Lecture 8: Case Study: UV - OH
1. Introduction
2. OH energy levels
Upper level
Lower level
3. Allowed radiative transitions
Transition notations
Allowed transitions
4. Working example - OH
Allowed rotational transitions from N"=13 in the A2Σ+←X2Π system
UV absorption of OH: A2Σ+ – X2Π (~300nm)
OH, a prominent flame emitter, absorber.Useful for T, XOH measurements.
2
1. Introduction
Selected region of A2Σ+←X2Π(0,0) band at 2000K
Steps in analysis to obtain spectral absorption coefficient
3
1. Introduction
1. Identify/calculate energy levels of upper + lower states
2. Establish allowed transitions
3. Introduce “transition notation”
4. Identify/characterize oscillator strengths using Hönl-London factors
5. Calculate Boltzmann fraction
6. Calculate lineshape function
7. Calculate absorption coefficient
Term energies
4
2. Energy levels
Separation of terms: Born-Oppenheimer approximation
G(v) = ωe(v + 1/2) – ωexe(v + 1/2)2
Sources of Te, ωe, ωexe Herzberg
Overall system : A2Σ+←X2Π
JFvGnTJvnE e ,,elec. q. no.
vib. q. no.ang. mom. q. no.
Electronic energy
Vibrationalenergy
Angular momentum energy (nuclei + electrons)
A2Σ+Te ωe ωexe X2Π
Te ωe ωexe
32682.0 3184.28 97.84 0.0 3735.21 82.21
in [cm-1]
Let’s first look at the upper state Hund’s case b!
Hund’s case b (Λ=0, S≠0) – more standard, especially for hydrides
5
2. Energy levels
Recall:
Σ, Ω not rigorously defined
N = angular momentum without spin
S = 1/2-integer values
J = N+S, N+S-1, …, |N-S|
i = 1, 2, …
Fi(N) = rotational term energy
Now, specifically, for OH?
The upper state is A2Σ+
6
2. Energy levels
For OH:
Λ = 0, ∴Σ not defined use Hund’s case b
N = 0, 1, 2, …
S = 1/2
J = N ± 1/2
F1 denotes J = N + 1/2
F2 denotes J = N – 1/2
Common to write either F1(N) or F1(J)
The upper state: A2Σ+
7
2. Energy levels
for pure case b
∴ the spin-splitting is γv(2N+1) function of v; increases with N
111
112
2
21
NNNDNNBNF
NNNDNNBNF
vvv
vvv
21 OHfor cm1.0constant splitting Av
Notes:
Progression for A2Σ+
“+” denotes positive “parity” for even N [wave function symmetry]
Importance? Selection rules require parity change in transition
γv(2N+1) ~ 0.1(5) ~ 0.5cm-1 for N2Compare with ∆νD(1800K) = 0.23cm-1
The ground state: X2Π (Λ=1, S=1/2)
8
2. Energy levels
Note:
1. Rules less strong for hydrides
2. OH behaves like Hund’s a @ low Nlike Hund’s b @ large N
at large N, couples more to N, Λ is less defined, S decouples from A-axis
3. Result? OH X2Π is termed “intermediate case”
L
Hund’s case a Hund’s case bΛ ≠ 0, S ≠ 0, Σ defined Λ = 0, S ≠ 0, Σ not defined
The ground state: X2Π
9
2. Energy levels
Notes:3. For “intermediate/transition cases”
where Yv ≡ A/Bv (< 0 for OH); A is effectively the moment of inertiaNote: F1(N) < F2(N)
