Transcript absorbance, a - srmbiotech25

INTRODUCTION
• Absorption Spectrophotometry in the ultraviolet and
visible region is one of the most oldest methods used
for quantitative analysis and structural elucidation.
• The ultraviolet and visible Spectrophotometry is
mainly used for quantitative analysis and used as a
auxiliary tool for structural elucidation.
Useful Terminology
 Photometer is an instrument which measuring the
ratio, or some function of the two of radiant power of
two electromagnetic beams.
 Colorimetry is the use of the human eye
to determine the concentration of colored species.
 Spectrophotometry is the use of instruments to make
the same measurements. It extends the range of
possible measurements beyond those that can be
determined by the eye alone. it use more sensitive
detectors.
Colorimetry
Visual Observations – Because colorimetry is based on
inspection of materials with the human eye, it is
necessary to review aspects of visible light.
Visible light is the narrow range of electromagnetic
waves with the wavelength of 400-700 nm.
ROY G. BIV= the mnemonic used to remember the colors of the visible spectrum.
Visible light is only a very small portion
of the electromagnetic spectrum.
Note: Frequency (υ) and Energy (E) are directly proportional
whereas Frequency (υ) and Wavelength (λ) are inversely proportional.
Electromagnetic Spectrum
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*
Electromagnetic radiation is characterized by its
wavelength, , Frequency,  and energy, E:
E = h= hc / 
c=
Where h = Planck’s constant & c = speed of light in a vacuum.
(a) longer wavelength, lower energy;
(b) shorter wavelength, higher energy.
Theory of spectrophotometry
• When a beam monochromatic light passes
through a transparent medium, part of the
light is absorbed and the transmitted beam
has a lower intensity than the intensity of the
incident beam.
• The TRANSMITTANCE, T of the solution is defined as
the ratio of the intensities of the transmitted
beam, P to the intensity, Po of the incident beam:
T = P/Po
• The ABSORBANCE, A of a solution is defined as
• Absorbance
A = log10 P0 / P
A = log10 1 / T
A = log10 100 / %T
A = 2 - log10 %T
A = -log10T
Since A is a logarithmic function, it is dimensionless.
Equation Summary
T= (I/Io) = 10-A
%T = (I/Io) x 100
A = -logT = log(1/T)
Sample Calculation
If %T = 95%, then
A = log(100/95) = log(1/.95) = -log(.95)
A = 0.02227
Note the scale for Absorbance: 9/10th of the scale is from 0-1 and 1/10th is from 1-2.
For this reason, the spectrometers have been calibrated in % Transmittance and
all readings will be taken in %Transmittance.
Lambert’s Law
• “When a beam of light is allowed to pass
through a transparent medium, the rate of
decrease of intensity with the thickness of
medium is directly proportional to the
intensity of the light”
- dI / dt α I
Or
- dI / dt = kI
• Integrating on both sides putting I=I0 When t=0, we get
In I0 / It = kt
or
It = I0 e-kt
Where k is the constant which depends on wavelength and
absorbing medium used.
• Converting to natural log
It = I0 10 -0.4343kt = I0 10 -Kt
Where K=k/2.3026 which is absorption coefficient, which is
defined as “the reciprocal of the thickness which is required to
reduce the light to 1/10 of its intensity”
I0 / It = 0.1=10-kt or Kt =1 or K α 1 / t
Beer’s Law
• Lambert’s law shows that there exists a
logarithmic relationship between the
transmittance and the length of the optical
path through the sample.
• Beers observed that a similar relationship
between the transmittance and the conc of a
solution. i.e “ the intensity of a beam of
monochromatic light decrease exponentially
with the increase in conc if the absorbing
substance arithmetically”
It = I0 e-k’c
It = I0 10 -0.4343k’t = I0 10 –K’t
Where k’ and K’ are constants and c is the
concentration of the absorbing substance.
Beer’s Law
•
•
•
•
•
It = I0 10 -act
or
log I0 / It = act
This is mathematical statement of Beer-Lambert law.
This is the fundamental for colorimetry and
Spectrophotometry.
Where c is the conc expressed in mole dm-3
t is the path length which is expressed in cm.
a is replaced with Є and is termed as the Molar
absorption coefficient or molar absorptivity (molar
extinction coefficient)
Beer-Lambert law
log I0 / It = act
A = act
A= Єct
Є=A/ct
Where A is absorbance (no units, since A = log10 P0 / P )
Є is the molar absorbtivity with units of L mol-1 cm-1
b is the path length of the sample - that is, the path length of the cuvette in
which the sample is contained. We will express this measurement in
centimetres.
c is the concentration of the compound in solution, expressed in mol L-1
Deviations from Beer’s Law
• Beer’s law states that a plot of absorbance versus
concentration should give a straight line passing
through the origin with a slope equal to ab. However,
deviations from direct proportionality between
absorbance and concentration are sometimes
encountered.
• These deviations are a result of one or more of the
following three things ; real limitations, instrumental
factors or chemical factors.
