Mean free path and collision cross section

AI Thread Summary
To calculate the mean free path for an ion with a radius of 9 x 10^-8 cm at various pressures, the relevant equations include λ = 1/Nσ and σ = πd². The user converted the radius to 0.9 nm and applied the equation mean free path = 7×10^-6 / (0.81 × P) for different pressures, but is uncertain about the correctness of this method. It is suggested that calculating moles per unit volume and using Avogadro's number could help determine gas number density, assuming nitrogen as the gas identity. The discussion highlights the need for clarity on the second equation's applicability for mean free path calculations.
Etox
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Homework Statement



I do need to calculate the mean free path for an ion with a radius of 9 x 10‐8 cm in pressures of 10‐8 Torr, 10‐6 Torr, 10‐4 Torr and 10‐2 Torr

Homework Equations



There are different equations I found but the most one is λ= 1/Nσ
N gas number density which I do not know how to calculate based on radius and pressure.
And σ which is collision cross section σ =πd2
Another equation I found = 7×10−6/ r coll2 ×P(torr)

The Attempt at a Solution


I did convert radius to nm (0.9 nm) and applied the second equation
mean free path= 7×10−6/ 0.81 ×P(10-8) and so on for the different pressure
but I am not sure if this attempt is wright

These calculations are to use in the ion free path (ion mobility) in mass spectrometer instruments.
 
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If you can calculate the moles per unit volume, then use Avogadro's number to convert to number density, would that do?
 
the identity of gas is not mentioned but I assume nitrogen.
would the second equation be possible in which needs only the pressure and cross section section to calculate the mean free path.
 
Etox said:
the identity of gas is not mentioned but I assume nitrogen.
would the second equation be possible in which needs only the pressure and cross section section to calculate the mean free path.

I'm afraid that I'm not familiar with the second equation that you gave, so I can't advise (not without doing a bit of research).
 
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