general introduction

1. Spectroscopy
Lecture
Microwave
spectroscopy
Dr. Kalpana V. Singh
Astt. Prof. Chemistry

2. Electromagnetic Radiations

3. Type of Radiation
Frequency Range
(Hz)
Wavelength Range Type of Transition
gamma-rays 1020-1024 <1 pm nuclear
X-rays 1017-1020 1 nm-1 pm inner electron
ultraviolet 1015-1017 400 nm-1 nm outer electron
visible 4-7.5x1014 750 nm-400 nm outer electron
near-infrared 1x1014-4x1014 2.5 µm-750 nm
outer electron
molecular vibrations
infrared 1013-1014 25 µm-2.5 µm molecular vibrations
microwaves 3x1011-1013 1 mm-25 µm
molecular rotations,
electron spin flips*
radio waves <3x1011 >1 mm nuclear spin flips*

4. Conversion Formula
• I meter = 102centimeter
• I meter = 103milimeter
• I meter = 106micrometer
• I meter = 109Nanometer
• I meter = 1012Picometer
• I meter = 1015
Femtometer
• I miillimeter = 103micrometer
• I micrometer = 103
Nanometer
• I Nanometer = 103Picometer

5. Electromagnetic field in the matter

6. Interaction of Radiation with Matter:

7. Born-Oppenheimer Approximation
• computation of the energy and the Wave function of an
average-size molecule is a formidable task
• For instance the benzene molecule consists of 12 nuclei and
42 electrons.
• The time independent Schrödinger equation, which must be
solved to obtain the energy and wavefunction of this
molecule, is a partial differential eigenvalue equation in 162
variables—the spatial coordinates of the electrons and the
nuclei.

9. Born-Oppenheimer Approximation
• The Born-Oppenheimer approximation is a way to simplify the
complicated Schrödinger equation for a molecule.
• The nucleus and electrons are attracted to each other with the same
magnitude of electric charge, thus they exert the same force and
momentum.
• While exerting the same kind of momentum, the nucleus, with a much
larger mass in comparison to electron’s mass, will have a very small
velocity that is almost negligible.
• Born-Oppenheimer takes advantage of this phenomenon and makes the
assumption that since the nucleus is way heavier in mass compared to the
electron, its motion can be ignored while solving the electronic
Schrödinger equation; that is, the nucleus is assumed to be stationary
while electrons move around it.
• The motion of the nuclei and the electrons can be separated and the
electronic and nuclear problems can be solved with independent wave
functions.

10. Born-Oppenheimer Approximation

11. Born-Oppenheimer Approximation
• In the first step of the BO approximation
the electronic Schrödinger equation is solved, yielding the
wave function depending on electrons only. For benzene this
wave function depends on 126 electronic coordinates. During
this solution the nuclei are fixed in a certain configuration,
• In the second step of the BO approximation this function
serves as a potential in a Schrödinger equation containing
only the nuclei—for benzene an equation in 36 variables.

12. Born-Oppenheimer Approximation
• Born–Oppenheimer name has also been attached to
molecular spectroscopy, because the ratios of the
periods of the electronic, vibrational and rotational
energies are each related to each other on scales in
the order of a thousand,
• the energy components are treated separately.

13. Rotational Spectroscopy
• Incident electromagnetic waves can excite the rotational
levels of molecules provided they have an electric dipole
moment.
• The electromagnetic field exerts a torque on the molecule.
The spectra for rotational transitions of molecules is typically
in the microwave region of the electromagnetic spectrum.

