# Compute the main free path of a hydrogen atom in interstellar space

Problem:
Interstellar space has an average temperature of about 10K and an average density of hydrogen atoms of about one hydrogen atom per cubic meter. Compute the main free path of a hydrogen atom in interstellar space. Take the diameter of an H atom to be 100 pm.

Here is what I did. I first calculated the density.

Density =(pressure)(Na)/RT=(1x10^-11)*(6.022E22)/(8.314*10) = 7.243 E10 m^-3
(I used 1x10^-11 for the pressure in space)

The I used this to find the MFP.

lamda = 1/(root(2)*density*pi*d^2) = 1/(root(2)*7.243E10*pi*(100x10^-12)^2)

So for the mean free path I got .310 nm.
Can someone verify if this is correct.

thanks
nertil

## Answers and Replies

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Your numbers don't seem reasonable. The density of heavy metals is on the order of 1 E4 kg/m^3 and for nearly empty space to have 1 million times that density doesn't seem right. And if there's only 1 atom per cubic meter, why would a hydrogen atom only travel 1 billionth of that before hitting another atom? It seems to me the density is already given. You have the volume taken up by one H atom, one cubic meter, and if you look up its mass, well divide the two and there's your density. A very small number, as it should be.

I think he used number density, not mass density. Then the value is explicitly told:
density=1/m^3. You don't need to calculate it from a (wrong) assumption for pressure.

OK, I haven't seen it done that way. So is a fraction of a nanometer a reasonable number, then?

Mean free path in space should be enormous, not nanometres.

Check out wikipedia: mean free path for maxwellian particles (low T) is given by (sqrt(2) n sigma)^(-1), where n is the number density, and sigma is the effective cross section. The cross section is just given by ~ pi r^2 (classical approximation), and the number density of 1/m^3.

Using n=1/m^3, and r=100e-12, I get the mean free path as 2.25 x 10^(19)m, which is about 2400 light years.

This is clearly not true. Why? The interstellar medium is ionised, so magnetic and electric fields play a role.

Yeah I get the same result, 2.25 x 10^(19) m. So this isn't the right answer then?

It is for this problem, as far as I can see. But it's not true in reality, for the reasons I mentioned.