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```Spectral intensity, Beer Lambert’s Law, Raman scattering
Electronic transition (ultraviolet, visible range)
Figure 1. shows two electronic states, vibration and rotation energy levels
superimposed to electronic states.
Potential energy vs. internuclear distance functions
The potential energy for a diatomic molecule is internuclear distance (xAB)
dependent.
When interatomic distance is identical with the bond distance, the molecule has
potential energy minimum.
Superposition principle
Onto each electronic state a vibrational level series is superimposed and onto
each vibrational state a rotational level series is superimposed.
In Figure 1. a possible electronic transition is shown (grey vertical arrow) and a
vibrational transition in the electronic ground state by (black vertical arrow).
An electronic transition may occur when a molecule absorbs photon with
energy equals to one of the possible energy differences between excited and
ground state, and transition dipole has a value different to zero.
1
Absorption of light, Beer-Lambert’s law
The light intensity functions
The flux of photons, Φ is defined as the number of photons, np transmitted
through a cross sectional area, q during time, t.
np

q t
If the source is monochromatic the equation can be transformed to
energy flux multiplying the number of photons by hν which gives
photon intensity:
I    h 
h  np
q t
.
1.
1a.
unit: Wm-2.
The incoming radiation intensity I0 decreases when transferring an optically
active sample. The outgoing radiation intensity, I equals the difference in
I = I0 – I a
2.
between incoming and absorbed intensities, accordig to the law of energy
conservation
Figure 2. The radiating power or intensity is reduced by an absorbing medium of
b optical path.
In photometry, we use relative intensities, dividing Eq. 1. by I0 we get
I
Ia
 1
I0
I0
3.
2
a quantity called transmittance.
I
T
I0
The percentage transmission, T%
4.
T % 100  T
Beer-Lambert law
Increasing the thickness l (optical pathlength) of an absorbing layer by dl the
transmitted intensity decreases by dI. Therefore, this derivative

dI
dl
is negative, and directly proportional a material coefficient ε, the outgoing
intensity, I and the concentration of light absorbing material c in the solution,.

dI
 I   c
dl
5.
Integrating this equation between limits
l
1

d
I



c
I 0 I
0 dl
I
the integrated function
 ln I  ln I 0     c  l
Rearranging and introducing the wavelength dependence
ln
I0
 A   c l
I
The function form:
A       c  l
6.
This form assign variables (asorbance, molar absorbance) dependence of other
variables. Be careful    is not a product!
The absorbance at constant wavelength is linearly dependent on the
concentration of absorbing material, optical path and specific absorbance (ε (λ)).
3
When concentration is given in mol dm-3 the coefficient ε is called molar
absorbance. The usual non SI unit of thickness is cm, and the unit of ε is
dm3mol-1cm-1.
ε depends on the nature of material, the matrix and the wavelength of radiating
source.
(Matrix refers to the components of a sample other than the analyte of interest.)
Intensity formulas
When the material is transparent: I = I0, T = 1 or T% = 100.
When the material absorbs all the incoming light: I = 0 or T% = 0
The logarithmic ratio of intensities called absorbance is frequently applied in
quantitative determinations.
I
7.
A  log10 0
I
The output data of spectrophotometers are transmittance, T transmittance
percentage, T% or absorbance, A.
Absorbance can be given in terms of transmittance,
1
A  log10   log10 T
T
7a.
or in terms of transmittance percentage
A  log10
100
 log10 100  log10 T %  2  log10 T %
T%
From intensity functions only the absorbance is directly proportional to the
concentration of analyte, therefore it is frequently used for concentration
determination.
The light absorbing components determine the absorbance of a multicomponent
system. E.g. dissolving iodine in CCl4 a violet coloured solution is obtained for
which the absorbance at 520 nm (the maximum of iodine absorption) is:
CCl
solution
iodine
A520
  520
 cCCl  l   520
 ciodine  l
4
4
At 520 nm the CCl4 is transparent thus  520  0
CCl 4
4
solution
iodine
A520
  520
 ciodine  l
At constant c and l a spectral band can be observed. By increasing concentration
a band series is obtained.
0,5
Concentration series
b = const.
 = const.
Absorbance
0,4
0,3
0,2
0,1
0,0
200
300
400
500
600
Wavelength / nm
Figure 3. concentration series of an absorption band centered at 520 nm.
Concentration dependence
0,5
Absorbance
0,4
max = 400 nm
b = const.
3
 = 833,3 dm /(mol cm)
0,3
0,2
0,1
0,0001
0,0002
0,0003
0,0004
0,0005
0,0006
0,0007
3
Concentration / mol/dm
Figure 4. Translation of Fig. 3 at a single wavelength. The linear concentration
dependence
5
We read absorbance data at λmax from band series and plot the concentration
dependence of absorbance.
As BL law is valid in the concentration range studied, we have a straight line in
Figure 4. As BL law contains no additive constant the ideal graph crosses the x
axis at the origin i.e. at A=0, c=0 point. Experimental error causes the value of
intercept to be different from zero.
Example 1.
The molar absorption coefficient of a substance dissolved in hexane is known to
be 855 M-1cm-1 at λ = 270 nm. Calculate the percentage reduction in intensity
when light of that wavelength passes through 2.5 mm of a solution of
concentration 3.25 10-3 M.
Solution
Excess datum: λ = 270 nm
l = 0.25 cm
c = 0.00325 M
ε = 855 M-1cm-1
Percentage reduction = T%
A   c l
I
A  lg 0 ,
I
T%
1
 A
100 10
I0
 10 A
I
T% 
.
100
10cl
T% 
100
 20.2
106.947
Example 2.
Riboflavine in dilute acetate solution shows a maximum in absorption spectrum
at 444 nm. Calculate the wavenumber in cm-1 and frequency of light at this peak.
Calculate the energy of corresponding electronic transition for one molecule and
one mole of riboflavin.
Data: h = 6.626 10-34 Js, c = 3.00 108 ms-1, NA = 6.022 1023.
λ = 444 10-9 m = 444 10-7 cm
~
1
 22522.5 cm-1
7
 444 10
c
3 108
 
