Transcript Principles of Spectroscopy

Quantum Physical Phenomena
in Life (and Medical) Sciences
Principles of spectroscopy
Péter Maróti
Professor of Biophysics, University of Szeged, Hungary
Suggested texts:
P. Maróti, G. Laczkó and L. Szalay: Medical Physics I-II, University of Szeged, 1988.
S. Damjanovich, J. Fidy and J. Szőlősi: Medical Biophysics, Semmelweis, Budapest 2006
P.W. Atkins: Physical Chemistry, Oxford University Press, Oxford, 5th Edition, 1994
Definition of the spectrum
Spectrum: distribution of a physical quantity according to energy.
Physical quantities can be e.g. emitted or absorbed light quanta (photons)
Energy can be represented by wavelength, frequency, wave number etc.
Example:
the absorption spectrum.
Selective absorption of
the light by visual
pigments of the cones
and rod in the human
eye. This is necessary to
the color vision.
http://insight.med.utah.edu/Webvision/imageswv/spectra.jpeg
1/21/2002
Vision
11
Electromagnetic spectrum
visible light
low energy
high energy
energy
frequency
radio
microwave
infrared
wavelength
ultraviolet
X-ray
Principal types of spectroscopies
Spectroscopy
Emission
Wavelength
Observed energy,
Quantum transition
< 0.1 nm
Gamma radiation coupled to
transitions in nucleus
0.01 – 10 nm
Resonance-reabsorption of
gamma radiation
0.01 – 10 nm
X-rays from reactions in
internal shells of the atom
Gamma (γ)
Mössbauer
Röntgen (X-ray)
Optical
Ultraviolet (UV)
200 – 300 nm
Visible
400 – 800 nm
Electron transitions
0.8 – 200 μm
Molecular vibrations
0.01 – 10 cm
Molecular rotations
Infrared (IR)
Raman
Microwave
Magnetic and
dielectric
EPR
NMR
Alternative current (AC)
Direct current (DC)
Electron spins
10 – 105 cm
Nuclear spins
1 cm - ∞
∞
Molecular conformations
Magnetic Resonance
Spectroscopy
Electron Spin Resonance (ESR)
or
Electron Paramagnetic Resonance (EPR)
and
Nuclear Magnetic Resonance (NMR)
NMR ~10-200 MHz
@ 4.7 T
EPR ~9.5GHz
@ 0.34 T
RT=0.002 kcal/mol
at 10oK
mI = -½
Energy
(n)
mI = ±½
E  hn 
B0 h
2
Radiowave
mI = +½
Energy of
transition
0
magnetic field strength (B0)
n =E/h
ms = -½
(E = hn)
RT= 0.593 kcal/mol
at 298oK
n - frequency Hz
Energy
(n)
ms = ±½
Microwave
n~ - wavenumber cm-1
- wavelength nm
n~ 
1


n
c
E  hn  gB0
ms = +½
0
magnetic field strength (B0)
EPR and NMR spectroscopy
1) EPR spectrum: absorbed electromagnetic
radiation as a function of the magnetic induction.
2) The magnitude of the applied magnetic induction is
smaller than in EPR but electromagnetic radiation of
larger frequency (microwave) is needed.
3) Spin-marker: compounds with stable and unpaired
electrons.
4) Kinetic measurements: to track translation and
rotation in the ms time domain.
1) NMR spectrum: absorbed electromagnetic
radiation as a function of the magnetic induction.
The area under the line (band) of the spectrum is
proportional to the number of the absorbing nuclea.
2) The position of the lines depends on the
interactions („chemical shift”): the structure of the
electron clouds modifies the local magnetic field
(induction) and therefore the condition for resonance
absorption is shifted. Excellent tool for
determination of the chemical structure.
The protons in three positions
offer three different groups of
lines.
CH2
quartet
Magnetic induction (gauss, 100 μT)
OH
CH3
singlet
triplet
Chemical shift
Magnetic Resonance Spectroscopy
Electron Paramagnetic Resonance (EPR) spectroscopy
1. Electrons have “spin”, - rotation of the charge about its axis generates a magnetic
field at each electron.
