Solutions Manual for Physical Principles of
Plasma Physics and Engineering, 3e by Alexander
Fridman, Lawrence Kennedy (All Chapters)
PART 1.
CHAPTER 2.
2.7.1. Electron Energy Distribution Functions. If electron energy distribution function is
Maxwellian and the electron temperature is Te = 1eV, (1) what is the mean velocity <v> of the electrons in
this case?, (2) what is the mean electron energy?, (3) which electron energy m<v> 2/2 corresponds to the
mean velocity?, (4) how can you explain the difference between the two “average electron energies” in
terms of standard deviation?
The mean electron velocity <v> corresponding to the Maxwell velocity distribution function:
f (v) = 4v 2 (m / 2Te ) 3 exp(−mv Te )
can be found after integration as: v = 8Te / m . Numerically at Te = 1eV,
v = 0.7 108 cm / sec . The mean electron energy in this case according to (2.1.2) is = 1.5 eV .
Electron energy corresponding to the mean velocity can be found as: m v = 4Te / , and
numerically at Te = 1eV gives the “average” electron energy about 1.3 eV. The difference between these
two “average” energies is due to the difference between mean square velocity and square of the mean
velocity, which corresponds to the variance or the square of the standard deviation of the velocity
distribution.
2.7.2. Ionization Potentials and Electron Affinities. Why are most ionization potentials (Table
2.1.1) greater than electron affinities (Table 2.1.2)?
1
, Ionization potentials are related to the electron-ion coulomb interaction energies, which are
usually greater than the electron-neutral coulomb interaction energies corresponding to electron affinities. It
should be noted however that the exchange interaction energies can make significant additional
contribution in both cases.
2.7.3. Positive and Negative Ions. Why is it very doable to produce the double- or multi-charged
positive ions such as A++ or A+++ in plasma, and impossible to generate in a practical gas discharge
plasmas the double or multi-charged negative ions such as A-- or A----?
Generation of the multi-charged positive ions such as A++ or A+++ obviously requires energies
greater than conventional ionization potentials (Table 2.1.1) but can be achieved at very high plasma
temperatures. Formation of the multi-charged negative ions such as A-- or A---- requires an electron
attachment to a negative ion. Such processes are suppressed in gas phase because the coulomb repulsion
energies (corresponding to the electron-ion interaction) are usually much stronger than the quantum
mechanical affinity corresponding to the exchange interaction.
2.7.4. Mean Free Path of Electrons. In which gases are the mean free path of electrons with
electron temperature Te = 1eV longer and why? (helium, nitrogen or water vapor at the similar pressure
conditions)?
Noble gases are characterized by the lowest electron-neutral elastic-collision cross-sections (no
significant long-distance polarization interaction). Therefore, at electron temperatures about Te = 1eV when
the elastic collisions dominate, the longest mean free path should be expected in the noble gases (in this
example, in helium) in accordance with Eq. (2.1.3).
2.7.5. Reaction Rate Coefficients. Recalculate the Maxwellian electron energy distribution
function into the electron velocity distribution function f(v) and then find an expression for reaction rate
coefficient Eq. (2.1.7) in the case when = 0 = const.
The Maxwell velocity distribution function is related to the Maxwell energy distribution function
(2.1.1) by the equation: f (v) dv = f ( ) d . Simple derivation results in the velocity distribution:
2
, f (v) = 4v 2 (m / 2Te ) 3 exp(−mv Te )
The reaction rate coefficient (2.1.7) can be found by integration:
k = 0 v f (v)dv = 0 v = 0 8Tm
2.7.6. Elastic Scattering. Charged particle scattering on neutral molecules with permanent
dipole momentum can be characterized by the following cross section dependence on
energy: ( ) = Const / . How does the rate coefficient of the scattering depend on temperature in this
case?
The scattering rate coefficient can be calculated in this case based on the Eq. 2.1.7 as:
const 1 1
k = ( )v f (v)dv = f ( )d = ,
T
which reveals the required temperature dependence.
2.7.7. Direct Ionization by Electron Impact. Using Eq.(2.2.8) for the general ionization
function f(x), find the electron energy corresponding to maximum value of direct ionization cross- section.
Compare this energy with the electron energy optimal for direct ionization according to the Thomson
formula.
Electron energy appears in the Eq. (2.2.6) only as x= , therefore simple differentiation of f(x)
I
in (2.2.8) gives the optimal x and therefore optimal electron energy to compare with x=2 for the Thompson
formula.
