Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Material Type: Lab; Class: Optical Spectroscopy; Subject: Physics; University: University of Illinois - Urbana-Champaign; Term: Unknown 1989;
Typology: Lab Reports
1 / 14
This page cannot be seen from the preview
Don't miss anything!
In this week’s lab, we will use a fluorometer in Prof. Clegg’s lab to study some issues involved with fluorescence measurements. The schematic design of a fluorometer is included below. This particular one has two emission monochromators (right and left sides – we will be using the right side monochromator).
When making quantitative fluorescence measurements, it is important to correct the measured spectra to account for losses throughout the system. Spectral corrections come in three forms: (1) Instrumental corrections (a) Excitation-side corrections (b) Emission-side corrections (2) Sample corrections
Instrumental Corrections Instrumental corrections on the excitation side are handled by the reference detector. This is done by splitting off a known fraction of the beam incident on the sample and measuring its intensity I 0 (λ) as a function of wavelength. The detected fluorescence Id(λ) is then normalized by dividing the measured intensity by the incident intensity at each wavelength. In effect, this assumes that the sample is excited with uniform, white-light excitation. It accounts for any variations or fluctuations in lamp power, as well as losses in the excitation monochromator. The excitation-side corrections are generally handled by the instrument by selecting to record in ratio or signal/reference mode.
The emission-side corrections , while similar in origin, must be handled differently by the instrument. In this case, we wish to know the probability p(λ) that an emitted photon will be successfully detected. Knowing this, we can back- track from the measured intensity to determine the fluorescence intensity I(λ) emitted by the sample. This correction will account for all losses in traveling from sample to detector including chromatic aberrations in the optics and the monochromator and detector efficiencies. These correction values are just the inverse of the relative probability of reaching the detector and triggering a response. They depend only on the emission wavelength λem. If the instrument has polarizers, it would also depend on the polarization of the emitted light. The correction is applied by multiplying the measured intensity at each wavelength by the appropriate value. In practice, the emission-side correction(s) are determining using either a calibrated lamp source or a series of known fluorophores with well-established emission spectra. In either case, the measured spectra (also known as technical spectra) are directly compared to the known source data to determine the probability curve.
We will measure the raw excitation and emission photon counts for the fluorescence spectra. Then, in your analysis, you will make the excitation side instrumental correction by dividing out the lamp excitation intensity measured by the reference detector. Then you will account for the
Fig. 1: Emission side spectral correction
Use the Agilent 8453 absorption spectrometer you’re familiar with to measure the absorption spectra. At a minimum, the spectra should include the excitation wavelength and the full range of wavelengths you used for the emission spectra. Make sure you place the cuvettes in the cuvette holder in the same orientation for each sample. (Also be aware that the cuvette holder tends to leave a mark along one side of the cuvette so it is best to keep track of this.) Save a spectrum on the computer. Remember to blank the sample to the solvent 0.1N NaOH. Record your data in the following table:
absorption fluorescence sample λmax [nm] A (^) max λem [nm] I (^) em [A.U.] 0.1N NaOH -- G F E D C B A
sample % glycerol λabs [nm] A (^) max A 620 λem [nm] I (^) em [A.U.] Cy5A 0 Cy5B 20 Cy5C 40 Cy5D 60
The numerator is proportional to the total fluorescence signal coming out of the sample and the denominator is proportional to the total absorption of the sample. You will calculate the relative quantum yield from:
where I (^) F,em,peak is the fluorescence intensity at the peak of the emission spectrum and Aλ,ex is the absorbance at the excitation wavelength.
To get the absolute value of the quantum yield, we will use the known value of 0.27 for the quantum yield of Cy5 in aqueous solution (0% glycerol). The formula to calculate the absolute yield based on this is:
where n is the index of refraction of the solvent. You will need the following data:
% glycerol n viscosity (cP) 0 % 1.3330 1. 20 % 1.3663 2. 40 % 1.39768 6. 60 % 1.426 21.
photons absorbed
photonsemitted F
ex
rel Fempeak F A
,
, , λ
rel
rel
n
n
2
1 2 2
2 1 2
1 φ
φ φ
φ = ⋅
Appendix: Correction of fluorescence spectra for the inner-filter effect.
This note summarizes the derivation of the corrections for the inner-filter effect. The figure at the end shows the light paths under consideration as they travel through the cuvette / sample. The points labeled 1 to 5 (in the ovals) correspond to steps 1 to 5 below. The remaining steps are manipulations of these equations.
The probability p(x,y,z) is a function of the emission optics. In the first approximation, it is equal to the fraction of the solid angle striking the first optic (lens or mirror) collecting the emission. This probability does not include the corrections for the monochromator and detector efficiency (handled by the software) or the sample corrections (which we are treating explicitly in this calculation). In principle, we might determine p(x,y,z) using ASAP to model the fluorimeter hardware.
