Regarding the cross sections Qen, Qeir they depend on the collision velocity c especially Qej. This is because the e-i coulombic interaction is soft", so a very fast electron can pass nearly undeflected near an ion, whereas a slow one will be strongly deflected. The complete theory yields an expression Q。=2.95×10 (T in Kelvin) here In A=-11.35+ 2In T(K)-=In P(atm) so that In A is usually around 6-12, and can even be taken as a constant(8) rough calculations For the neutral hydrogen atoms, the collisions are fairly hard and one can use the approximation Q=2×10°m Numerical Example Consider Hydrogen at P=l atm. For T >4000K, diatomic H2 is not present anymore (H,->2H). So ionization is from atomic hydrogen, H, for which V=13.6 volts, s=2.42×1021Te (m3) We also find 7.34×1027 6210√T(m/s) σ=2821×10na/v 16.522, Space P pessan Lecture 10 Prof. Manuel martinez Page 5 of 12Regarding the cross sections Qen, Qei, they depend on the collision velocity c , e especially Qei. This is because the e-i coulombic interaction is “soft”, so a very fast electron can pass nearly undeflected near an ion, whereas a slow one will be strongly deflected. The complete theory yields an expression Qei  2.95 ×10-10 ln Λ (m2 ) (17) T (T in Kelvin) where 1 ln Λ  -11.35 + 2ln T(K) - ln P(atm) (18) 2 so that ln Λ is usually around 6-12, and can even be taken as a constant (~8) in rough calculations. For the neutral hydrogen atoms, the collisions are fairly “hard”, and one can use the approximation 2 Q  2 ×10-19m (19) eH Numerical Example Consider Hydrogen at P=1 atm. For T ≥ 4000K , diatomic H2 is not present anymore (H2 → 2H) . So ionization is from atomic hydrogen, H, for which V = 13.6 volts, i q = 1, i q = 2, n so that 3 2 157,800 - e S = 2.42 ×1021 T T (m-3). We also find 7.34 ×1027 (m-3 n= ), T c = e 6210 T (m/s), σ = 2.821×10-8 n ν e e (ν = νei + ν ). e en 16.522, Space Propulsion Lecture 10 Prof. Manuel Martinez-Sanchez Page 5 of 12