2.7.8. Comparison of Direct and Stepwise Ionization. Why does the direct ionization make a
dominant contribution in non-thermal electric discharges while in thermal plasmas stepwise ionization is
more important?
The degree of electronic excitation in thermal plasma is usually much higher than in non-thermal
plasma, which explains the high effectiveness of the stepwise ionization in this case.
3
, 2.7.9. Stepwise Ionization. Estimate a stepwise ionization reaction rate in Ar at electron
temperature 1 eV, assuming quazi-equilibrium between plasma electrons and electronic excitation of
atoms, and using Eq.(2.2.19).
Based on the Eq. (2.2.19), ratio of stepwise and direct ionization rate coefficients can be calculated
as:
k is (Te ) I 7
( ) 2 = (16eV / 1eV ) 3.5 1.6 10 4
k i (Te ) Te
The direct ionization rate coefficient k i (Te ) in Ar at electron temperature 1 eV can be found
−15
from the Eq. (2.2.11) as 2 10 cm3 / sec . Therefore, the stepwise ionization rate coefficient kis (Te ) in
Ar at electron temperature 1 eV can be estimated as 3 10 −11 cm3 / sec .
2.7.10. Electron Beam Propagation in Gases. How does the propagation length of a high
energy electron beam depend on pressure at fixed temperature? How does it depend on temperature at a
fixed pressure?
As seen from (2.2.22) and (2.2.23), the propagation length is inversely proportional to the gas
density ( 1 / n0 ). Taking into account the equation of state ( p = n0T ), the electron beam propagation
length decreases proportionally to pressure at fixed temperature, and increases proportionally to
temperature at constant pressure.
2.7.11. Photoionization. Why can the photoionization effect play the dominant role in
propagation of both non-thermal and thermal discharges in fast flows including supersonic ones? What is
the contribution of photoionization in the propagation of slow discharges?
Propagation of the fast flow discharges, and especially supersonic flow discharges, cannot be
supported by transfer of active and energetic heavy particles as well as temperature (which simplify the
ionization conditions), because the propagation of the heavy particles is limited by velocities much lower
than speed of sound. Contribution of photoionization in such systems can be crucial, because it is not
4
Plasma Physics and Engineering, 3e by Alexander
Fridman, Lawrence Kennedy (All Chapters)
PART 1.
CHAPTER 2.
2.7.1. Electron Energy Distribution Functions. If electron energy distribution function is
Maxwellian and the electron temperature is Te = 1eV, (1) what is the mean velocity <v> of the electrons in
this case?, (2) what is the mean electron energy?, (3) which electron energy m<v> 2/2 corresponds to the
mean velocity?, (4) how can you explain the difference between the two “average electron energies” in
terms of standard deviation?
The mean electron velocity <v> corresponding to the Maxwell velocity distribution function:
f (v) = 4v 2 (m / 2Te ) 3 exp(−mv Te )
can be found after integration as: v = 8Te / m . Numerically at Te = 1eV,
v = 0.7 108 cm / sec . The mean electron energy in this case according to (2.1.2) is = 1.5 eV .
Electron energy corresponding to the mean velocity can be found as: m v = 4Te / , and
numerically at Te = 1eV gives the “average” electron energy about 1.3 eV. The difference between these
two “average” energies is due to the difference between mean square velocity and square of the mean
velocity, which corresponds to the variance or the square of the standard deviation of the velocity
distribution.
2.7.2. Ionization Potentials and Electron Affinities. Why are most ionization potentials (Table
2.1.1) greater than electron affinities (Table 2.1.2)?
1
, Ionization potentials are related to the electron-ion coulomb interaction energies, which are
usually greater than the electron-neutral coulomb interaction energies corresponding to electron affinities. It
should be noted however that the exchange interaction energies can make significant additional
contribution in both cases.
2.7.3. Positive and Negative Ions. Why is it very doable to produce the double- or multi-charged
positive ions such as A++ or A+++ in plasma, and impossible to generate in a practical gas discharge
plasmas the double or multi-charged negative ions such as A-- or A----?
Generation of the multi-charged positive ions such as A++ or A+++ obviously requires energies
greater than conventional ionization potentials (Table 2.1.1) but can be achieved at very high plasma
temperatures. Formation of the multi-charged negative ions such as A-- or A---- requires an electron
attachment to a negative ion. Such processes are suppressed in gas phase because the coulomb repulsion
energies (corresponding to the electron-ion interaction) are usually much stronger than the quantum
mechanical affinity corresponding to the exchange interaction.
