VIBRATIONAL SPECTRA
• COMBINING TWO ATOMS TO A MOLECULE
RAISES TWO ISSUES.
– (i)
REPULSIVE FORCES DUE TO ELECTRONS IN
SHELLS AND POSITIVE CHARGES IN THE
NUCLEI
– (ii)
ATTRACTIVE FORCES BETWEEN THE
NUCLEI AND THE ELECTRONS IN THE
SHELLS
THESE RESULT IN THE FORMATION OF A BOND AT
AN OPTIMUM NUCLEAR DISTANCE (req).
VIBRATIONAL SPECTRA
• THE EQUILIBRIUM DISTANCE RESULTS IN
MINIMUM ENERGY SINCE BOTH REPULSIVE
AND ATTRACTIVE FORCES HAVE TO BE
BALANCED.
THIS CAN BE COMPARED TO BEHAVIOUR OF A
SPRING WHICH CAN BE COMPRESSED OR
EXTENDED LIKE A SPRING OBEYING THE
HOOKE’S LAW.
HOOKE’S LAW
• HOOKE’S LAW CAN BE STATED AS FOLLOWS;
F=-f(r-req), WHERE f IS THE RESTORING FORCE,
k IS THE FORCE CONSTANT, r IS THE
INTERNUCLEAR DISTANCE.
• THE KINETIC ENERGY OF THE SYSTEM
BEHAVES PARABOLIC WHICH IS; E=1/2f(r-req).
THIS IS A SIMPLE HARMONIC OSCILLATOR
MODEL.
NEWTON’S EQUATION
• AS STATED, THE RESTORING FORCE (F) IS
GIVEN AS; F = -f(x2-x1); IN NEWTON’S
LAWS
2
2
d
x2 =-f(x -x )
m d x1 =-f(x2-x1): +m
2 1
2
dt
dt 2
• THE SOLUTION TO THIS EQUATION IS GIVEN
BY:
f
1
 0  2

VIBRATIONAL ENERGY
• IN QUANTUM MECHANICS, THE POTENTIAL
ENERGY OF A VIBRATING PARTICLE IS GIVEN
BY: 1
2
• V(x)= f ( x2  x1 )
2
• WHERE x=x2-x1 IS THE DISPLACEMENT
• USING THE SCHRODINGER’S EQUATION,
d 2  8 2 
1 2

( E  fx )  0
2
2
dx
h
2
VIBRATIONAL ENERGY
• THE SOLUTION OF THE EQUATION LEADS TO THE
VIBRATIONAL ENERGY WHICH CAN BE
REPRESENTED BY;
1
Ev  hoo (  );  0,1,2,3....
2
• WHERE v IS THE VIBRATIONAL QUANTUM
NUMBER AND γ IS FREQUENCY FOR A HARMONIC
OSCILLATOR o IS THE FUNDAMENTAL
FREQUENCY
NEWTON’S EQUATION
• IN TERMS OF WAVE NUMBER UNITS THEN THE
EQUATION IS DIVIDED BY SPEED OF LIGHT AND h
Ev
1 ṽ cm-1
•
1
Ev (cm )   (v  )
•
hc
2
• FROM THIS EQUATION,
1
1
• IF v=0, E0 (cm )  ṽ cm-1
2
3
1
• IF v=1, E1 (cm ) 
ṽ cm-1
2
IMPLICATIONS
• THE ENERGY DIFFERENCE BETWEEN EACH
VIBRATIONAL LEVELS IS 1 cm-1
• WHEN v=0, Eo=1/2ṽ cm-1. THIS MEANS THAT A
MOLECULE IS NEVER AT REST BUT VIBRATES
EVEN WHEN AT REST RELATIVE TO EACH
OTHER.
• THIS ENERGY IS CALLED ZERO POINT ENERGY.
• ITS DEPENDS ON THE CLASSICAL FREQUENCY.
SELECTION RULES
• FOR A VIBRATION TO BE ALLOWED; (1) THERE
MUST A CHANGE IN DIPOLE MOMENT,
• (2) TRANSITIONS SHOULD OBEY Δv=±1
• (3)THE FREQUENCY EMITTED OR ABSORBED
RADIATION MUST SATISFY BOHR’S EQUATION
WHICH IS ΔE=hν
• VIBRATIONAL SPECTRA ARE NOT USUALLY
SINGLE LINES DUE TO SUPERIMPOSITION OF
ROTATIONAL SPECTRA.