22/122222
22/122221
14421
1414211
NNDYYNNBNF
NNDYYNNBNF
vvvv
vvvv
For large N
NNBF
NNBF
v
v
22
2
221 11
0121 2221 NNNBFF v
For small NBehaves like Hund’s a, i.e., symmetric top, with spin splitting ΛA
Behaves like Hund’s b, with small (declining) effect from spin
The ground state: X2Π
10
2. Energy levels
Notes:4. Some similarity to symmetric top
Showed earlier that F1 < F2
N
3
2
1
J
5/2
3/2
1/2
J
7/2
5/2
3/2F1: J = N + 1/2 F2: J = N – 1/2
Ω = 3/2 Ω = 1/2
Te = T0 + AΛΣFor OH, A = -140 cm-1
Te = T0 + (-140)(1)(1/2), Σ = 1/2+ (-140)(1)(-1/2), Σ = -1/2
∆Te = 140 cm-1
Not too far off the 130 cm-1 spacing for minimum J
130
Hund’s a → 2|(A-Bv)| Recall: Hund’s case a has constant difference of 2(A-Bv) for same J
F(J) = BJ(J+1) + (A-B)Ω2
(A–B)Ω2 ≈ -158.5Ω2
(A for OH~ -140, B ~ 18.5), Ω = 3/2, 1/2 Ω = 3/2 state lower by 316 cm-1
Actual spacing is only 188 cm-1, reflects that hydrides quickly go to Hund’s case b
The ground state: X2Π
11
2. Energy levels
Notes:5. Role of Λ-doubling
Showed earlier that F1 < F2
icid
diid
ciic FFJJJFFJJJFF
11
Fic(J) – Fid(J) ≈ 0.04 cm-1 for typical J in OH
c and d have different parity (p)
Splitting decreases with increasing N
N
3
2
1
J
5/2
3/2
1/2
J
7/2
5/2
3/2F1: J = N + 1/2 F2: J = N – 1/2
Ω = 3/2 Ω = 1/2
p
+–
–+
+–
p–+
+–
+–
Now let’s proceed to draw transitions, but first let’s give a primer on transition notation.
Transition notations
General selection rules Parity must change + → – or – → + ∆J = 0, 1 No Q (J = 0) transitions, J = 0 → J = 0 not allowed
12
3. Allowed radiative transitions
Full description: A2Σ+ (v')←X2Π (v") YXαβ(N" or J")
where Y – ∆N (O, P, Q, R, S for ∆N = -2 to +2)X – ∆J (P, Q, R for ∆J = -1, 0, +1)α = i in Fi'; i.e., 1 for F1, 2 for F2
β = i in Fi"; i.e., 1 for F1, 2 for F2
"or " JNXY
Notes:1. Y suppressed when ∆N = ∆J2. β suppressed when α = β3. Both N" and J" are used
Strongest trans.e.g., R1(7) or R17
Example: SR21: ∆J = +1, ∆N = +2F' = F2(N')F" = F1(N")
Allowed transitions
13
3. Allowed radiative transitions
Allowed rotational transitions from N"=13 in the A2Σ+←X2Π system
12 bands possible (3 originating from each lambda-doubled, spin-split X state) Main branches: α = β; Cross-branches: α ≠ β Cross-branches weaken as N increases
F1(13)F1c(13) F1d(13)
State or level
a specific v",J",N",andΛ-coupling
Allowed transitions
14
3. Allowed radiative transitions
Allowed rotational transitions from N"=13 in the A2Σ+←X2Π system
Notes: A given J" (or N") has12 branches (6 are strong; ∆J = ∆N) + ↔ – rule on parity F1c–F1d ≈ 0.04N(N+1) for OH for N~10, Λ-doubling is ~ 4cm-1, giving clear separation If upper state has Λ-doubling, we get twice as many lines!
