Saha Ionization Formula | Physics Optional Notes for UPSC PDF Download

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integers.  The law of conservation of mass and the laws of definite proportions and multiple 
proportions led to the formulation of atomic theory of matter.  
For equilibrium at constant temperature and volume, and for given quantities of the various 
reactants, the condition of chemical equilibrium reduces to  
                    ? ?? ?? ?? ?? = 0
??                                                                             (13.2) 
where ?? ?? is the chemical potential of the ?? th reactant.  Chemical potential is a thermodynamical 
variable; it is basically Gibbs free energy per molecule.  It turns out that Equation (13.2) is an 
essential condition for chemical equilibrium.  If we use the relation stated in Module 12, ?? =
?? ????
 , 
http://www.physicallensonthecell.org/chemical-potentialthis relation reduces further to 
                    ? ?? ?? ?? ?? = 0
?? .                                                                             (13.3) 
Suppose now, following Saha, that the process of ionization of an ?? -times ionized can be 
considered a chemical reaction, written symbolically as 
                    ?? ?? = ?? ?? +1
+ ?? .                                                                      (13.4) 
Then the application of Equation (13.3) gives us 
                    ?? ?? = ?? ?? +1
+ ?? ?? .                                                                     (13.5) 
If the particles are nonrelativistic, then the energies of the particles are given by 
                   ?? ?? =
?? 2
2?? ?? , 
                  ?? ?? ,?? = 
?? 2
2?? ?? + ?? ?? ,?? ,                                                                   (13.6) 
              ?? ?? +1,?? =
?? 2
2?? ?? +1
+ ?? ?? +1,?? + ?? ?? . 
Here ?? ?? is the ionization potential (from the ground state) of the ?? -times ionized atom, ?’s are the 
excitation potentials and the masses obey the relation  
                  ?? ?? = ?? ?? +1
+ ?? ?? . 
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integers.  The law of conservation of mass and the laws of definite proportions and multiple 
proportions led to the formulation of atomic theory of matter.  
For equilibrium at constant temperature and volume, and for given quantities of the various 
reactants, the condition of chemical equilibrium reduces to  
                    ? ?? ?? ?? ?? = 0
??                                                                             (13.2) 
where ?? ?? is the chemical potential of the ?? th reactant.  Chemical potential is a thermodynamical 
variable; it is basically Gibbs free energy per molecule.  It turns out that Equation (13.2) is an 
essential condition for chemical equilibrium.  If we use the relation stated in Module 12, ?? =
?? ????
 , 
http://www.physicallensonthecell.org/chemical-potentialthis relation reduces further to 
                    ? ?? ?? ?? ?? = 0
?? .                                                                             (13.3) 
Suppose now, following Saha, that the process of ionization of an ?? -times ionized can be 
considered a chemical reaction, written symbolically as 
                    ?? ?? = ?? ?? +1
+ ?? .                                                                      (13.4) 
Then the application of Equation (13.3) gives us 
                    ?? ?? = ?? ?? +1
+ ?? ?? .                                                                     (13.5) 
If the particles are nonrelativistic, then the energies of the particles are given by 
                   ?? ?? =
?? 2
2?? ?? , 
                  ?? ?? ,?? = 
?? 2
2?? ?? + ?? ?? ,?? ,                                                                   (13.6) 
              ?? ?? +1,?? =
?? 2
2?? ?? +1
+ ?? ?? +1,?? + ?? ?? . 
Here ?? ?? is the ionization potential (from the ground state) of the ?? -times ionized atom, ?’s are the 
excitation potentials and the masses obey the relation  
                  ?? ?? = ?? ?? +1
+ ?? ?? . 
 