APPLICATIONS OF VIBRATION SPECTRA
• HOMONUCLEAR ATOMS THAT CANNOT
CHANGE IN DIPOLE MOMENT CANNOT BE
ACTIVE IN INFRARED SPECTROSCOPY. APPLIED
TO MOLECULES THAT CAN HAVE A CHANGE IN
DIPOLE MOMENT WHEN RADIATION IS
ABSORBED OR EMITTED
• TO OBTAIN BOND STRENGTH OR MASSES OF
ATOMS IN A MOLECULE
ANHARMONIC OSCILLATOR
• REAL MOLECULES DO NOT OBEY EXACTLY
HOOKE’S LAW OR LAWS OF SIMPLE HARMONIC
MOTION.
• THIS IMPLIES REAL BONDS THOUGH ELASTIC WILL
NOT OBEY LAWS OF SIMPLE HARMONIC MOTION.
• THIS IS BECAUSE BONDS HAVE LIMITS OF
EXTENTION OR STRETCHING.
• WHEN ARE OVER STRETCHED TO EXCEED THE
ELASTIC LIMIT, THE BOND CAN BREAK
ANHARMONIC OSCILLATOR
• WHEN THE MOLECULES BREAKS THEN IT
MEANS IT HAS DISSOCIATED INTO ITS ATOMS.
• ALTHOUGH SMALLER COMPRESSIONS AND
EXTENSIONS MAY LEAD TO PERFECT ELASTIC
BEHAVIOUR, LARGER AMPLITUDES WILL LEAD
MAY LEAD GREATER COMPLICATIONS.
• THIS ILLUSTRATED IN A DIAGRAM BELOW
WITH SHAPE OF ENERGY CURVE FOR BOTH
SHM AND ANHARMONIC OSCILLATOR.
ANHARMONIC OSCILLATOR
• FOR ANHARMONIC OSCILLATOR, THE POTENTIAL
ENERGY DERIVED EMPIRICALLY BY P.M.MORSE
(1929) NORMALLY REFERED TO AS MORSE
POTENTIAL.
• MATHEMATICALLY, MORSE POTENTIAL FUNCTION
IS GIVEN AS: E(x)=De(1-exp(-βx))2, WHERE De IS
THE DISSOCIATION ENERGY OF A MOLECULE
RECKONED FROM THE LOWEST POINT IN THE
CURVE, β IS RELATED TO THE FORCE CONSTANT
AND ANHARMONICITY CONSTANT AND IT IS
EQUAL TO β=ṽ
ANHARMONIC OSCILLATOR
2 c
)
• EQUAL TO β = ṽ (
De
2
1
2
• THE VIBRATIONAL ENERGY STATE OF AN
ANHORMONIC OSCILLATOR IS OBTAINED BY
SUBSTITUTING THE MORSE POTENTIAL
FUNCTION IN THE SCHRODINGER WAVE
EQUATION
ANHARMONIC OSCILLATOR
• THE VIBRATIONAL ENERGY STATE CAN
OBTAINED BY SUBSTITUTING THE MORSE
POTENTIAL FUNCTION IN THE
SCHRODINGER’S EQUATION.
• FOR A ONE-DIMENSIONAL OSCILLATOR
Ev=(v+½)hγ-xe(v+½)2hγ OR IN TERMS OF WAVE
NUMBERS
ԑ=Ev/hc=(v+½)ṽ-xeṽ(1+½) (cm-1) xe IS THE
ANHARMONICITY CONSTANT.
ANHARMONIC OSCILLATOR
• THE ANHARMONICITY CONSTANT xe IS
USUALLY SMALL AND POSITIVE AND EQUAL TO
xe=ṽ/4De.
• VIBRATIONAL ENERGY LEVELS ARE NOT
EQUALLY SPACED BUT NARROW AS THE
VIBRATIONAL QUANTUM NUMBER
INCREASES.
