where a,=nfm(ck o,(chmc2dc a 8 kTe nm=m ncdc The ionization cross-section o (c)in zero below c Using a Maxwellian form for fm(c), we find easily Ge=euo, udu and the energy spent by secondaries in ionization(p u. time and volume) is then n,nmCeo, (u++em) Similarly, the energy spent in excitation is nnnmCedexcUexc The energy balance is therefore(dividing by n throughout) PuPa. (Wo-Em)=npuplo, (vD Nu++6m)+Gexc(VD Vexd +nmcb. (u++Em)+excUexc) This can be solved for - p U++e+Exc 0. exc -9)-+- which is a function of Te for a fixed Vo. Hence n is also a function of Te. This is because, gi excOM(1- Lecture 16 Prof. manuel martinez- Sanchez Page 5 of 8∞ 2 where σ c c σ ( )4π dc c + = 1 fm ( ) + c c n e ∫ m o ∞ 8 kTe 2 and ce = , nm = ∫ fm 4π dc c π me o The ionization cross-section σ t (c) in zero below c + = 2eU + . Using a Maxwellian me form for fm ( ) c , we find easily ∞ σ t = ∫ e−u u σ + ( ) ⎛ E ⎞ du u ⎜ ⎜u = ⎟ ⎝ kTe ⎟ ⎠ +u and the energy spent by secondaries in ionization (p.u. time and volume) is then c n n eσ + (U + + ε m ) n m Similarly, the energy spent in excitation is n e m c n n σ excUexc The energy balance is therefore (dividing by nn throughout) υ σ (V υ σ (V )(U + + ε m ) + σ ( D )U V exc ] + c n e [σ (U + + ε m ) + σ excUexc n ] p p + D − ε m ) = n p p [ + D exc m + np This can be solved for : nm U + + ε m + Uexc σ exc np = ce σ + nm (V ) − σ (V ) σ exc (V ) υ p D + D D (V − ε m ) σ T K (U + + ε m D ) − σ + σ + σ + * which is a function of Te for a fixed VD. Hence ε p is also a function of Te. This is because, given ε* p ε p = 1 − exp[− m C  (1 − η )] o u 16.522, Space Propulsion Lecture 16 Prof. Manuel Martinez-Sanchez Page 5 of 8