2. Electrons in orbitals with two electrons are spin-paired, - one with ms +½ (, a),
one with ms -½ (, ), - so that the spins and magnetic fields cancel ( ).
3. Most molecular bonds are formed by coalescence of atomic orbitals so as to
satisfy the lower energy state arising from spin-pairing. Most molecules therefore
have all orbitals occupied by magnetically “silent” electron pairs.
4. If an electron is added to or subtracted from such a molecule (by reduction or
oxidation) an unpaired electron is introduced which is not spin coupled, and
therefore acts as a magnet. Such molecules are said to be paramagnetic.
5. Some bonds have unpaired electrons, giving them an inherent paramagnetic
property.
6. In EPR spectroscopy, we take advantage of this paramagnetic property to measure
properties of the the paramagnetic center, and its interaction with other local
magnets. These interactions provide information about local structure, including
local environment, distances, angles, polarity, etc.
Electron Paramagnetic Resonance
The photon energy for resonance depends on
the applied field. The field aligns the spins, to
give a splitting of the electron energy levels,
with a population in the ms +½ state that is
greater than that in the ms -½ state. Absorption
of a photon at the resonance energy flips the
spins. The net absorbance is due to transitions
from ms +½ to ms -½ levels.
Relaxation
No field
External field applied
ms = -½
Energy
(n)
ms = ±½
E  hn  gB0
ms = +½
0
magnetic field strength (B0)
Absorption of microwaves
changes the energy level so that
the small fraction of spins in the
lower energy state are flipped into
the opposite orientation.
Saturation occurs when flips in
both directions occur with equal
probability. The power needed to
saturate depends on the relaxation
time of the spin transition.
The nomenclature for EPR
The general principles of EPR and NMR are essentially the same, but with the
differences in energies.
However, the nomenclature is different, mainly for historical reasons. The resonance
condition for EPR is described by the equations:
E  gB0ms
E  hn  E 1  E 1  gB0
2
2
Here, g, the dimensionless Landé g-value, describes the resonance energy. Values
around 2.003 are found for simple free-radical systems. The term  (written also in
form of μB) is the Bohr magneton:
 = eh/(4 me) = 927.4 x 10-26 JT-1 in SI units, or
n = /h = e/(4 me) = 13.996 x 109 Hz T-1 in frequency units, or
eV =  /(e/C) = 5.7884 x 10-5 eV T-1 in electron volt units, where (e/C) is the electron
charge, 1.602 x 10-19 C.
Nuclear Magnetic Resonance (NMR) spectroscopy
1. The atomic nucleus is made up of protons (+ charge) and neutrons (neutral).
2. Like electrons, protons and neutrons (or nucleons) are quantum mechanical
entities, and their energetic properties can be described by operators, wave
functions, and quantum numbers.
3. Both protons and neutrons have a spin ½, with spin quantum number, mI,
which can have values of  (up, +½, or a), or  (down, -½, or ).
4. Total nuclear energy levels are lower if the constituent protons and neutrons are
spin-paired, so that for most nuclei, the spins (and magnetic fields) cancel.
Protons are spin-paired with protons, neutrons with neutrons.
5. Isotopes that have an odd number of either protons or neutrons have a net spin,
according to the Table below:
Number of protons
Number of neutrons
Nuclear spin, (I)
Even
Even
0
Even
Odd
1/2
or 3/2 or 5/2 …
Odd
Even
1/2
or 3/2 or 5/2 …
Odd
Odd
1 or 2 or 3 …
Nuclear Magnetic Resonance
In the absence of an external magnetic
field, the spins of the nuclei are arranged
more or less at random. When a magnetic
field is applied, the spins align with the
field, just like bar magnets. However,
because the energy involved is so small,
they can flip direction relatively easily.
mI = -½
Energy
(n)
mI = ±½
E  hn 
B0 h
2
mI = +½
0
magnetic field strength (B0)
The consequence is that the energy level of the nucleus splits in a magnetic field,
as shown on the right. The population of spins in the mI +½ (a) state is (very
slightly) higher than in the mI -½ () state, because the energy is lower.