SUMMARY OF ROTATING AND
VIBRATING DIATOMIC MOLECULES
SYSTEM MODEL
POTENTIAL ENERGY
FUNCTION
QUANTIZED ENERGY
LEVELS
RIGID ROTOR
V=0
Ej=BhcJ(J+1)
B=EJ/8π2hc
NON-RIGID ROTOR
V=½fx2
EJ=Bhc J(J+1)-Dhc
J2(J+1)2
HARMONIC OSCILLATOR V=½fx2
ANHARMONIC
OSCILLATOR
V=De(1-exp(-βx))2
Ev=(v+½)hγ
Ev=(v+½)hγ-xe(v+½)2hγ
NEW SELECTION RULES
FROM THE TABLE THE POTENTIAL FUNCTION CAN
REPRESENTED BY V(x)=f1x2+f2x3.....
BY INCLUDING THE CUBIC TERM IMPLIES THAT THE
SELECTION RULES FOR TRANSITON HAVE TO BE
MODIFIED.
HENCE Δv=±1 AND ALSO ±2, ±3, ±4, ............
THIS MEANS THE LINES ARE NOT EQUALLY SPACED
AND SO WEAK TRANSITIONS CAN OCCUR AT ±2,
±3, ±4, ............ THESE ARE CALLED OVERTONES.
ANHARMONIC OSCILLATOR
• THE ENERGY FOR AN ANHARMONIC
OSCILLATOR IS GIVEN BY:
Ev=(v+½)hν-xe(v+½)2hν. OR
Ev/hc=ṽ(v+½)-xeṽ(v+½)2 cm-1.
WHERE THE ANHARMONICITY CONSTANT IS
POSITIVE AND IS GIVEN BY: xe=ṽ/4De
EXAMPLES
• CONSIDER TRANSITIONS OF Δv=±1 WHERE THERE
IS A TRANSITION FROM Eo → E1.
Eo=hνo(v+½)-hνoxe (v+½)2, v=0
→Eo=½hγo - h νoxe (½)2
→E1= 3/2 hνo – (3/2)2hνoxe v=1
ΔE=E1-Eo=3/2 hνo–(3/2)2hνoxe-[½hνo-hνoxe(½)2]
=½hνo- hνoxe(½)2
ΔE=hvo-2xehov=hvo(1-2xe) OR IN TERMS OF WAVE
NUMBERS; ԑ=ΔE/hc=2ṽo(1-2xe)
EXAMPLES
• FOR TRANSITIONS FROM Eo→E2 THIS IS Δv= ±2
AND HENCE ΔE=2hvo(1-3xe) OR IN WAVE
NUMBERS; ΔE/hc=2ṽo(1-3xe)cm-1
• FOR Δv= ±3 Eo→E3 ΔE=3hvo(1-4xe)
• ΔE/hc=3ṽo(1-4xe)cm-1
• FOR Δv= ±4 Eo→E4 ΔE=4hvo(1-5xe)
• ΔE/hc=4ṽo(1-5xe)cm-1
• ASSIGNMENT: WORK OUT E0 TO EN WHERE N IS
5,6,7,8,9,10 AND CHANGE E0 TO E1,2,3 .
• ESTABLISH ANY RELATIONSHIP
EXAMPLES
• THE FREQUENCY ASSOCIATED WITH THESE
TRANSITIONS ARE APPROXIMATELY A WHOLE
MULTIPLE OF THE FUNDAMENTAL FREQUENCY
OF THE HARMONIC OSCILLATOR ṽo AND ARE
CALLED OVERTONES.
THUS Δv=±1, →ν= ṽo (1-2xe), FOR Δv=±2, ±3,
±4,.... ν0 → 2=2ṽo(1-3xe), ν0 → 3= 3ṽo (1-4xe),
ν0 → 4= 4ṽo (1-5xe).
IMPLICATIONS
• TRANSITION FROM v=0→1 IS THE MOST
INTENSE AND IS CALLED THE FUNDAMENTAL
FREQUENCY.
• AT ROOM TEMPERATURE MOST MOLECULES
ARE IN THE LOWEST VIBRATIONAL LEVEL.
• IN PURE VIBRATIONS, v=1, 2, 3, .... NOT
USUALLY STRONG EVEN IF Δv=±1.