 6.756 1014 s-1
9
 444 10

1

For one molecule E  h   6.626 10
-34
6
 6.756 1014  4.476 1019 J
19
For one mole Emolar  E  N A  4.476 10
 6,02 1023  269455.2 J/mol
Example 3
A 2 mm path-length cell is filled 0.01 M benzene. The wavelength is set to
maximum wavelength of benzene, 256 nm. At this wavelength the transmitted
intensity was 48% of incident intensity. Suppose the solvent absorbance can be
neglected. What is the molar absorbance and absorbance of benzene at this
wavelength?
c = 0.01 M
l = 0.2 cm
excess datum: 256 nm
I = 0.48·I0
I
  lg 0.48  0.3187
I0
A 0.3187
 
 159.35
cl 0.01 0.2
A   lg
Acid-base equilibrium of weak acid indicators
Ionisation (dissociation) equilibrium of a weak acid:
HA ↔ H+ + AThe equilibrium constant at dT = 0 and dp = 0 is
H  A 
K


8.
HA 
Both the undissociated weak acid, [HA], and its dissociated anion, [A-], absorbs
light at different wavelength (see Fig. 5.).
7
absorption band of
A form
0,9
0,8
absorption band
of HA form
0,7
Abszorbance
0,6
pH = 4,0
pH = 8,5
0,5
0,4
pH = 6,5
0,3
0,2
0,1
0,0
-0,1
350
400
450
500
550
600
isobestic
650
700
750
 / nm
Figure 5. Dissolved in a pH=4 buffer the weak acid is undissociated, in a
pH=8.5 buffer solution it is in pure ionized form. At medium pH = 6.5 both
components are present with appreciable intensity. Equation 8. can be
transformed by introducing dissociation degree, α
[c] = α·c0
The α fraction of the total indicator concentration (c0) is dissociated.
The limits for α are: 0    1 .
α
K  H  
1 α
9.
Turning to common logarithm
lg K  lgH    lg
α
1 α
Introducing pH
8
 
 lg K   lg H   lg
pK  pH  lg
α
1 α
α
1 α
or
10.
Setting the pH of weak acid solution by a buffer the fraction of dissociation can
When α = 0.5, from Eq. 10., pK  pH
If interactions among light absorbing components of a solution can be neglected,
the measured absorbance at λ = constant is a sum of component absorbances.
 Ai  A1  A2  ...  An  ε1c1l  ε2c2l  ...  εn cn l
11.
Turning back to the dissociation equilibrium applying the maximum wavelength
of A- (see Fig. 5.).
The BL law for the two species
AHA  εHA  c0  l
AA  εA  c0  l

-
A  εHA 1  α c0l  εA α  c0l
-
A  AHA 1  α   AA- α
12.
and for α
α
A  AHA
and
AA  AHA