Resonance conditions at NMR
When photons of the “right” energy are absorbed, the spin of the nucleus flips between
the two states. If we look at the diagram, we can see that the energy gap between the
two states is dependent on the strength of the applied magnetic field (magnetic
induction, B0). As a consequence, photons are absorbed which have a particular energy
at a particular field. This is what is meant by resonance.
E  hn 
B0 h
2
Here,  is the gyromagnetic ratio for the nucleus, B0 is the magnetic induction, and h is
the Planck’s constant.
When we come to discuss pulsed NMR, we will need to refer to circular motion. Here
the preferred frame of reference is that of the circular motion of precession. We define
the precession or Larmor frequency:
0  2 n    B0
Some consequences of the energy scale.
The transitions leading to NMR absorption have energies in the radio frequency
range, depending on nucleus () and the strength of the magnetic field generated by
the magnet.
NMR machines are rated by the frequency at which the proton is in NMR
resonance for the magnet they are built around, so we have 200 MHz, 500 MHz,
750 MHz and even 1 GHz NMR spectrometers. To achieve the higher fields, high
electrical currents are needed, which can be achieved using superconducting coils,
- these are generally called superconducting magnets (costing $M).
Since  = 2n = B0, for a 200 MHz machine we need a magnet generating 4.7
Tesla; for a 500 MHz machine we need 11.74 Tesla, etc.
As we increase the energy gap (increase frequency), the small differences of energy
for transitions of nuclear magnets in different environments are also increased, and
our NMR spectrum will be better resolved. In addition, we also increase the
population difference for the two spin states as we increase the energy (E/RT is
increased), as discussed in the next slide.
These factors make a big difference in the amount of time needed to generate a data
set, - for example in solution of the structure of a protein.
As we have seen, the energy of electromagnetic waves is generally expressed in
frequency, but energy scales are all related through Planck’s constant and the speed
of light, so we can express these energies in J/mol, eV, or any other energy units.
For a particular temperature, any transition can be described in terms of an
equilibrium between two states. For example, when we flip the energy level of a
proton in a magnetic field, we have two states,  or a, and  or , separated by E,
calculated as above. Let us use Na and N to represent the relative populations in the
two states. We can represent the equilibration between these two states by Na  N .
Then the ratio of these populations is given by the Boltzman distribution:
Na / N = e E / RT
Remembering that the energies of the transitions associated with NMR (and those of
EPR) are both very much less than RT, we can see that only a very small excess of
states will be in the lower energy level. For the proton at 500 MHz, at ~25oC, since E
= hn, and Planck and Avagadro constants are 6.626.10-34 J.s and 6.02.1023 mol-1
respectively, energies in the RF range (~500 MHz) are 0.2 J/mol.
Na / N = 1.00008
It is these one in ten thousand spins that are available to absorb a photon to
provide an NMR signal.
Molecular vibrations
Vibrational spectroscopy
Frederick William Herschel
(~1800)
He discovered the non-visible
(infrared) part of the spectrum of the
sun.
Elektromagnetic spectrum
X-RAY
0.2 nm
ULTRA-VIOLET
VISIBLE
2 nm
400-800 nm
INFRARED
NEAR IR
wavelength (cm)
wavenumber (cm-1)
7.8·10-5 to 3·10-4
12820 to 4000
MICROWAVE
3 mm-20 cm
RADIO
10 m-30 km
Mid IR FAR IR
3·10-4 to 3·10-3
4000 to 400
3·10-3 to 3·10-2
400 to 33
Information about the molecules
Overtones and combinations
of the normal vibrations
Basic (normal)
vibrations
Rotations
3. Particles (atoms and molecules)
in parabolic potential well
The energy profile of a harmonic oscillator is parabolic.
The particle is freely moving inside the parabolic
potential well.
Semiclassic treatment
The total energy of the particle is E  1 k x02  1 m v 2
2
2
L  2  x0
which is equal to the kinetic energy between the walls.
Those energy levels are allowed which can be covered
with standing waves
n
n
h
Possible velocities of the particle
Possible energies of the particle
1
n2 h2
k
2
E n  m vn 

2
8m 2En
E n
mn

4
n  hn

nh
nh
k


4 x0 m 4 m 2 E n
vn 
where
n  0,1,2,3,...
