LASER Induced Fluorescence of Iodine Eðlisefnafræði 5 – 30. mars 2006 Ómar Freyr...
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Transcript of LASER Induced Fluorescence of Iodine Eðlisefnafræði 5 – 30. mars 2006 Ómar Freyr...
LASERLASER Induced Induced Fluorescence of IodineFluorescence of Iodine
Eðlisefnafræði 5 – 30. mars 2006Eðlisefnafræði 5 – 30. mars 2006
Ómar Freyr Sigurbjörnsson Ómar Freyr Sigurbjörnsson
Introduction
Iodine is the heaviest common halogen and exists as a solid at room temperature in sublimation equilibrium with its vapor.
Its vapor has the appearance of a violet gas, indicating a visible absorption. This absorption corresponds to a spin-forbidden transition from the lowest vibrational levels of the singlet electronic ground state to high vibrational levels of a triplet excited state.
B 3u X 1g
By laser excitation, the reverse process of fluorescence is studied
The experimental setup
Experimental
Iodine crystals were inserted inside the fluorescence cell and the cell was then vacuated and heated
The monochromator was set to a wavelength of 630 nm and slidt width of 2000 m for the indirect absorption measurement by scanning the Dye-Laser and recording the amount of fluorescence
The fluorescence spectra was then recorded by scanning the monochromator 1 nm/min and slit width of 500 m for a fixed Laser excitation wavelength
Getting a strong signal was difficult and the slit width could not be smaller, resulting in lower resolution
Optimally the scanning speed should be slower
Iodine absorption spectrum
Generally an absorption spectrum can mostly give us information on the physical characteristics of excited states of molecules
Alternatively, emission spectra contain information on the ground state
Measuring directly the absorption spectrum of a molecule using Laser light is impractical The light is so strong that almost no difference is detectable due to
molecular absorption So the absorption spectrum of Iodine is measured indirectly by
measuring the amount of fluorescence as a function of laser wavelength
Recorded Iodine fluorescence as a function of Laser wavenumber
120
100
80
60
40
20
Rel
ativ
e I
nte
nsity
1704017030170201701017000cm
-1
Rotational peak (17010.2 cm-1
) selected for Laser excitation
588,2 586,5 nm
Iodine absorption spectrum
Rotational structure is observed due to the vibrational transition v’=16 v’’=2 (B X) as determined by the UV/VIS absorption spectrum recorded and assigned in an experiment performed in Physical Chemistry 2
Laser wavelength of 17010,3 cm-1 (587,88 nm) is selected to excite iodine molecules to the specific rotational and vibrational state, then the fluorescence to the ground state vibrational levels is recorded
Iodine fluorescence spectra
The expected Iodine fluorescence spectra should have peaks corresponding to different vibrational levels of the ground state
Also, each vibrational peak should be rotationally split in two due to the fact that for each transition, the selection rule J= ± 1 applies
Recorded Iodine fluorescence spectra
633 nm 593 nm
160
140
120
100
80
60
40
20
0
Rel
ativ
e in
tens
ity
169001680016700166001650016400163001620016100160001590015800 [cm
-1]
v´´= 3
v´´= 4
v´´= 5v´´= 6
v´´= 7
Analysis of measured and calculated peak positions
A transition between two electronic states is described by the equation (neglecting rotational contribution)
(v",v’ ) = el + G(v’) - G(v")
v’’+½ measured [nm]
measured [cm-1]
calculated [cm-1]
3,5 593,5 16850 16892
4,5 600,7 16645 16682
5,5 608,7 16429 16474
6,5 615,8 16238 16267
7,5 623,9 16028 16061
Where G(v) = e(v+ ½) - ee (v + ½)2
By plotting measured against v’’+½ the constants e and ee can be determined
16800
16700
16600
16500
16400
16300
16200
16100
[c
m-1
]
7654V''+1/2
y = 1,0714x2
- 216,89x + 17596
measured [cm-1]
Huber & Herzberg [cm-1]
e216,9 214,50
ee1,07 0,614
Results
Discussion The rotational split mentioned before can be seen in the recorded
spectrum shown earlier If the data and resolution are good enough, analysis can be
performed to assign the rotational quantum number J’ and rotational constants B for the ground vibrational states
The following equation describes the relationship E(J’,V’’) = 2(2J’+1)Be’’ - e’’ 2(2J’+1)(v’’+ ½) By plotting E as a function of v’’+ ½ and using known values for
Be’’ & e’’, the constant J’+1 can be found and consequently other B values
Unfortunately my recorded spectrum does not yield satisfactory results by such treatment
Discussion
There are a few things that could explain this Problems with signal strength and noise Scan speed and slit width to high Some data points from laser pulses are lost/do
not register on the computer so the data aquisition is not completly continous
More peaks and more scans are required for a thorough treatment
I wish to thank Mr. Wang for his help in performing this experiment