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

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SUMMARY

The mean free path (MFP) of a hydrogen atom in interstellar space, calculated using a density of one hydrogen atom per cubic meter and a diameter of 100 pm, results in a value of approximately 2.25 x 1019 meters, or about 2400 light years. This calculation employs the formula λ = 1/(√2 * n * σ), where n is the number density and σ is the effective cross section. However, the discussion highlights that this theoretical value does not account for the ionized nature of the interstellar medium, which introduces additional complexities such as magnetic and electric fields that affect particle interactions.

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gazepdapi1
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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
 
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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.
 

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