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Instrumental Analysis Sheet#9 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) 1|Page Compound (II) Instrumental Analysis Sheet#9 Dr.Emad Hamdan Done by:Noor Aswad 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). 2|Page Instrumental Analysis Sheet#9 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) 3|Page 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 4|Page 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). 5|Page