• REASON: LOW POPULATION IN THOSE LEVELS
IMPLICATIONS
• TRANSITIONS AT HIGHER VIBRATIONAL LEVELS
i.e. v=1, 2, 3, 4, .. ARE MORE LIKELY TO BE SEEN
AT HIGHER TEMPERATURES SINCE THE RISE IN
TEMPERATURE WILL INCREASE THE POPULATION
IN THE HIGHER LEVELS.
• TRANSITIONS FROM SUCH LEVELS WILL GIVE RISE
TO WEAKER BANDS THAN FROM v=0→1 AND
WITH LOWER FREQUENCIES.
• BANDS ARISING DUE TO TEMPERATURE INCREASE
ARE KNOWN AS HOT BANDS.
OVERTONES AND HOT BANDS
• OVERTONE BANDS WHERE Δv=2, 3, 4... HAVE
LESS INTENSITY THAN THE FUNDAMENTAL
FREQUENCY
2nd hot band
1st hot band
Fundamental
1st Overtone
2nd Overtone
IMPLICATIONS CONT’D
• Δv=±1: FUNDAMENTAL FREQUENCY(0→1).
MOST INTENSED LINE WITH ENERGY
ΔE=hvo(1-2xe) OR
ṽo(1-2xe) cm-1.
• Δv=±2: 1st OVERTONE BAND (0→2). SMALLER
INTENSITY WITH ENERGY DIFFERENCE OF
ΔE=2hvo(1-3xe) OR 2ṽo(1-3xe) cm-1.
• Δv=±3: 2nd OVERTONE BAND (0→3) OF SMALLER
INTENSITY WITH ENERGY:
• ΔE=3hvo(1-4xe) OR 3ṽo(1-4xe)cm-1.
WORKED EXAMPLES
• (1) CO HAS AN INTENSE BAND AT 2143 cm-1
AND A WEAK BAND AT 4260cm-1. CALCULATE
THE FUNDAMENTAL FREQUENCY, FORCE
CONSTANT, AND ANHARMONICY CONSTANT.
• SOLUTION:
• THE INTENSE BAND IS THE FUNDAMENTAL
LINE AND WEAKER ONE IS THE 1st OVERTONE
HENCE THE FUNDAMENTAL AND FIRST OVER
TONE EQUATIONS WILL BE APPLIED.
EXAMPLES
• v= ṽo(1-2xe) AND v1=2ṽo(1-3xe).
2143= ṽo(1-2xe) 4260=2ṽo(1-3xe)
2143= ṽo-2ṽoxe .............................................(1)
4260=ṽo2-6ṽoxe OR 2130=ṽo -3ṽoxe ...............(2)
MULTIPLYING EQN (1) BY 3 AND EQN (2) BY 2
6229=3ṽo - 6ṽoxe ...........................................(3)
4260=2ṽo - 6ṽoxe.............................................(4)
SUBSTRATING EQN (4) FROM EQN (3) THEN WE
EXAMPLES
• ṽo =2169 cm-1 THEN xe IS OBTAINED FROM
AS 2143=2169-2x2169xe=2169-4338xe WHICH GIVES
26=4338xe→xe=0.006.
THE REDUCED MASS IS EQUAL µ=m1m2/m1+m2
THE ATOMIC MASSES OF C AND O ARE 12 AND 16
RESPECTIVELY. HENCE:
12x16x1.66057x10-27/12+16=1.1387x10-26 Kg
FORCE CONSTANT
f=4π2ṽoc2µ2=4π2x(2169)2x3x108x1.1387x10-26 =190.6
Nm-1.
EXAMPLES
• PROBLEM: AN I.R. SPECTRUM OF HCl SHOWS
AN INTENSE BAND AT 2886 AND WEAKER
BAND AT 5668cm-1. CALCULATE THE
FOLLOWING: (i) FUNDAMENTAL ABSORPTION
FREQUENCY (ii) THE ANHARMONICITY
CONSTANT (iii) DISSOCIATION ENERGY ON THE
BASES OF MORSE FUNCTION.
• SOLUTION: THE BAND FREQUENCY ARE
ROUGHLY 1:2 i.e. 2886 : 5668 cm-1.
EXAMPLES
• HENCE; THE FUNDAMENTAL FREQUENCY IS
v=0 →v=1 AND THE FIRST OVERTONE IS FROM
v=0 →v=2
• v= ṽo(1-2xe) AND v1=2ṽo(1-3xe) ARE USED.