Real Limitations
• Beer’s law is successful in describing the absorption
behavior of dilute solutions only ; in this sense it is a
limiting law. At high concentrations ( > 0.01M ),the
average distance between the species responsible for
absorption is diminished to the point where each
affects the charge distribution of its neighbors.
• This interaction, in turn, can alter the species’ ability
to absorb at a given wavelength of radiation thus
leading to a deviation from Beer’s law.
Limitations (cont)
• Deviations also arise because e is dependent upon
the refractive index of the solution. Thus, if
concentration changes cause significant alterations in
the refractive index h of a solution, departures from
Beer’s law are observed.
Chemical Deviations
• Chemical deviations from Beer’s law are caused by
shifts in the position of a chemical or physical
equilibrium involving the absorbing species.
• A common example of this behavior is found with
acid/base indicators.
• Deviations arising from chemical factors can only be
observed when concentrations are changed.
eg:dichromate ions on dilution.
Cr2O72- + H2O -- 2HCrO4- - 2H+ + 2CrO4 2(orange color)
(yellow color)
Instrumental Factors
• Unsatisfactory performance of an instrument
may be caused by fluctuations in the powersupply voltage, an unstable light source, or a
non-linear response of the detector-amplifier
system.
• There are at least six conditions that need to be
fulfilled in order for Beer’s law to be valid. These are:
– The absorbers must act independently of each other;
– The absorbing medium must be homogeneous in the interaction
volume
– The absorbing medium must not scatter the radiation - no turbidity;
– The incident radiation must consist of parallel rays, each traversing the
same length in the absorbing medium;
– The incident radiation should preferably be monochromatic, or have at
least a width that is narrower than that of the absorbing transition;
and
– The incident flux must not influence the atoms or molecules; it should
only act as a non-invasive probe of the species under study. In
particular, this implies that the light should not cause optical
saturation or optical pumping, since such effects will deplete the lower
level and possibly give rise to stimulated emission.
Instrumentation
• Source: A continuous source of radiant energy covering the
region of spectrum in which the instrument is designed to
work.
• Filter or Monochromator: Both filter and monochromator
allow the light of the required wavelength to pass through but
absorb the light of other wavelengths.
• Sample cells: All instrument must contain a container for the
sample.
• Detector: It is used for measuring the radiant energy
transmitted through the sample.
Radiation sources
• Requirements of a radiation sources
– It must be stable.
– It must be of sufficient intensity for the
transmitted energy to be detected at the end if
the optical path.
– It must supply continuous radiation over the
entire wavelength region in which it is used.
Tungsten filament lamp
• It is most common source of visible radiation.
• Its construction is similar to the household
lamp. But the tungsten wire is heated at
controlled atm.
• Constant power supply is needed to achieve
constant radiant energy.
Drawback of tungsten lamp
• Major portion of its radiant energy in the near
IR region of the spectrum, i.e only 15% at the
operating temp of about 2725o c and only 1 %
at 1725o c.
• If the temp is increased to 2850o c and above
will increase the total energy output and shift
the wavelength of maximum intensity to
shorter wavelength but the life of the lamp
will be reduced.
• Tungsten lamp is the most satisfactory for
lamp filaments but the carbon arc is used
when a more intense source is required.
• The energy from a tungsten lamp can be used
above 375 nm.
Filters and Monochromators
• A Source is generally emits the continuous
spectra.
• We need a device to select the narrow bands
from wavelengths of the continuous spectra.
• For that selection we have
– Filters and
– Monochromators.
FILTERS
• A light filter is device that allow light of the
required wavelength to pass but absorbs light
of other wavelengths wholly or partially.
• Filter are two type
– 1) absorption filters and
– 2) interference filters
Absorption filter
• It is a solid sheet of glass on which the color
pigments are dispersed.
• Absorption filters are classed as either cut off
or band pass filters.
Color Wheel
(ROYGBIV)
Complementary colors lie across the diameter on the color
wheel and combine to form “white light”, so the color of
a compound seen by the eye is the complement of the
color of light absorbed by a colored compound; thus it
completes the color.
Observed Color of
Compound
Color of Light
Absorbed
Approximate
Wavelength of Light
Absorbed
Green
700 nm
Blue-green
600 nm
Violet
550 nm
Red-violet
530 nm
Red
500 nm
Orange
450 nm
Yellow
400 nm
Observed Color of
Compound
Color of Light
Absorbed
Approximate
Wavelength of Light
Absorbed
Green
Red
700 nm
Blue-green
Orange-red
600 nm
Violet
Yellow
550 nm
Red-violet
Yellow-green
530 nm
Red
Green
500 nm
Orange
Blue
450 nm
Yellow
Violet
400 nm
Interference filters
• These filters works on the phenomena of
interference.
• It has a semi-transparent metal film is
deposited on the glass plate.
• Then the metal film is coated with thin layer of
dielectric material like MgF2.
• Followed by another metal film and then glass
plate for support.
Interference filters
Monochromators
• It is separate the narrow wavelength more
specifically then filters.