Note the identical form of the corrections for the excitation and emission wavelengths. All that is required to apply these corrections is information about the cuvette dimensions ( Li ), the detected volume ( Δi ), and the corrected absorption spectrum for the sample ( αi ).
i i
i i i i i
α
α α β (^) [10]
i i i
At low concentrations ( αF << 1 ), Nd#^ is directly proportional to the concentration of fluorophore (this must be true). At high concentrations ( αF >> 1 ), Nd#^ goes to zero (all the light is absorbed before reaching the center of the cuvette). Thus, somewhere between these end points, the detected signal must have a maximum value.
x
F
max max
LAMP
DETECTOR
1_d^2 N (^0) 3_d 3 N 2_d^2 N (^) x A
4_d 3 N (^) F
5_d^3 N (^) d
absorbed transmitted d 2 Nx+dx
0 x x+dx Lx
0
Ly
y
y+dy
ΔY
ΔX
N 0 = I 0 Δ y Δ z
Nd = Id Δ x Δ z
Detected fluorescence
confined to the volume
Δ x by Δ y by Δ z about the
center of the cuvette
LAMPLAMP
DETECTORDETECTOR
1_d^2 N (^0) 3_d 3 N 2_d^2 N (^) x 3_d 3 NA (^) A
4_d 3 N (^) F
5_d^3 N (^) d
absorbed transmitted dd 2 2 NNx+dxx+dx
0 xx x+dxx+dx Lx
0
Ly
y
y+dyy+dy
ΔY
ΔX
NN 00 = I= I 00 ΔΔ yy ΔΔ zz
NNdd = I= Idd ΔΔ xx ΔΔ zz
Detected fluorescence
confined to the volume
Δ x by Δ y by Δ z about the
center of the cuvette
Emission Wvl Excitation Emission ExcitationStdError EmissionStdErr 480 54024.9 2675.2 148.1 87. 482 53927.8 1747.3 233.63 45. 484 53823.9 2290.3 271.02 37. 486 54448.4 3267.6 265.13 40. . . where excitation is the number of counts at the reference photomultiplier tube, emission is the number of counts at the emission photomultiplier tube. Whether you’re measuring an excitation or an emission wavelength, your raw spectrum is given by the emission column.
For each spectra, you need to do the following:
( ) ( ) 2 2
ln(10) ( ) ln(10)^ (^ ) 2 2 10 10 ln(10) ( ) ln(10) ( ) sinh (^) sinh (^2 )
ex em
ex x em^ y A A x y in fil corr ex x em y x (^) y
A A L L Em Em A A L (^) L
λ λ
λ λ
λ λ
⎛ ⎞ ⎛ ⎞ ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ − −
⎛ (^) Δ ⎞⎛^ Δ ⎞ ⎜ ⎟ ⎜^ ⎟ ⎜ ⎟⎜^ ⎟ = (^) ⎜ ⎟ ⎜ (^) ⎛ (^) Δ ⎞ ⎟ ⎛ (^) Δ ⎞ ⎜ (^) ⎜ ⎟ ⎟⎜^ ⎜ ⎟⎟ ⎜ ⎟ (^) ⎜ ⎜ ⎟⎟ ⎝ ⎝^ ⎠⎠ (^) ⎝ ⎝ ⎠⎠ where Δx and Δy are from the empirically determined detection volumes for your fluorometer. For our case, Δx = 1 cm and Δy = 0.1 cm works well. A(λex ) is the absorption at the excitation wavelength 470nm and A(λem) is the absorption at the emission wavelength you are scanning (i.e, for an emission spectrum, you need A(λem) is a function of the emission wavelength you are scanning. Note that your emission spectra start at 480 nm but the absorption spectrum starts at a lower wavelength. You should truncate off the data for the appropriate data points when you are doing your correction (although you’ll need to use the OD at 470 nm for A(λex ).
Turn in a plot of the emission spectra with the instrumental corrections. Use one plot for all the samples.
Turn in a plot of the emission spectra with instrumental and sample corrections. Use one plot for all the samples.
The fluorescein concentrations are the same in all 6 samples. Make a plot of the raw and the corrected emission peak value vs. phenolphthalein concentration (just use the OD at the absorption peak of phenolphthalein around 550 nm as the x-axis). How good are your corrections?
7) Excitation spectrum of fluorescein (G). Here, we’ll just use this excitation spectrum to take a look at the effect of excitation side corrections. Make a plot of the raw excitation spectrum, and the excitation spectrum where you make excitation side correction by dividing Em by Ex. Put both spectra on the same plot (either make a left and right axis plot or renormalize one of the spectra).
Follow the instructions in the experiment handout to calculate the quantum yield. For our purpose here, the only correction you need to do for the fluorescence spectra is to divide the Emission column by the excitation column to get Iem.
9). Calculate the absolute quantum yields for the Cy5 samples and prepare a plot versus the solvent viscosity. Present your table of data from p.7 of the experiment handout. Briefly explain why the quantum yield increases with increasing viscosity.
10). The factor of n^2 in the quantum yield calculation arises because the fluorescent rays are refracted as they exit the sample into air on their way toward the detector. Briefly explain why a larger index solvent leads to less light reaching the detector (all other things being equal). This is easy to derive using Snell’s Law, but I am not asking for the full derivation.
11). In the calculation of quantum yield, is it correct to use only the peak emission intensity? What about all the other photons of different wavelengths? What should actually be used in this calculation? Why were we able to use just the peak intensity in this case?