2886= ṽo(1-2xe) 5668=2ṽo(1-3xe)
SOLVING THESE EQUATIONS GIVES THE
FOLLOWING; ṽo =2990 cm-1, xe=0.0174.
THE ANHARMONIC CONSTANT IS RELATED TO
MORSE POTENTIAL AS xe=ṽ/4De
EXAMPLES
• De= ṽo/4xe= 2990/4x0.0174= 6.626x10-34 x
2990 x 3 x 108/4 x 0.0174= 8.54 x 10-20 J= 514
kJmol-1.
VIBRATIONAL-ROTATIONAL SPECTRA
• VIBRATIONAL LEVELS CONTAIN CLOSELY
SPACED ROTATIONAL LEVELS.
• ANY CHANGE IN THE ENERGY OF THE
VIBRATIONAL LEVELS AFFECTS THAT OF THE
ROTATIONAL LEVELS.
• THEREFORE IF A MOLECULE IS VIBRATING, THE
MOLECULE WILL ALSO UNDERGO ROTATION
SINCE THE VIBRATION AND ROTATION WILL BE
MOVING SIMULTANEOUSLY.
VIBRATIONAL-ROTATIONAL SPECTRA
• CONSIDER A DIATOMIC MOLECULE VIBRATING
AND ROTATING AND ASSUME THE MOTIONS ARE
INDEPENDENT.
• IT WILL HAVE A VIBRATIONAL ENERGY AND
ROTATIONAL ENERGY COMBINED AS FOLLOWS;
EVT=hνo(v+½)+J(J+1)h2/8π2I OR hνo(v+½)+BhcJ(J+1),
WHERE v=0,1,2,3...J=0,1,2,3...
THE ENERGY OF EACH STATE CAN BE OBTAINED BY
INSERTING THE APPROPRIATE VALUES OF v AND J.
VIBRATIONAL-ROTATIONAL SPECTRA
E
•
5
4
3
2
1
J’=0
5
4
3
2
1
J”=0
V’=1
V”=0
VIBRATION-ROTATION CONT’D
• SELECTION RULES FOR VIBRATIONAL-ROTATIONAL
TRANSITIONS ARE THE SAME AS IN THE SEPARATE
TRANSITIONS, i.e. Δv=±1 AND ΔJ=±1.
• CONSIDER A TRANSITION FROM A LOWER STATE
AT LEVEL v” AND J” TO AN UPPER STATE v’ AND J’.
THE CHANGE IN ENERGY IS GIVEN BY:
ΔE=hνo(v’-v”)+Bhc[J’(J’+1)-J”(J”+1)].........(a)
SINCE Δv=1 AND ΔJ=1, THEN WE HAVE J’-J”=1
• ΔEVT=hνo+2Bhc(J”+1); J”=0, 1, 2, 3,.............(b)
• ΔE/hc=ṽo +2B(J”+1) cm-1
VIBRATION-ROTATION CONT’D
• FOR v=1 AND ΔJ=-1, THEN J’-J”=-1
EVT=hνo-2BhcJ”, J”=1,2,3,..... J” CANNOT BE ZERO.
IF EQUATIONS (a) AND (b) ARE COMBINED, THEN;
EVT=hνo+m.2hc B.
m=±1, ±2, ±3,...
WHERE m IS REPLACING J”+1 IN (a) FOR POSITIVE
VALUES OF ΔJ=+1 AND IN EQUATION (b) FOR J’
FOR NEGATIVE VALUES OF ΔJ=-1.
m CAN NEVER BE ZERO.
FREQUENCIES (cm-1) IS GIVEN AS;ΔE/hc=ṽo+m.2B
(cm-1)
VIBRATION-ROTATION CONT’D
• FROM THE ABOVE EQUATION THE LINES ARE
EQUALLY SPACED WITH WIDTH Δṽ=2B(cm-1).
• THESE LINES WILL FALL INTO TWO BRANCHES,
THOSE WITH m VALUES NEGATIVE J WHERE
J=-1 ARE CALLED P-BRANCH AND THOSE
WITH m WITH POSITIVE VALUES WHERE J=+1
CORRESPOND TO THE R-BRANCH.
