or the form I think they have used and why?

*Note the above question in block quotes aboveis not problem 10-7 but is a generalisation of it.

Edit

Here is how I get my result. Since all balls are radius $d/2$ there centres need to be within a distance of $d$ if they are going to collide. This means that if any stationary ball is within the area $2dL$ of the moving ball they will collide. Where $L$ is the distance travelled by and we have $2d$ since the centre of the stationary ball can be a distance $d$ either side of the moving for it to collide. When there is on average one ball in this area the area is given by: $$2dL=\frac{1}{n}$$ Meaning that our mean free path is: $$\lambda=\frac{1}{2dn}$$

• can you give a more detailed analysis of your calculation? Jan 13 '16 at 9:30
• @BruceLee I have edited the question to include this detail Jan 13 '16 at 10:17
• I agree with your result, and so does this answer. I'm now skeptical of your textbook's answer - they do sometimes get it wrong (rarely, but sometimes). Jan 13 '16 at 12:40