Mean free path and effective collision radius

[bon]

Homework Statement

Beam of silver atoms are passing through air at a temp 273K and a pressure of 1 Pa. The beam is attenuated by a factor of 2.72 in a distance of 10^-2 m. Find the mean free path of the silver atoms and estimate the effective collision radius..

Homework Equations

The Attempt at a Solution

Ok so taking the probability of the collision between x and x+dx to be 1/a e ^-1/a x where 1/a is the mean free path, i worked it out to be 0.01m...but this seems huge? Why should it be about 1 cm?

Also, to work out the effective collision radius, do i just use pi r^2 = collision cross section (usually denoted by sigma), where r is the effective collision radius?... I get this to be about 2.91 x 10^-10 m...but is this right?

 

[zhermes]

All of the above looks right. Remember air isn't very dense, so silver atoms can have a pretty large mean free path.