• Monochromators have the following parts
– An Entrance slit
– A dispersing element (Prism or grating) and
– An Exit slit
• Replica grating are cheaper than prisms.
• Main disadvantage of the gratings is that they
produce more then one order of diffraction.
• For instance, the second order of 400 mμ may
interfere with the first order of 800 mμ.
• This can be removed my employing filters in
front of the entrance slit to absorb interfering
radiation.
Grating
• A Grating consist of large number if parallel
lines ruled on a highly polished surface such as
alumina.( 15,000 – 30,000 lines per square
inch)
• Grating are very difficult to prepared.
Therefore replica gratings are prepared from
an original grating using an epoxy resin.
• The main advantage of prisms is that they
undergo dispersion giving wavelengths which
do not overlap.
• But the main disadvantage is that they give
Non-linear dispersion.
• On the other hand gratings give linear
dispersion but they suffer from an overlap of
spectral orders.
• For this reason, filters are employed to reduce
the radiation of different orders and stray
radiation.
Cells
DETECTORS
• PHOTOVOLTANIC CELL
• PHOTOTUBES
• PHOTOMULTIPLIER
PHOTOVOLTANIC CELL
Photocell or Phototubes
Photoemissive cell
Photomultiplier tube
Power supply
• The power supply serve a triple function
– It decrease the line voltage to the instrument’s
operating level with a transformer.
– It convert alternating current to direct current
with a rectifier if direct current is required by the
instrument.
– It smooths out any ripple which may occur in the
line voltage in order to deliver a constant voltage
to the source lamp and instrument.
Visual comparators
Intensity: For light shining through a colored solution, the
observed intensity of the color is found to be dependent on both
the thickness of the absorbing layer (pathlength) and the
concentration of the colored species.
• There are four techniques for color
comparison in quantitative visual colorimetry.
– Multiple standard method,
– Duplication method,
– Dilution methods, and
– Balancing method-uses Klett Bio colorimeter or
Dubosque colorimeter
Multiple standard method
• The unknown solution is taken in a 50/100 ml
Nessler tube and made up to the mark.
• The color of the unknown solution is
compared with the series of known amount of
the known substance.
• The conc of the unknown will be equal to the
conc of known if the color matches exactly.
←Side view
←Top view
(a.k.a. Bird’s eye view)
For One Color: A series of solutions of a single color
demonstrates the effect of either concentration or pathlength,
depending on how it is viewed.
Visual Colorimetry
←Ratio used
←Purple produced
For more than one color: the ratio of an unknown mixture
can also be determined by matching the shade of the color to
those produced from known ratios.
In this example, the ratio of a mixture of red and blue can be
determined visibly by comparing the mixture to purples
produced from known ratios of red and blue.
Duplication method
• The unknown solution is taken in one Nessler
tube followed by the development of color
with appropriate reagent, and made up to the
mark.
• In another tube the solution of developing
reagent is taken and made up a little lower
than the mark on the Nessler tube.
• From the burette the std solution is added to
this tube until the color matches the unknown
solution.
Dilution methods
• Two Nessler tubes of equal diameter, height
and made out same quality of glass are used.
• To the first tube we add a standard solution
and to the second with the unknown conc
solution.
• Light from the same source is allowed to pass
through each cell and the emergent beams
are the compared.
• The more concentrated solution is
progressively diluted until the intensity of the
emerging light from the both the cells become
equal.
• At this point the concentrations per unit
volume of solution in both the tubes (std and
the unknown) should also be the same.
• As the two beams are equal in intensity, the
absorbance A is equal in each case.
A=εc1t1
A=εc2t2
from the above equations, we get
εc1t1 = εc2t2
But
t1 = t2
c1 = c 2
Therefore
• As the total quantity of material in the sample
remains equal to the original concentration x
original volume of sample.
c1 V2 = c(original) V1
------------- original
• but the final conc of the sample C1 is equal to
the conc of the std, therefore
c(original) =c1 V2 / V1 or
Conc of std x volume of solution after dilution
Original volume of solution
Balancing method
• Dubosque colorimeter is used.
• By varying the length of the light oath in two
solutions the color of the unknown is matched
with the std.
Intensity: When the product of the concentration and the
pathlength of any two solutions of a colored compound are the
same, the same intensity or darkness of color is observed.
Duboscq
visual
colorimeter
Adjustable
Path Lengths
problems
• 40 ml of the 0.5 M Fe3+ std is diluted with 25
ml of water in order to match the intensity of
the unknown. Calculate the Fe3+ conc of the
unknown solution.
• 50 ml of the unknown sample is diluted to a
total volume of 60 ml in order to match the
intensity of the original std. the final conc of
the unknown is 0.04 M. find out the original
conc of the unknown.
Single beam instrument
Double beam instrument
Advantage of double beam
• It is not necessary to continually replace the
blank with the sample or to zero adjust at
each wavelength as in the single beam units.
• The ration of the power of the sample and
reference beams is constantly obtained and
used.
• B’cos if the previous two factor the double
beam system leads itself to rapid scanning
over a wide wavelength region.