2
L
Energy levels of the harmonic oscillator
Uniform ladder of spacing h·ν
Solution from the Schrödinger equation:
En = h·ν(n + ½), where n = 0, 1, 2, ...
Conclusions:
(a) The energy levels of the oscillator are
equidistant: ΔE = h·ν.
(b) The selection rule: Δn = ± 1.
Transitions are allowed between adjacent
states only (in first order). Transitions
among not neigboring levels are forbidden.
(c) The minimum energy of the oscillator
(n = 0) differs from zero: E0 = ½ hν. The
zero-point energy is not zero, because of
For typical molecular oscillator, the zerothe uncertainty principle: the particle is
point energy is about
confined, its position is not completely
E0 = 3·10-20 J = 100 meV
uncertain, and therefore its momentum,
The Boltzmann-energy at room temperature and hence its kinetic energy, cannot be
exactly zero.
is ½ kBT = 25 meV
Infrared (IR) spectroscopy
The energy levels are associated with vibrations of bonds in molecules.
Each bond type will absorb IR light at one wavelength only, but there are
different modes of vibrations:
Bond
Wavelength (μm)
(a) valence vibrations and
C – N (aliphatic)
3.3 – 3.7
(b) straching vibrations.
C – H (aromatic)
3.2 – 3.3
All (arbitrarily complex) vibrations
of a molecule can be
decomposed into the sum of
normal modes.
C–O
9.55 – 10.0
C – O (aldehydes)
5.75 – 5.80
S–H
3.85 – 3.90
Example: normal modes of vibrations of CO2
Normal vibrations
antiszimmetric
szimmetric
R
R
H
R
R
R
H
R
H
R
H
R
stretch
H
H
R
scissoring
R
H
H
in plane bending
rocking
H
R
H
R
H
H
out of plane bending
wavenumber, 1/λ
Measuring methods
IR
Reflexion
IR Transmission
IR
Absorption
IR Reflexion
IR Transmission
Detector
IR light
IR light
Detector
Advantages of the IR
spectroscopy
Non-invasive determination of the sucrose content of the blood
Advantages of the IR
spectroscopy
Remote sensing and measurement of hazardous materials
IR light source
Sample
Grating
Detector
Fourier transform (FT) techniques in the infrared (IR)
spectrometers (FTIR)
The heart of the FTIR spectrometer is a Michelson
interferometer which works by splitting the beam from the
sample into two and introducing a varying path difference x
into one of the two arms. When the two components
recombine, there is a phase difference between them, and
they interfere either constructively or destructively
depending on the extra path that one has taken. The
detected signal oscillates as the two components
alternately come into and out of phase as the path
difference is changed. If the radiation has wavenumber k
(=1/λ), the detected signal varies with x as
I ( x)  I (k )  1  cos2 k  x)
The problem is to find I(k), the
variation of intensity with
wavenumber, which is the spectrum
we require, from the record of values
I(x). The step is a standard technique
of mathematics: Fourier
transformation.
The interferometer converts the presence of a particular
component in the signal into a variation in intensity of the
radiation reaching the detector. An actual signal consists of
radiation spanning a large number of wavenumbers, and
the total intensity at the detector is the sum of all their
oscillating intensities


I ( x)  I ( k )  1  cos(2  k  x)d k
known
0
unknown
Raman spectroscopy
About 1 in 107 of the incident
laser photons collide with the
molecules in the sample, give
up some of their energy to the
molecular vibrations, and
emerge with a lower energy
(Stokes radiation).
Other incident photons may
collect energy from the
molecular vibrations, and
emerge as higher-frequency
(anti-Stokes) radiation.
Basic properties:
An advantage of Raman spectroscopy over infrared
spectroscopy is that the radiation can be entirely in
the visible range, so the complications arising from
needing to select a range of infrared-transparent
sample cells are avoided.
-The shifts in frequency are
small (good monochromator is
needed) and
-- the intensities of the lines
are low (high laser energy is
needed).
Comparison of Raman and IR
spectroscopies:
advantages and disadvantages
The two methods utilize complementer properties of the same molecule.
