Download L09 Instru Spectrofluorometery

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Dr.Emad Hamdan Done by:Noor Aswad
Spectrofluorometery
Fluorescent molecules: molecules that have the ability to emit light.
Fluorescent molecules first absorb light as in UV- the molecular
absorption. In UV we said that If we provide the electron with the
amount of energy ▲E (E2- E1) then the electron will jump to E2 , but
the electron won’t remain in E2 (unstable), it’ll return to E1 (after
fractions of second), because the electron prefers the most stable
state, and it will lose the energy as heat, BUT here in Fluorescent
molecules, when the electron return to its ground state, it will lose at
least some of the energy as light (emit light).
Type of emitted
energy (when going
back to ground state)
Presence of the
excitation step
(absorption) in the
beginning
-
Molecular absorption
(UV)
All of the energy
emitted as heat
yes
Atomic
spectrophotometery
All of the energy
emitted as light
(because there are no
vibrational
excitations – no
bonds)
yes
Spectrofluorometery
At least part of the
energy emitted as
light
yes
Not all the molecules have the ability to emit light.
What kinds of molecules are more likely to lose at least some of the energy acquired in form
of light? Those who are more rigid or more loose?
The more rigid molecules because they have no ability to do vibrations and rotations --- not all
the energy is emitted as heat.
Typical exam:
Compound(I) is not fluorescent but if we make little medication on the structure, we’ll have
compound(II) that is highly fluorescent.
Compound (I)
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Compound (II)
Instrumental Analysis
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So, today we are talking about measuring emission, which must be proportional to the
concentration of the fluorescent compound, so this makes us able to know the concentration of
the compound when we do the measurement of its emission.
Wavelengths
In spectrofluorometery, in general, the emitted light has lower energy than that of the
absorbed light. So, in general the wavelength of the emitted light is longer than the
wavelength of the absorbed light (inverse proportionality).
-
The range of wavelengths we use here in this technique is the same as that we use in UV
and visible spectrophotometery (200-800nm).
Conclusion: the absorbed light in UV range (shorter wavelength) may be emitted as light in
visible range (longer wavelength) and this is the most common case. In other cases: maybe both
of them (absorbed+emitted) in UV range or both in visible range.
Is it the same case of the spontaneous decay of radioactive elements (isotopes) like uranium
and plutonium? Is it as the same as what we said about fluorescent molecules?
In fluorescent molecules, we must give the molecules certain amount of energy (light must be
absorbed- excitation) to emit the light BUT in radioactive elements, they emit light
SPONTANEOUSLY; we don’t have to supply light for absorption first.
What are the qualifications required to have a fluorescent molecule?
1- Chromophore. To have the ability to absorb light first (excitation step).
2- Rigid molecule. To insure that not all of the energy will be lost as heat.
Spectrofluorometer
We said that fluorescent molecules first they must
absorb light, so we must have light source. Light
source is made from xenon (Xe) or mercury (Hg).
Usually we have xenon. Why we have changed the
light source from Deuterium despite that we are
using the same wavelength range of UV and visible
spectroscopy? Because in spectrofluorometery we
need the strongest possible intensity of light to get
the highest possible sensitivity. Here we measure the
absolute intensity of emitted light which will depend
on the intensity of the original light source. (Here
sensitivity is directly proportional to the intensity of light).
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Instrumental Analysis
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Dr.Emad Hamdan Done by:Noor Aswad
In UV, we didn’t give the intensity of light that importance because the sensitivity was not
directly proportional to the intensity of light source, we were measuring ratios (absorbance=
𝑰°
log 𝑰𝒕).
Warning: because of the high intensity of light, don’t look at this light source because it will hurt
your eyes.
Conclusion: fluorescence is 10 to 100 times more sensitive than UV. (e.g.-if 100 times- if the
minimum conc. measured of a substance in UV is 10μg, then the minimum conc. measured of
this substance in fluorescence is .1μg). (10 to 100 times depending on the fluorescence yield).
Fluorescence yield (Φ): amount that expresses the intrinsic ability of molecule to fluoresce. As
the fluorescence yield increase, the fluorescence will be higher, so we’ll detect the compound
at better sensitivity (i.e. smaller concentrations can be measured).
Compounds that have zero fluorescence yields, this means that these compounds have no
fluorescence.
Continuing in the basic design of spectrofluorometer, the incident light from the light source
will go to the excitation monochromator (to have a specific wavelength). Then the light will go
through the sample cell then to emission monochromator.