 
Notice that we have assumed the same momentum for all the particles.  This does not matter 
because ultimately we are going to integrate over all momenta.  This is because we are interested 
in the relative numbers of the three types of particles of all kinetic energies.  Using Equations 
(13.6) in the expression for the particles per unit volume in the momentum range ?? and ?? + ???? 
(refer to Module 12), that is, 
                   ???? (?? ) = ?? .
4?? ?? 2
????
h
3
?? ?? ?? -?? 2
2?????? /
                                          (13.7) 
and integrating over all momenta, we get for ?? -times and (?? + 1)-times ionized particles per unit 
volume: 
                    ?? ?? = ?? ?? (?? )?? ?? ?? (
2?? ?? ?? ????
h
2
)
3/2
                                                 (13.8) 
                ?? ?? +1
= ?? ?? +1
(?? )?? ?? ?? +1
(
2?? ?? ?? +1
????
h
2
)
3/2
?? -?? ?? /????
.                           (13.9) 
 Dividing Equation (13.9) by (13.8) and multiplying by ?? ?? , we get 
                 
?? ?? +1
?? ?? ?? ?? = ?? ?? (?? -?? ?? )
?? ?? +1
(?? )
?? ?? (?? )
(
?? ?? +1
?? ?? )
3/2
?? -?? ?? /????
                                (13.10) 
after we have used Equation (13.5).  In these equations ?? ?? (?? ) is the partition function given by 
                     ?? ?? (?? ) = ? ?? ????
?? -?? ????
/????
?? ,                                                (13.11) 
where ?? ’s are statistical weights and ?? ’s are the excitation potentials.    
Equation (13.10) is the Saha’s ionization formula.  The use of partition function shows that 
summation has been taken over all the excitation levels of the ?? -times and (?? + 1)-times ionized 
atoms.  Saha’s formula in the form derived above holds even when the electrons are degenerate.  
In case the electrons are non-degenerate, their number per unit volume is given by, following 
Equations (13.8) and (13.9), 
                       ?? ?? = 2(?? ?? ?? )(2?? ?? ?? ???? )
3/2
                                             (13.12) 
since the statistical weight of an electron is 2?? + 1 = 2 with ?? = ½.  Substituting the above 
equation in right hand side of Equation (13.10), we get 
                   
?? ?? +1
?? ?? ?? ?? =
2?? ?? +1
(?? )
?? ?? (?? )
(
2???????? h
2
)
3/2
?? -?? ?? /????
.                                   (13.13) 
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integers.  The law of conservation of mass and the laws of definite proportions and multiple 
proportions led to the formulation of atomic theory of matter.  
For equilibrium at constant temperature and volume, and for given quantities of the various 
reactants, the condition of chemical equilibrium reduces to  
                    ? ?? ?? ?? ?? = 0
??                                                                             (13.2) 
where ?? ?? is the chemical potential of the ?? th reactant.  Chemical potential is a thermodynamical 
variable; it is basically Gibbs free energy per molecule.  It turns out that Equation (13.2) is an 
essential condition for chemical equilibrium.  If we use the relation stated in Module 12, ?? =
?? ????
 , 
http://www.physicallensonthecell.org/chemical-potentialthis relation reduces further to 
                    ? ?? ?? ?? ?? = 0
?? .                                                                             (13.3) 
Suppose now, following Saha, that the process of ionization of an ?? -times ionized can be 
considered a chemical reaction, written symbolically as 
                    ?? ?? = ?? ?? +1
+ ?? .                                                                      (13.4) 
Then the application of Equation (13.3) gives us 
                    ?? ?? = ?? ?? +1
+ ?? ?? .                                                                     (13.5) 
If the particles are nonrelativistic, then the energies of the particles are given by 
                   ?? ?? =
?? 2
2?? ?? , 
                  ?? ?? ,?? = 
?? 2
2?? ?? + ?? ?? ,?? ,                                                                   (13.6) 
              ?? ?? +1,?? =
?? 2
2?? ?? +1
+ ?? ?? +1,?? + ?? ?? . 
Here ?? ?? is the ionization potential (from the ground state) of the ?? -times ionized atom, ?’s are the 
excitation potentials and the masses obey the relation  
                  ?? ?? = ?? ?? +1
+ ?? ?? . 
 