1 α 1
A A
A  AHA
 A
AA  AHA AA  AHA



Substitute for α and 1- α to Eq. 10 we get
9
13.
pK  pH  lg
A  AHA
AA-  A
14.
containing measured data: A, AHA, AA- and pH.
From analysis of Eq. 12 pK and consequently K can be determined.
1.Applying pH = 8.5 buffer solution (see Fig. 5.), the weak acid is totally
dissociated, α = 1, and from Eq. 12. A  AA2.Applying pH = 4 buffer solution, the weak acid is totally undissociated, α = 0,
and from Eq. 12. A  AHA
3. For medium pH Eq. 12. can be applied as it is.
By knowing the pH, measuring AA-, A and AHA the equilibrium constant can be
determined.
This part of lecture series is a task in practical semester.
Raman scattering, Raman intensities
History
Raman scattering is named after Indian physicist
C. V. Raman who discovered it in 1928, though predictions had been made of
such an inelastic scattering of light as far back as 1922. The importance of this
discovery was recognised even then, and for his observation of this effect
Raman was awarded the 1930 Nobel Prize in Physics. This was and remains the
shortest time from a discovery to awarding of the Prize.
Rayleigh scattering
Most light passing through a transparent substance undergoes Rayleigh
scattering. This is an elastic effect, which means that the light does not gain or
lose energy during the scattering. Therefore it stays at the same wavelength.
10
In Rayleigh scattering a photon interacts with a molecule, polarising the electron
cloud and raising it to a “virtual” energy state. This is extremely short lived (on
the order of 10-14 seconds) and the molecule soon drops back down to its ground
state, releasing a photon. This can be released in any direction, resulting in
scattering. In the sense of energy conservation, the energy of emitted photon is
the same as the incident one.
Fig. 1. Energy level diagram of scattering processes
Raman effect
Raman scattering is different in that it is inelastic. The light photons lose or gain
energy during the scattering process, and therefore increase or decrease in
wavelength respectively. In Raman scattering, the energies of the absorbed and
emitted photons are different.
The Raman effect comprises a very small fraction, about 1 in 107, of the incident
photons.
Stokes scattering
If the molecule is promoted from a ground to a virtual state and then drops back
down to a (higher energy) vibrational state then the scattered photon has less
energy than the incident photon, and therefore a longer wavelength. This is
called Stokes scattering.
Generally only the lowest vibrational level is occupied to any extent and the
transition v =1 ← v = 0 with rotational states superimposed gives the strongest
Raman band.
The Stokes Raman frequency is displaced to a lower frequency from the
Rayleigh scattered photon by a frequency equal to the vibrational frequency.
11
anti-Stokes scattering
If the molecule is in a vibrational state to begin with and after scattering is in its
ground state then the scattered photon has more energy, and therefore a shorter
wavelength. This is called anti-Stokes scattering.
An anti-Stokes line can be observed when photon collides a molecule in excited
state, and a higher energy photon is emitted. That lines are less intense because
of the population density of excited vibrational state is smaller than that is for
ground state.
Frequency shifts
Rayleigh line (most intensive):
 incident   scattered
Stokes shift
 vibration   incident  scattered
anti-Stokes shift
 vibration   scattered  incident
A Stokes and anti-Stokes line belonging to the same vibration mode are equally
displaced from Rayleigh line.
The intensity of Raman scattering is inversely proportional to the fourth power
of wavelength (IRaman ≈ 1/λ4).
Selection rule:
Recalling the definition equation of polarizability (defined in an earlier lecture
of this course)
μinduced  α  E
The polarizability, α, represents the ability of an applied electric field, E, to
induce a dipole moment, μind in an atom or molecule.
For example, large atoms such as xenon have a strong polarizability because
their electron clouds — distant from the xenon nucleus — are relatively easy to
distort with an applied electric field. Helium atoms, which are smaller and more
compact, have a small polarizability.
Polarizabilities for atoms are isotropic (that is, the same in all directions),
whereas polarizabilities for molecules may vary with position about the
molecule, depending on the molecule’s symmetry.
12
Fig. 2. Schematic representation of the anisotropic polarisability of a molecule
At a molecule’s equilibrium nuclear geometry, the polarizability is some value,
α0. At some distance, Δr, away from the molecule’s equilibrium geometry, the
instantaneous polarization α is given by
1.
the derivative
1a.
represents the change in polarizability with change in position. If that derivative
equals zero, the entire second term is zero and there will be no Raman
scattering.
Thus, we have as a gross selection rule: a molecular motion will be Ramanactive only if the motion occurs with a changing polarizability.
The schematic representations of the general set up of the monochromator based
dispersion Raman spectrometer is given in
13
Fig. 3. The experimental setup of a Laser Raman spectrometer. Signal detection
is done perpendicular to the direction of monochromatic beam.
Raman spectrum of CCl4
The position of peaks of Stokes and anti Stokes are symmetrical
Fig. 4. Laser source: argon ion laser, producing laser radiation of 488 nm.
14
Group frequencies
When light is scattered from a molecule or crystal, most photons are
elastically scattered. The scattered photons have the same energy
(frequency) and, therefore, wavelength, as the incident photons.
However, a small fraction of light (approximately 1 in 107 photons) is
scattered at optical frequencies different from, and usually lower than,
the frequency of the incident photons. The process leading to this
inelastic scatter is termed the Raman effect. Raman scattering can occur
with a change in vibrational, rotational or electronic energy of a molecule.
If the scattering is
elastic, the process is called Rayleigh scattering. If it’s not elastic, the
process is called Raman scattering. Raman scattering (or the Raman
effect) was discovered in 1928 by V. C. Raman who won the Nobel prize
for his work. If the substance being studied is illuminated by
monochromatic light, for example from a laser, the spectrum of the
scattered light consists of a strong line (the exciting line) of the same
frequency as the incident illumination together with weaker lines on
either side
shifted from the strong line by frequencies ranging from a few to about
3500 cm-1. The lines of frequency less than the exciting lines are called
Stokes lines, the others anti-Stokes lines Raman spectroscopy is very
important practical tool for quickly identifying molecules and minerals. A
Raman spectrometer was deployed on the Viking landers in 1972 and in
15
other missions. Raman spectroscopy also has important scientific
applications in studying molecular structure. In this experiment we will
study both kinds of applications
16
```