VIBRATION-ROTATION CONT’D
ΔJ=+1
ΔJ=-1
6
5
4
3
2
1
J’=0
V=1
6
5
4
3
2
1
J”=0
V=0
R-branch
2B
4B
P-branch
2B
ASYMMETRY IN VIBRATIONALROTATIONAL SPECTRA
• EARLIER, IT WAS POINTED OUT THAT SPACING
BETWEEN ROTATIONAL SPECTRA ARE THE SAME.
BUT THIS IS NOT TRUE IN REAL MOLECULES.
• IN NON-RIGID ROTOR THERE ARE CENTRIFUAL
STRETCHING OF THE BOND NORMALLY
OBSERVED IN HIGH J-VALUES.
• CHANGE IN CENTRIFUGAL STRETCHING CAUSES
A CHANGE IN MOMENT OF INERTIA, I.
• THIS ALSO IMPLIES A CHANGE IN INTERNUCLEAR
DISTANCE ro.
ASYMMETRY IN VIBRATIONALROTATIONAL SPECTRA
• WHEN ROTATION AND VIBRATION ARE
CONSIDERED TOGETHER, THE FOLLOWING WILL
COME TO PLAY;CONTRIBUTIONS FROM
ROTATIONS AND VIBRATIONS, ANHARMONICITY,
CENTRIFUGAL DISTORTIONS, ROTATIONVIBRATION INTERACTIONS.
• THESE WILL HAVE AN EFFECT ON THE OVERAL
SPECTRUM AND THE MODIFIED EQUATON IS
GIVEN BELOW.
ASYMMETRY IN VIBRATIONALROTATIONAL SPECTRA
• EVT=(v+½)hνo+BhcJ(J+1)-(v+½)2hνoxeDhcJ2(J+1)2-α(v+½)J(J+1). THE PARAMETERS
νo= 1/2π(f/µ)½; B=h/8π2Ic; D=4B2C2/ ν2o OR
4B3/ν2o I=µr2e, xe=ṽ/4De.
• α= IS THE ROTATIONAL-VIBRATIONAL
INTERACTION PARAMETER. IT’S VALUE IS VERY
SMALL AND CAN BE NEGLECTED.
ASYMMETRY IN VIBRATIONALROTATIONAL SPECTRA
• IN TERMS OF WAVE NUMBERS;
ṽ=(v+½)ṽo-(v+½)2ṽoxe+BJ(J+1)-DJ2(J+1)2 .
USING THE SELCTION RULE Δv=±1 AND ΔJ=±1,
THE FOLLOWING CASES CAN BE DERIVED.
(i) TRANSITIONS INVOLVING ONLY CHANGE IN
ROTATIONAL QUANTUM NUMBER J WITH
ΔJ=±1 AND v=0;
ṽJ→J+1=EVT/hc=2B(J+1)-4D(J+1)3
ASYMMETRY IN VIBRATIONALROTATIONAL SPECTRA
• D IS THE CENTRIFUGAL DISTORTION
CONSTANT DEPENDS ON TWO FACTORS;
(i) THE MOMENT OF INERTIA. THE LARGER
THE MOMENT OF INERTIA, THE SMALLER, THE
MAGNITUDE OF B IMPLYING A SMALLER
DISTORTION CONSTANT. THEREFORE LIGHT
MOLECULES WILL HAVE VERY SMALL
DISTORTION CONSTANT.
ASYMMETRY IN VIBRATIONALROTATIONAL SPECTRA
• THE WEAKER THE INTERATOMIC BOND OR
THE HEAVIER THE ATOMS, THE LOWER THE
VIBRATIONAL FREQUENCY THEREFORE THE
MOLECULE IS LIKELY TO BE AFFECTED BY
CENTRIFUGAL DISTORTION
• ii. TRANSITIONS INVOLVING CHANGE IN
VIBRATIONAL QUANTUM NUMBER (Δv=±1,
ΔJ=0). EVT=(v+½)hνo-(v+½)2hνoxe.