Allowed transitions
15
3. Allowed radiative transitions
Allowed rotational transitions from N"=13 in the A2Σ+←X2Σ+ system
Note:1. The effect of the parity selection rule in reducing the number of
allowed main branches to 42. The simplification when Λ=0 in lower state, i.e., no Λ-doubling
Complete steps to calculate absorption coefficient
16
4. Working example - OH
1. Identify/characterize oscillator strengths using Hönl-London factors
2. Calculate Boltzmann fraction
3. Calculate lineshape function(narrow-band vs broad-band)
4. Calculate absorption coefficient
Absorption coefficient
0121
21 exp1cm,
kTh
e
fNcm
ek
0.0265 cm2/s
#/cm3 in state 1,a
a NNN 1
#/cm3 of species (OH), kTpA
Fractional pop. in state 1
ϕ[s] = (1/c) ϕ[cm]
To do: evaluate f12, N1/Na
Step 4 Step 5
Absorption oscillator strength
17
4.1. Oscillator strengths
1"2notation shorthandin or
1"2'"
'"'"'"
'"'"'"',',',',',",",","v,"
JSqff
JSqff
JJvvnnJJ
JJvvnnJvnJn
elec. vib. spin ang. mom. Λ-doubling
elec. osc. strength F-C factor H-L factor
'"vvf = band oscillator strength
(v',v") fv'v"
(0,0) 0.00096(1,0) 0.00028
Notes: qv"v' and SJ"J' are normalized
1'
'" v
vvq
2Xfor 4"
''" 121"2
elgJ
JJ SJS
this sum includes the S values for all states with J"
1 for Λ = 0 (Σ state), 2 otherwise
For OH A2Σ+–X2Π
Is SJ"J' = SJ'J"? Yes, for our normalization scheme! From g1f12 = g2f21, and recognizing that 2J+1 is the ultimate (non removable) degeneracy at
the state level, we can write, for a specific transition between single states
In this way, there are no remaining electronic degeneracy and we require, for detailed balance, that and
Do we always enforce for a state? No! But note we do enforce (14.17)
and (14.19)
where, for OH A2Σ←X2Π, (2S+1) = 2 and δ = 2.
When is there a problem? Everything is okay for Σ-Σ and Π-Π, where there are equal “elec. degeneracies”, i,e., g"el =
g'el. But for Σ-Π (as in OH), we have an issue. In the X2Π state, gel = 4 (2 for spin and 2 for Λ-doubling), meaning each J is split into 4 states. Inspection of our H-L tables for SJ"J' for OH A2Σ←X2Π (absorption) confirms ΣSJ"J' from each state is 2J"+1. All is well. But, in the upper state, 2Σ, we have a degeneracy g'el of 2 (for spin), not 4, and now we will find that the sum of is twice 2J'+1 for a single J' when we use the H-L values for SJ"J' for SJ'J". However, as there are 2 states with J', the overall sum as required by (14.19)
18
4.1. Oscillator strengths
1'2
'1'21"2
"1"2 "'"'
'"v'v"
J
SqfJJSqfJ JJ
vvelJJ
el
1"2'
'" JSJ
JJ
"''" JJJJ SS
"
"'J
JJS
v"v'v'v",'" qqff elel
121"2'
'" SJSJ
JJ
121'2"
'" SJSJ
JJ
41'2"
"' JSJ
JJ
Absorption oscillator strength for f00 in OH A2Σ+–X2Π
19
4.1. Oscillator strengths
Source f00
Oldenberg, et al. (1938) 0.00095 ± 0.00014
Dyne (1958) 0.00054 ± 0.0001
Carrington (1959) 0.00107 ± 0.00043
Lapp (1961) 0.00100 ± 0.0006
Bennett, et al. (1963) 0.00078 ± 0.00008