 
Notice that we have assumed the same momentum for all the particles.  This does not matter 
because ultimately we are going to integrate over all momenta.  This is because we are interested 
in the relative numbers of the three types of particles of all kinetic energies.  Using Equations 
(13.6) in the expression for the particles per unit volume in the momentum range ?? and ?? + ???? 
(refer to Module 12), that is, 
                   ???? (?? ) = ?? .
4?? ?? 2
????
h
3
?? ?? ?? -?? 2
2?????? /
                                          (13.7) 
and integrating over all momenta, we get for ?? -times and (?? + 1)-times ionized particles per unit 
volume: 
                    ?? ?? = ?? ?? (?? )?? ?? ?? (
2?? ?? ?? ????
h
2
)
3/2
                                                 (13.8) 
                ?? ?? +1
= ?? ?? +1
(?? )?? ?? ?? +1
(
2?? ?? ?? +1
????
h
2
)
3/2
?? -?? ?? /????
.                           (13.9) 
 Dividing Equation (13.9) by (13.8) and multiplying by ?? ?? , we get 
                 
?? ?? +1
?? ?? ?? ?? = ?? ?? (?? -?? ?? )
?? ?? +1
(?? )
?? ?? (?? )
(
?? ?? +1
?? ?? )
3/2
?? -?? ?? /????
                                (13.10) 
after we have used Equation (13.5).  In these equations ?? ?? (?? ) is the partition function given by 
                     ?? ?? (?? ) = ? ?? ????
?? -?? ????
/????
?? ,                                                (13.11) 
where ?? ’s are statistical weights and ?? ’s are the excitation potentials.    
Equation (13.10) is the Saha’s ionization formula.  The use of partition function shows that 
summation has been taken over all the excitation levels of the ?? -times and (?? + 1)-times ionized 
atoms.  Saha’s formula in the form derived above holds even when the electrons are degenerate.  
In case the electrons are non-degenerate, their number per unit volume is given by, following 
Equations (13.8) and (13.9), 
                       ?? ?? = 2(?? ?? ?? )(2?? ?? ?? ???? )
3/2
                                             (13.12) 
since the statistical weight of an electron is 2?? + 1 = 2 with ?? = ½.  Substituting the above 
equation in right hand side of Equation (13.10), we get 
                   
?? ?? +1
?? ?? ?? ?? =
2?? ?? +1
(?? )
?? ?? (?? )
(
2???????? h
2
)
3/2
?? -?? ?? /????
.                                   (13.13) 
 
 
Here ?? is the reduced mass of the (?? + 1)-times ionized atom.  It equals ?? ?? to a very good 
approximation.  It is usual in astrophysics to express ?? ?? in terms of the electron pressure ?? ?? =
?? ?? ???? .  We then have  
                 
?? ?? +1
?? ?? ?? ?? =
2?? ?? +1
(?? )
?? ?? (?? )
(
2????
h
2
)
3/2
 (???? )
5/2
?? -?? ?? /????
.                            (13.14) 
This is the form in which Saha’s ionization equation is better known.   In the form more useful to 
us, we write this equation as  
                  log(
?? ?? +1
?? ?? ?? ?? ) = log (
2?? ?? +1
(?? )
?? ?? ) + 2.5 log ?? -
5040
?? ?? ?? - 0.48,  (13.15) 
where the ionization potential is expresses in eV and electron pressure is measured in dyne/cm
2
. 
Using this equation, we can get the fraction of atoms ionized at various temperatures and at a 
given electron pressure.  The electron pressure may vary from a few dyne/cm
2
 in the 
atmospheres of cool stars of classes K and M to a few hundred dyne/cm
2 
in the atmospheres of 
hot stars of classes A and B.                        
The primary influence on the degree of ionization is, however, the temperature.  The pressure, 
within the values mentioned above, only slightly modifies the degree of ionization.  But, as we 
shall see a little later, even this slight modification plays an important role in classifying stars on 
the basis of their luminosity.