Electron impact ionization cross section

In this case, it's the density of the gas, not the density of the whole system.In summary, the conversation discusses finding the ionization cross section in a mass spectrometer for the gas Argon, with a desired value in the range of 2x10^-16 cm^2. The equation for ionization cross section is provided, but the difficulty lies in finding the gas density, N. Suggestions for finding N using the ideal gas law and Avogadro's constant are discussed, but it is ultimately determined that the number density needs to be the number density of the gas, not the whole system, in order to get the desired cross section value.
  • #1
gothloli
39
0

Homework Statement


I want to find the ionization cross section in a mass spectrometer for the gas Argon.
The value obtained should be in the 2x10-16 cm2 range.

Homework Equations


Q = ionization cross section
I = K(V,B)xNxQxdxIe
Where I = (0.17 x 10-11, K(V,B) =1, d = 0.1 cm, and Ie = 5 x 10-5 Amperes, the trouble is finding N. N is the gas density, but I'm not sure what units it's in.

The Attempt at a Solution


To[/B] find N, you could use the ideal gas law, PV = nRT, The pressure is known as 1.44x10-3
To find N I said that N = P(Avogadro's constant)/ RT
where R = 8.314x106, for it to be in cm3, and T is room temperature in kelvins.
The value I get from this is 9.5x10-19, which is too small. Is there something I'm missing here, I need to get in the 10-16 range.
 
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  • #2
gothloli said:

Homework Statement


I want to find the ionization cross section in a mass spectrometer for the gas Argon.
The value obtained should be in the 2x10-16 cm2 range.

Homework Equations


Q = ionization cross section
I = K(V,B)xNxQxdxIe
Where I = (0.17 x 10-11, K(V,B) =1, d = 0.1 cm, and Ie = 5 x 10-5 Amperes, the trouble is finding N. N is the gas density, but I'm not sure what units it's in.

The Attempt at a Solution


To[/B] find N, you could use the ideal gas law, PV = nRT, The pressure is known as 1.44x10-3
To find N I said that N = P(Avogadro's constant)/ RT
where R = 8.314x106, for it to be in cm3, and T is room temperature in kelvins.
The value I get from this is 9.5x10-19, which is too small. Is there something I'm missing here, I need to get in the 10-16 range.

What are your pressure units? Useful Rule of Thumb: Number density = 3.0 x 10^16/(cc-Torr)
 
  • #3
The pressure is in Pa, in torr it is 1.07 x 10-5, how do you get that value for the number density?
 
  • #4
gothloli said:
The pressure is in Pa, in torr it is 1.07 x 10-5, how do you get that value for the number density?
n = PV/RT


P = 1/760 atm
V = 1cm^3 = 10^-3 l
R = 0.0821 l-atm/K-mol
T = 298 K (25 C)

n = 5.38 x 10^-8 mole/cc-Torr => 3.24 x 10^16 /cc-Torr (I misremembered the 3.0, it should be 3.2)

So, with your P, you should have 3.47 x 10^11/cc
 
  • #5
I calculated, and I get that value for the number density as well. N = P/kT, if you use Boltzmann constant. But with this value, the cross section is 5.71 x 10-18, which is more like the cross section for Ar2+, and not Ar+
 
  • #6
To get the right value, I had to use the partial pressure of the gas, not the whole pressure in the system. So changing the pressure works.
 
  • #7
gothloli said:
To get the right value, I had to use the partial pressure of the gas, not the whole pressure in the system. So changing the pressure works.
Good. The number density needs to be the number density of the thing you are measuring.
 

FAQ: Electron impact ionization cross section

What is electron impact ionization cross section?

Electron impact ionization cross section is a measure of the likelihood that an electron will collide with an atom or molecule and cause it to lose an electron, resulting in ionization.

How is electron impact ionization cross section measured?

Electron impact ionization cross section is typically measured through experiments in which electrons are directed at a sample and the resulting ionization is measured.

What factors affect electron impact ionization cross section?

The electron impact ionization cross section of an atom or molecule is influenced by a variety of factors, including the energy of the incident electron, the ionization potential of the atom or molecule, and the angle of collision.

Why is understanding electron impact ionization cross section important?

Understanding electron impact ionization cross section is crucial for many areas of science, including astrophysics, atmospheric chemistry, and plasma physics. It can also be used in practical applications such as designing particle accelerators.

How can electron impact ionization cross section be calculated?

There are various theoretical models and computational methods that can be used to calculate electron impact ionization cross section. These include the Born approximation, the distorted wave approximation, and the more advanced R-matrix method.

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