Golden, et al. (1963) 0.00071 ± 0.00011
Engleman, et al. (1973) 0.00096
Bennett, et al. (1964) 0.0008 ± 0.00008
Anketell, et al. (1967) 0.00148 ± 0.00013
Absorption oscillator strength
20
4.1. Oscillator strengths
Hönl-London factors for selected OH transitions
Transition SJ"J'/(2J"+1) ΣF1(J) ΣF2(J) Σ[F1(J)+F2(J)]Q12(0.5) 0.667 0 2 2Q2(0.5) 0.667R12(0.5) 0.333R2(0.5) 0.333P1(1.5) 0.588 2 2 4P12(1.5) 0.078P21(1.5) 0.392P2(1.5) 0.275Q1(1.5) 0.562Q12(1.5) 0.372Q21(1.5) 0.246Q2(1.5) 0.678R1(1.5) 0.165R12(1.5) 0.235R21(1.5) 0.047R2(1.5) 0.353P1(2.5) 0.530 2 2 4P12(2.5) 0.070P21(2.5) 0.242P2(2.5) 0.358Q1(2.5) 0.708Q12(2.5) 0.263Q21(2.5) 0.214Q2(2.5) 0.757R1(2.5) 0.256R12(2.5) 0.173R21(2.5) 0.050R2(2.5) 0.379
Transition SJ"J'/(2J"+1) ΣF1(J) ΣF2(J) Σ[F1(J)+F2(J)]P1(3.5) 0.515 2 2 4P12(3.5) 0.056P21(3.5) 0.167P2(3.5) 0.405Q1(3.5) 0.790Q12(3.5) 0.195Q21(3.5) 0.170Q2(3.5) 0.814R1(3.5) 0.316R12(3.5) 0.131R21(3.5) 0.044R2(3.5) 0.402P1(9.5) 0.511 2 2 4P12(9.5) 0.016P21(9.5) 0.038P2(9.5) 0.488Q1(9.5) 0.947Q12(9.5) 0.050Q21(9.5) 0.048Q2(9.5) 0.950R1(9.5) 0.441R12(9.5) 0.035R21(9.5) 0.014R2(9.5) 0.462
1. We seek the fraction of molecules in a single state for which
2. In general,
3. Electronic mode
21
4.2. Boltzmann fraction
1"2'
'" JSJ
JJ
re
kTii
QQQQQegNN i
v
/ //
OH 2
42
2
e
e
g
g
12 Sge 0,20,1
N"
J"=N"-1/2
J"=N“+1/2
c
dc
dA “state”
ΣSJ"J'=2J"+1 for each state
# of rot. levels produced by spin splitting & Λ-doubling = 4 for 2Π
nee
ee
kTnhcTSQ
QkTnhcTSN
nN
/exp12
//exp12
Elec. level
Note:hund’s (a) includes AΩ2
the sum of this over all levels is 1
4. Vibrational mode
5. Rotational mode (hund’s (b))
22
4.2. Boltzmann fraction
v
v
kThcGQ
QkThcGnN
nN
/vexp
//vexpv,
v Again, each of these →1 when summed
Nr
r
kTNhcFNQ
QkTNhcFNnN
NnN
/exp12
//exp12v,,v,
rr
r
vr
TQ
khcB
Tfor
/
Note: don’t use F1+F2(N) here; until we add spin splitting Now what about fraction of those with N in a given J?
1212
12,v,
,,v,
SNJ
NnNJNnN
≈ ½ for OH as expected
Since # of states in N is (2N+1)(2S+1)ϕ, while # of states in J is (2J+1) ϕ
1,,v,,,,v,
JNnN
pJNnN(fraction with spectral parity)
6. Combining
23
4.2. Boltzmann fraction
Note: 1. The fraction in a given state is 1/4 of that given by rigid rotor!2. Always know Σ(Ni/N) = 1, both in total and for each mode
separately.
re
ie
QQQ
NFGnTkThcJ
JNnNpJNnN
NnNJNnN
nNNnN
nNnN
NnN
NN
NpJNnN
v
1
vexp12
,,v,,,,v,
,v,,,v,
v,,v,v,
,,,v,
Proper Fi now!
We have 1 loose end to deal with:narrow-band and broadband absorption measurement.
(i.e., the Boltzmann fraction in state 1)
Narrow-band absorptionMeasured quantity
with
thus, if Tν (e.g. Tν0) is measured, and
if L, p, 2γ, T, f12 are known
then can solve for Nl
24
4.3. Narrow-band vs broad-band absorption measurement
LSIIT 120 exp/
kTh
e
fNcm
eS exp1121
2
12
Oscillator strength for transition
kTplN
NN l
ll
of dens. no. tot.1
Boltzmann fraction of species l = Fv",J",…(T)
species bd.