• OR ṽv→v+1=ṽo-2(v+1)ṽoxe
ASYMMETRY IN VIBRATIONALROTATIONAL SPECTRA
• TRANSITIONS INVOLVING ΔJ=+1 AND Δv=0 ARE
OBSERVED IN MICROWAVE REGION, WHILE Δv=
±1 AND ΔJ=0 OCCUR IN INFRARED BUT CANNOT
BE OBSERVED BECAUSE ΔJ=0 IS NOT ALLOWED
FOR DIATOMIC MOLECULES.
iii. TRANSITIONS INVOLVING SIMULTANEOUS
CHANGE IN BOTH v AND J WHILE D AND xe AND α
ARE NEGLECTED. EVT=(v+½)hνo+BhcJ(J+1) OR
ṽ=ṽo(v+½)+BJ(J+1). TRANSITIONS FOR v=0 AND
v=1 FALL INTO BRANCHES GIVEN BY:
ASYMMETRY IN VIBRATIONALROTATIONAL SPECTRA
• ṽR=ṽo+2B(J+1) WHEN ΔJ=+1
R-BRANCH
THESE LINES WILL OCCUR AT ṽR=ṽo+2B, ṽR=ṽo+4B
ṽR=ṽo+6B ETC.
AT ΔJ=-1, WE HAVE THE P-BRANCH WHICH ALSO
OCCUR AT ; ṽP=ṽo-2B(J+1).
LINES WILL OCCUR AT ṽP=ṽo-4B, ṽP=ṽo-6B ETC.
THESE LINES LIE BELOW AND ABOVE THE
FUNDAMENTAL FREQUENCY ṽo .
OBSERVATION OF SPECTRAL LINES
• FOR DIATOMIC MOLECULES, P AND R ARE
OBSERVED WHEN THE DIPOLE OSCILLATES
PARALLEL TO THE BOND AXIS KNOWN AS THE ΣMODE WITH A SELECTION RULE ΔJ=±1
• IN LINEAR POLYATOMIC MOLECULES, APART
FROM THE PARALLEL OSCILLATIONS, THERE ARE
PERPENDICULAR OSCILLATIONS TO THE BOND
AXIS WITH SELECTION RULE ΔJ=0 AND ΔJ=±1,
SUCH THAT WHEN ΔJ=0 AND Δv=1, THEN ṽQ=ṽo,
J→J. THIS TRANSITION IS THE Q-BRANCH.
BRANCHES
R-BRANCH
P-BRANCH
PARALLEL
P-BRANCH
Q-BRANCH
R-BRANCH
PERPENDICULAR
OBSERVATION OF SPECTRAL LINES
• IN ROTATIONAL-VIBRATIONAL SPECTRA OF
LINEAR MOLECULES, THERE ARE THREE
BRANCHES OF LINES THAT ARISE, P, Q AND R.
• P-BRANCH: THESE LINES ARISE WHEN Δv=1
AND ΔJ=-1; e.g. J=0→J-1 AND THE OVERALL
ENERGY CHANGE IS ṽ=ṽo-2BJ. THE LINES IN
THIS SECTION ARE DISPLACED BY 2B, 4B,
6B..IN THE LOWER FREQUENCY RANGE.
OBSERVATION OF SPECTRAL LINES
• Q-BRANCH: THESE SET OF LINES ARISE WHEN
Δv=1 AND ΔJ=0 AND THE FREQUENCY IS GIVEN
BY ṽJ→J=ṽo. FOR ALL VALUES OF J AND THIS
CORRESPONDS TO J→J TRANSITION WHICH IS A
SINGLE LINE.
• R-BRANCH: THESE ALSO ARISE FROM Δv=1 AND
ΔJ=+1; e.g. J→J+1. THE OVERALL ENERGY
CHANGE IS ṽ=ṽo+2B(J+1). THE LINES ARE SPACED
AT 2B, 4B, 6B, ......TO HIGHER FREQUENCY THAN
THE FUNDAMENTAL FREQUENCY.
FORMATION P, Q, R BRANCHES
5
4
3
2
1
J=0
Upper
vibrational level
v=1
5
4
3
2
1
J=0
Lower
vibrational level
v=0
P-BRANCH
Q-BRANCH
R-BRANCH
USES
• NOTE THAT THE SEPARATION BETWEEN P AND
R-BRANCHES GIVE THE ROTATIONAL
CONSTANT IN WHICH THE BOND LENGTH CAN
BE OBTAINED.