The next table summarize the two steps:
Type of
monochromator
Spectrum
(in
fluorescence,
always we have
two kinds of
spectra and two
kinds of lambda
maxes)
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Absorption step
excitation monochromator
Emission step
emission monochromator
--- excitation spectrum
band spectrum (because of
associated vibrations and
rotations)with a lambda max at the
wavelength that has the highest
absorption (lambda max of
excitation)
--- emission spectrum
Band spectrum with a lambda max at wavelength
that has the maximum emission
(lambda max of emission)
- Note that the lambda max of emission is
longer than the lambda max of
absorption
Instrumental Analysis
Sheet#9
Dr.Emad Hamdan Done by:Noor Aswad
Then to measure the intensity of the monochromatic emitted light, we’ll use a photocell
*detector*(the photocell in UV designed to measure I° and It and take the ratio but here in
spectrofluorometer it will measure the intensity of the monochromatic emitted light).
Note the design of the spectrofluorometer:
The angle between the incident light and the photocell is 90° (but in UV all things were at the
same line). This is important here to avoid the interference from the incident light (It). we just
want to measure the emitted light. But how this design would prevent the interference of the
incident light despite the fact that the light moves in all directions? Light moves in all directions
from its source but after the light moves from the source, if we put one pore (slit) to allow light
to move through, then the light just will move in that direction (only straight line- no other
direction) and this is what happens with the incident light. The emitted light will go in all
directions because the source of it is the sample cell (sample solution itself). This is a trick to
differentiate between the incident light and the emitted light.
Applications of spectrofluorometery:
1- Quantitatively.
2- Qualitatively (identification). Fluorescence has higher identification power than the UV
because we have more parameters in spectrofluorometery (more identifiers: lambda
max of emission is extra here).
- If two compounds have the same lambda max of excitation (e.g.280nm) and the same
absorptivity, BUT very unlikely (but not impossible) for both to have the same emission
profile, lambda max of emission, and the same fluorescence yield. So here, fluorescence
increased the identification power.
Mostly we use it for quantitative analysis (more common) but we can use it for
qualitative analysis. (Even in plasma)
What about selectivity and sensitivity of this method?
Sensitivity, we’ve talked about this at the beginning of this lecture but to summarize:
We said that in general, fluorescence is more sensitive than UV, and the reason of this
that in spectrofluorometery we measure the intensity of emitted light that is
proportional to the intensity of light source (i.e. when we increase the intensity of light
source, we’ll increase the sensitivity of the instrument and to get this we used xenon
light) but this in not applicable on UV (because we measure ratio).
How can we use spectrofluorometery in quantitative analysis (to know unknown
concentrations)?
As in UV, from the calibration curve, we plot the concentration against the intensity of
light emitted.
However, be attention to one thing here, that is, practically, the intensity of emitted
light is not reproducible, because the intensity of emitted light may be affected by the
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Instrumental Analysis
Sheet#9
Dr.Emad Hamdan Done by:Noor Aswad
intensity of light source (lamp that we use and its voltage), so for the same solution we
measure, readings are not stable, and this is a problem (we didn’t face such a problem
in UV because we were measuring a ratio).
To solve or to minimize this problem we we’ll do the next trick:
We’ll start measuring the highest concentration of the compound prepared (standard
solution) for the calibration curve (e.g. 80μg/ml) and the machine will give you for this
an absolute intensity, then we’ll require from the machine to make this value the
median (the machine will have this function), then the machine will understand to relate
all the next readings to this median value (i.e. it’ll take the ratio instead of taking the
absolute intensity and this will solve our problem of variability of the absolute intensity
values). Also, we’ll make the machine assume that this value is 100% (also this is a
function you can find in the machine). This is like taring of the balance or making a blank
in UV but here this won’t make the reading= zero but instead this’ll make the reading =
100% .
PLEASE DON’T CONFUSE:
We will set two values on the spectrofluorometer:
1- The blank (zeroing of the solvent without analyte) as in UV.
2- The highest concentration to assume it as 100%.
Then we’ll continue the calibration curve, and find the
equation. The slope is the intrinsic fluorescence of the
molecule (this is like the absorptivity in UV).
In the exam you’ll have the equation and you’ll find the
concentration.
Ideally, the y-intercept is zero. If it was positive then there is
interference of other matter that has fluorescence or from
Conc.
the machine. If it was negative then there is a matter that grabs this fluorescence.
Note: it is not the most important thing to have the highest intensity of light; BUT THE MOST
IMPORTANT THING is to have the SAME percent of the signal every time you measure (e.g.
30% of the single).
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