2i
ii X
Quantity usually sought
Let’s look at the classical (old-time) approach, pre 1975 Broadband absorption
25
4.3. Narrow-band vs broad-band absorption measurement
ν
Instrument broadening (no change in eq. width)
1 line
Eq. width WJ"J' (cm-1)
Integrated area is called: integrated absorbance, or eq. width
line JJ
JJline
dLK
dTdAWW
'"
'"
exp1
1
0 '"'" ,2ln2exp12ln
dxaxVLKWD
JJD
JJ
xD
2ln2
(for 1 line from 1 state)
Transform variables
iJJJJ PSK '"'"
Let’s look at the classical (old-time) approach, pre 1975 Broadband absorption Requires use of “curves of growth”
26
4.3. Narrow-band vs broad-band absorption measurement
ν
1 line
Eq. width WJ"J' (cm-1)
0 '"'" ,2ln2exp12ln
dxaxVLKWD
JJD
JJ
Procedure: measure WJ"J', calculate ∆νDand a, infer KJ"J', convert KJ"J' to Nspecies
Note:1. Simple interpretation only in optically
thin limit,
2. Measured eq. width is indep. of instrument broadening!
3. Before lasers, use of absorption spectroscopy for species measurements require use of Curves of Growth!
LfNmc
eLKW
dLKW
JJJJ
JJJJ
1212
2
'"'"
'"'" 11
Curve of growth
Consider spectral absorption coefficient of the (0,0)Q1(9) line in the OH A2Σ+–X2Π system, at line center.
27
4.4. Example calculation (narrow-band)
λ~309.6nm, ν~32300cm-1, T = 2000K, ∆νC = 0.05cm-1
Express kν as a function of OH partial pressure
4'"
v'v"9 1009.9947.000096.01"21
JSff JJ
Q
Oscillator strength (using tables)
Lineshape factor (narrow-band)
s1004.1or cm13.317.0cm05.0K2000
cm25.0K2000 1001
1
a
C
D
0'"",,,v,
221
scm10651.2cm JJ
a
Jna fN
NkTPk kTPN aa /
Consider spectral absorption coefficient of the (0,0)Q1(9) line in the OH A2Σ+–X2Π system, at line center.
28
4.4. Example calculation (narrow-band)
λ~309.6nm, ν~32300cm-1, T = 2000K, ∆νC = 0.05cm-1
Express kν as a function of OH partial pressure
0'"",,,v,
221
scm10651.2cm JJ
a
Jna fN
NkTPk kTPN aa /
Population fraction in the absorbing state
0.0193.08390 .9200 25.0
0.7529.6
287.0264.0
41
K66.26//K2313exp20
287.0/K2660exp
40exp
/5.9exp1"2/0exp/0exp 15.01
TTT
QkThcFJ
QkThcG
QkThcT
NN
rve
e
a
f c
Consider spectral absorption coefficient of the (0,0)Q1(9) line in the OH A2Σ+–X2Π system, at line center.
29
4.4. Example calculation (narrow-band)
λ~309.6nm, ν~32300cm-1, T = 2000K, ∆νC = 0.05cm-1
Express kν as a function of OH partial pressure
0'"",,,v,
221
scm10651.2cm JJ
a
Jna fN
NkTPk kTPN aa /
atmatmcm177
s1004.11009.9%93.1atmcm1066.3atm
scm10651.2cm
1
1043
182
21
a
a
P
Pk
LkII exp0Beer’s Law
59% absorptionfor L = 5cm, XOH = 1000ppm, T = 2000K, P = 1atm
Selected region of OH A2Σ+←X2Π (0,0) band at 2000K
30
4.4. Example calculation (narrow-band)
Notes:
Lines belonging to a specific branch are connected with dashed or dotted curve
Thicker dashed lines –main branches; thin dotted lines – cross branches
Bandhead in R branches if Bv'<Bv"; Bandhead in P branches if Bv'>Bv“
Note bandhead in RQ21branch
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