2.7.4. Mean Free Path of Electrons. In which gases are the mean free path of electrons with
electron temperature Te = 1eV longer and why? (helium, nitrogen or water vapor at the similar pressure
conditions)?
Noble gases are characterized by the lowest electron-neutral elastic-collision cross-sections (no
significant long-distance polarization interaction). Therefore, at electron temperatures about Te = 1eV when
the elastic collisions dominate, the longest mean free path should be expected in the noble gases (in this
example, in helium) in accordance with Eq. (2.1.3).
2.7.5. Reaction Rate Coefficients. Recalculate the Maxwellian electron energy distribution
function into the electron velocity distribution function f(v) and then find an expression for reaction rate
coefficient Eq. (2.1.7) in the case when = 0 = const.
The Maxwell velocity distribution function is related to the Maxwell energy distribution function
(2.1.1) by the equation: f (v) dv = f ( ) d . Simple derivation results in the velocity distribution:
2
, f (v) = 4v 2 (m / 2Te ) 3 exp(−mv Te )
The reaction rate coefficient (2.1.7) can be found by integration:
k = 0 v f (v)dv = 0 v = 0 8Tm
2.7.6. Elastic Scattering. Charged particle scattering on neutral molecules with permanent
dipole momentum can be characterized by the following cross section dependence on
energy: ( ) = Const / . How does the rate coefficient of the scattering depend on temperature in this
case?
The scattering rate coefficient can be calculated in this case based on the Eq. 2.1.7 as:
const 1 1
k = ( )v f (v)dv = f ( )d = ,
T
which reveals the required temperature dependence.
2.7.7. Direct Ionization by Electron Impact. Using Eq.(2.2.8) for the general ionization
function f(x), find the electron energy corresponding to maximum value of direct ionization cross- section.
Compare this energy with the electron energy optimal for direct ionization according to the Thomson
formula.
Electron energy appears in the Eq. (2.2.6) only as x= , therefore simple differentiation of f(x)
I
in (2.2.8) gives the optimal x and therefore optimal electron energy to compare with x=2 for the Thompson
formula.
2.7.8. Comparison of Direct and Stepwise Ionization. Why does the direct ionization make a
dominant contribution in non-thermal electric discharges while in thermal plasmas stepwise ionization is
more important?
The degree of electronic excitation in thermal plasma is usually much higher than in non-thermal
plasma, which explains the high effectiveness of the stepwise ionization in this case.
3
, 2.7.9. Stepwise Ionization. Estimate a stepwise ionization reaction rate in Ar at electron
temperature 1 eV, assuming quazi-equilibrium between plasma electrons and electronic excitation of
atoms, and using Eq.(2.2.19).
Based on the Eq. (2.2.19), ratio of stepwise and direct ionization rate coefficients can be calculated
as:
k is (Te ) I 7
( ) 2 = (16eV / 1eV ) 3.5 1.6 10 4
k i (Te ) Te
The direct ionization rate coefficient k i (Te ) in Ar at electron temperature 1 eV can be found
−15
from the Eq. (2.2.11) as 2 10 cm3 / sec . Therefore, the stepwise ionization rate coefficient kis (Te ) in
Ar at electron temperature 1 eV can be estimated as 3 10 −11 cm3 / sec .
2.7.10. Electron Beam Propagation in Gases. How does the propagation length of a high
energy electron beam depend on pressure at fixed temperature? How does it depend on temperature at a
fixed pressure?
As seen from (2.2.22) and (2.2.23), the propagation length is inversely proportional to the gas
density ( 1 / n0 ). Taking into account the equation of state ( p = n0T ), the electron beam propagation
length decreases proportionally to pressure at fixed temperature, and increases proportionally to
temperature at constant pressure.
2.7.11. Photoionization. Why can the photoionization effect play the dominant role in
propagation of both non-thermal and thermal discharges in fast flows including supersonic ones? What is
the contribution of photoionization in the propagation of slow discharges?
Propagation of the fast flow discharges, and especially supersonic flow discharges, cannot be
supported by transfer of active and energetic heavy particles as well as temperature (which simplify the
ionization conditions), because the propagation of the heavy particles is limited by velocities much lower
than speed of sound. Contribution of photoionization in such systems can be crucial, because it is not
4
