Transcript LASER flúrljómun joðs

LASER Induced
Fluorescence of Iodine
Eðlisefnafræði 5 – 30. mars 2006
Ómar Freyr Sigurbjörnsson
Introduction
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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 3P0+u  X 1Sg+
By laser excitation, the reverse process of
fluorescence is studied
The experimental setup
Experimental
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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 mm 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 mm 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
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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
-1
Rotational peak (17010.2 cm )
selected for Laser excitation
120
Relative Intensity
100
80
60
40
20
17000
17010
17020
17030
17040
-1
cm
588,2
586,5 nm
Recorded Iodine fluorescence as a function of Laser wavenumber
Iodine absorption spectrum
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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
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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 DJ= ± 1
applies
Recorded Iodine fluorescence spectra
v´´= 3
160
140
v´´= 4
Relative intensity
120
100
80
v´´= 5
v´´= 6
60
v´´= 7
40
20
0
15800
633 nm
15900
16000
16100
16200
16300 16400
-1
 [cm ]
16500
16600
16700
16800
16900
593 nm
Analysis of measured and
calculated peak positions
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A transition between two
electronic states is
described by the
equation (neglecting
rotational contribution)
(v",v’ ) = el + G(v’) - G(v")
v’’+½ lmeasured measured calculated
[nm]
[cm-1]
[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) = we(v+ ½) - wece (v + ½)2
By plotting measured against v’’+½ the constants we
and wece can be determined
Results
we
wece
16800
16700
measured
[cm-1]
Huber &
Herzberg [cm-1]
216,9
214,50
1,07
0,614
16600
2
16500
-1
[cm ]
y = 1,0714x - 216,89x + 17596
16400
16300
16200
16100
4
5
6
V''+1/2
7
Discussion
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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
 DE(J’,V’’) = 2(2J’+1)Be’’ - ae’’ 2(2J’+1)(v’’+ ½)
 By plotting DE as a function of v’’+ ½ and using known values for
Be’’ & ae’’, 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
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There are a few things that could explain this
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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
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I wish to thank Mr. Wang for his help in
performing this experiment