Differences between IR and Raman methods
Raman
IR
1
It is due to the scattering of light by
the vibrating molecules.
It is the result of absorption of light by
vibrating molecules.
2
The vibration is Raman active if it
causes a change in polarisability.
The vibration is IR active if there is a change
in dipole moment during the vibration.
3
The molecule need not possess a
permanent dipole moment.
The vibration concerned should have a
change in dipole moment due to that
vibration.
4
Water can be used as a solvent.
Water cannot be used due to its intense
absorption.
5
Sample preparation is not very
elaborate sample can be almost in
any state.
Sample preparation is elaborate
Gaseous samples can rarely be used.
6
Gives an indication of covalent
character in the molecule.
Gives an indication of ionic character in the
molecule.
7
Cost of instrumentation is very high
Comparatively inexpensive.
Complexity and complementarity of the infrared (IR) and
the Raman spectra
wavenumber, 1/cm
Hoop: out-of-plane
motion of H atoms
streching of the methyl
group
rocking of the methyl
group
Single bond
vibrations
deformation of the
methyl group
Double bond
vibrations
Electron Transitions
to rebrush your mind
see the previous chapters on
„Electron in hyperbolic and rectangular shape of
potential wells”
and
for additional information
see the next slide-show on
„Visible absorption and fluorescence
spectroscopy: Luminescence”
Problems for Seminar
1. An electron is confined to a molecule of length 1.0 nm. What is (a) its
minimum energy and (b) the minimum excitation energy from that state?
2. Estimate a typical nuclear excitation energy by calculating the first excitation
energy of a proton (mass of m = 1.673·10-27 kg, the Planck’s constant is h =
6.626·10-34 J·s) confined to a region roughly equal to the diameter of a
nucleus, 1·10-14 m.
3. Calculate the zero-point energy of a harmonic oscillator consisting of a
particle of mass 2.33·10-26 kg and force constant 155 N/m.
4. For a harmonic oscillator consisting of a particle of mass 1.33·10-25 kg, the
difference in adjacent energy levels is 4.82·10-21 J. Calculate the force
constant of the oscillator.
5. Calculate the minimum excitation energies of (a) a pendulum of length 1.0 m
on the surface of the Earth, (b) the 33 Hz quartz crystal of a watch and (c)
the bond between two oxygen atoms in the O2 molecule, for which the force
constant is k = 1177 N/m.
Problems for Seminar
6. Determine the wavefunction and the probability distribution of the harmonic
oscillator in the first excited state. Compare these functions with those of the
ground state.
7. Calculate the longest and the shortest wavelength transitions in the Balmer
series of atomic hydrogen.
8. Do the same with the Lyman and Paschen series of the H atom.
9. Calculate the saparation between the two lowest levels for an oxygen
molecule in a one-dimensional container of length 5.0 cm. At what value of n
does the energy of the molecule reach 1/2·kBT at 300 K, and what is the
separation of this level from the one immediately below?
10. To a crude first approximation, a π-electron in a linear polyene may be
considered to be a particle in a one-dimensional box. The polyene β-carotene
contains 22 conjugated C atoms, and the average internuclear distance is 140
pm. Each state up to n = 11 is occupied by two electrons. Calculate the
separation in energy between the ground state and the first excited state in
which one electron occupies the state with n = 12. What is the wavelength of
the radiation required to produce a transition between these two states?
Problems for Seminar
11. The inverse value of the reduced (effective) mass of a diatomic molecule is
the sum of the inverse values of the individual masses of the atoms mA and mB:
1/μ = 1/mA + 1/mB. The table contains the infrared absorption wavenumbers (in
cm-1) of the following diatomic molecules:
H35Cl
H81Br
HI
CO
NO
2990
2650
2310
2170
1904
Calculate the force constants of the bonds and arrange them in order of
increasing stiffness.
12. An Ar atom rotates in a circle around a fixed axis (z) with orbital angular
momentum quantum number ml = 2. Its energy of rotation is Erot = 2.47·10-23 J.
Calculate the radius (r) of the rotation. (The angular momentum about the axis
(Jz = r·p) is quantized: Jz = ml ·(h/2π).)