Download Boltmann/Saha Equation Examples

Boltmann/Saha Equation Examples
March 4, 2011
Boltzmann Equation: Describes how energy levels within atoms are populated....look
at one individual atom, electron may be jumping around but consider atoms as a statistical
ensemble (all at once) and they follow the following:
The Boltzmann equation gives ratios of level populations as a function of temperature:
gj
Nj
= e−(Ej −Ei )/kT
Ni
gi
g1/g2 are statistical weights that take into account degeneracy of energy states - more
than one state having the same energy...classically NO but works in quantum mechanics!!
(j, i = 2, 1; Ej-Ei = ∆E)
EXAMPLE:
A) For a gas of neutral hydrogen (gn =2n2 ), at what temperature is the number of atoms in
the first excited state only 1% of the number of atoms in the ground state?
B) At what temperature is the number of atoms in the first excited state equal to 10%
of the number in the ground state?
C) For a gas of neutral hydrogen, at what temperature will equal numbers of atoms have
electrons in the ground state (n=1) and the first excited state (n=2)? What is the energy
required to excite the electron from the ground state to n=2?
D) As T approaches ∞, what is the predicted distribution of electrons in each orbital according to the Boltzmann equation? Will this be the distribution that actually occurs? Why
or why not?
————
SAHA EQUATION
Similar to Boltmann eq, but describes ionization states.....
2Qr+1
ne N (Xr+1 )
=
N (Xr )
Qr
1
2πmr kT
h2
3/2
e−χi /kT ,
Q’s are partitiion functions (stat mech) accounting for both probabliity and degeneracy
of states
χi is Ionization potential (energy required to ionize from ground state)
where ne is the electron number density,N (Xn ) are the numbers of atoms in ionization state r
and r+1, mr is the reduced mass of the electron. Since mr is approximately me , by convention
and for ease of use we can replace mr with me . Also, when considering stellar atmospheres,
it is often convenient to use Pe , the electron pressure, in place of ne . Pe and ne are related
by the ideal gas law:
Pe = ne kT
Consequently, the Saha equation takes the following alternative form:
2kT gr+1
N (Xr+1 )
=
N (Xr )
Pe g r
2πmr kT
h2
3/2
e−χi /kT
From the study of stellar atmospheres, we know that Pe ranges from 0.1 N/m2 for cooler
stars to 100 N/m2 for hotter stars.
a) What is the ionization energy of hydrogen? Give the number and explain in words.
b) Consider your results from the Boltzmann problems above as well as your answer to a).
Would you expect a significant number of Hydrogen atoms to be ionized at T=10000 K?
Why or why not?
c) Now use the second version of the Saha equation to calculate the fraction of atoms that
would be ionized in a stellar atmosphere of pure hydrogen at T=8000K. What about an
atmosphere at T=12000K? Assume the electron pressure is a constant, Pe =20 N/m2 .
d) Does your answer for d) change your thinking for part c). Explain why or why not.
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BOLTZMANN/SAHA PROBLEM SOLUTION 1. a) For neutral hydrogen, we use the
Boltzmann equation (see hw 3 solutions). For N1 = N2 , Nj /Ni = 1, and gn = 2(n)2 and
T=85,386 K (hot!). The energy required to excite an electron from n=1 to n=2 is E2 - E1
= -3.4eV + 13.6eV = 10.2 eV.
b) As stated in the problem, the ionization energy of hydrogen is the energy required to remove the electron from the ground state - effectively a transition from n=1 to n=inf, which
simply corresponds to the energy of the ground state: E1 =χi =13.6 eV.
c) According to the Boltzmann equation, at T=85,000 K, only half of the atoms have been
excited to n=2 (requiring an energy of 10.2 eV). On the other hand, it requires even more
energy (13.6 eV) to ionize hydrogen. Consequently, based solely on the answers for a) and
b) a significant number of H atoms would not be expected to be ionized at T=10,000 K.
d) Use the Saha Equation with the following values for the constants: k=1.38 × 10−23 J/K,
h=6.63×10−34 Js, mr = 0.9995me ≈ me = 9.11×10−31 kg, χi = 13.6eV = 2.18×10−18 J, and
Pe = 20 N/m2 . The challenge of this problem is what to do with those pesky statistical
weights? Recall that statistical weight is a measure of the degeneracy of a particular energy
state. So, we are asking how many different states j are there with the same energy Ej ? In
this problem, we are comparing the only possible ionization states of Hydrogen: neutral and
singly ionized (HII). Since a hydrogen ion is just a proton, there is only one possible energy
state so gr+1 = 1. Neutral hydrogen has gr =gn =2n2 but based on your answer for part a),
it is safe to assume that at T=8000K and T=12000K most of the atoms can be found in
the ground state. (Note that you have just made use of the Boltzmann equation to provide
required input to the Saha equation!)
At T=8000 K, NHII /NH = 0.025
At T=12000 K, NHII /NH = 50.5
The fraction of ionized atoms is
NHII
NHII /NH
NHII
=
=
Ntotal
(NH + NHII )
1 + NHII /NH
(1)
so at T=8000, approximately 2% of hydrogen atoms are ionized. Conversely, at T=12000K
98% of hydrogen atoms are ionized.
e) Clearly, ionization occurs at a much lower temperature than predicted by the boltzmann
equation and our earlier analysis is incorrect. The reason for the error is that the argument
described above does not take into account the likelihood of ionization from energy states
n>1.
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