14. Rotational Spectroscopy
• .

15. ROTATIONAL SPECTROSCOPY
DIATOMIC MOLECULE AS A RIGID ROTATOR ……………
 A Rotating diatomic molecule whose
nuclei are separated by a definite mean distance may
be treated as a rigid rotator with free axis of rotation

16. ROTATIONAL SPECTROSCOPY
• DIATOMIC MOLECULE AS A RIGID ROTATOR ……………
• Consider a diatomic molecule in which masses m1
and m2 of atoms A and B are joined by a rigid bar
whose length is … r..
•
•
• r1 r2
• r
m1 m2

17. ROTATIONAL SPECTROSCOPY
– DIATOMIC MOLECULE AS A RIGID ROTATOR
r= r1+r2…………(1.)

18. ROTATIONAL SPECTROSCOPY
• Molecule rotates end over end around bond length…r…Moment of inertia
around center of mass is given by..
I=m1r1
2+m2r2
2………(2.)
• As system is balanced about its centre of gravity, we may write
m1r1=m2r2……………(3.)
substituting equation 3. in 2.
I=m2r2r1+m1r1r2
I=r1r2(m1+m2)……………..(4.)
From equation 1. and 3.
m1r1=m2r2=m2(r-r1)……….(5.)
m1r1+m2r1=m2r
r1(m1+m2)=m2r
r1=m2r/(m1+m2)……………(6.)
similarly r2=m1r/(m1+m2)…(7.)

19. ROTATIONAL SPECTROSCOPY
• Substituting values of r1 and r2 from equation 6,7 in equation 2.
I=m1m2
2/(m1+m2)2Xr2+m1
2m2/(m1+m2)2Xr2
I=m1m2
2+m1
2m2/(m1+m2)2Xr2
I=m1m2(m1+m2)/(m1+m2)2Xr2
I=m1m2/(m1+m2)Xr2
But m1m2/(m1+m2)=µ Reduced mass
I=µr2……………(8.)

20. ROTATIONAL SPECTROSCOPY
• Kinetic energy of rigid rotator is
• K.E.=½ m1v1
2 + ½ m2v2
2………….(9.)
• Where v1 and v2 are linear velocities of masses m1 and m2
molecule is rotating so it has angular velocityω as the molecule is rigid
rotator r1 and r2 remain constant
K.E.=½m1ω2r1
2+ ½m2ω2r2
2
• K.E.=½ ω2(m1r1
2+m2r2
2)
• I=m1r1
2+m2r2
2
• K.E.=½ω2I………………………..(10.)

21. ROTATIONAL SPECTROSCOPY
• Angular momentum L of rotating molecule from classical mechanics is
given by
• L=Iω……………………..(11.)
• Angular momentum is quantised i.e it is an integral multiple of ĥ/2π and is
given by
• L= I𝜔 = 𝐽(𝐽 + 1) .
ℎ
2𝜋
…………………..(12)
J ……is rotational quantum number having values..J=0,1,2,3,4………….etc.
𝐿2
=𝐼2
𝜔2
= J(J+1) .
ℎ2
4𝜋2………………………………….(13)
• there is no potential energy.
• Total energy of rotating molecule would be given as…
• EJ=
1
2
Iω2 or ω2 =
2EJ
𝐼
SUBSTITUTING VALUES IN EQ 13
• 𝐼2
.
2EJ
𝐼
= J(J+1) .
ℎ2
4𝜋2
• EJ = J(J+1) .
ℎ2
𝐼 8𝜋2………………………………(14)

22. Rotational spectroscopy
• ROTATIONAL ENERGY OF RIGID ROTATOR
• EJ = J(J+1)
• Writing in terms of Wave function
• Ej = hυ= h.
𝐶
λ
= hcū
• Ū = Substituting value of Ej from eq. 14
• Ū =
• .
1
𝐶
=B
• εj =BJ (J+1)

23. Selection rules for rotational
spectrum
• Gross Selection Rule:
• For a molecule to exhibit a pure
rotational spectrum it
• must posses a permanent dipole
moment.
• (otherwise the photon has no
means of interacting –
• SPECI FIC SELECTION RULE
• ΔJ = ± 1

24. Frequency of spectral lines
• Consider there is a transition from rotational level J´ to that of higher
quantum number J
• ∆ Ej=Ej-Ej´ if j´=0 and j=1
• Ū = B (1+1)-0
• Ū = 2B cm-1
• Absorption line will appear at 2B cm-1