Optical Spectroscopy Instrumentation Design
187
Integrate two regions of the spectrum:
•
The excitation spectrum from the background (zero level) before the peak to the same
•
background after the peak to give the total number of photons per second in the wavelength range of the peak, and therefore the number of excitation photons that are not
absorbed by the sample. L3
The photoluminescence emission spectrum from the background (zero level) before
•
the peak to the same background after the peak to give the total number of photons
per second in the wavelength range of the peak, and therefore the number of emission
photons that are emitted by the sample. P3
Store the spectrum and the two integrated photons per second signal rates,
•
L3 and P3.
When the light hits the sample in measurement 3, some light is absorbed, and A and a
fraction 1 – A are reflected or transmitted. The light that is not absorbed is reflected by the
integrating sphere inner surface and a fraction of this unabsorbed light is reabsorbed by the
sample, μ. It is not possible to know exactly the behavior of the sample in terms of reflectance, transmission, and refractive index when it is illuminated so we can only evaluate the
effect of secondary absorption in terms of the photons absorbed and photoluminescence
photons emitted. This is the reason two measurements are taken, that is, to simulate the secondary absorption and assume that the samples and sphere have the same manner regardless of where the excitation light comes from. Therefore, measurements 2 and 3 need to be
compared to determine the ratio of reabsorbed photons, μ. At the same time, some light,
1 – μ, is not reabsorbed and this excitation light leaves the sphere in an identical manner to
measurement 1. Thus, LB LA
µ
=
−
(
)
1
.
The absorption ratio, A, is determined from measurement 3. LA photons are in the excitation and A LA
×
are absorbed, while (
)
1− ×
A LA are reflected and μ of these reflected
photons are reabsorbed by the sample. Thus, LC LA
A
µ
=
−
−
(
)(
)
1
1
.
Assuming that absorption is independent of excitation wavelength (assuming a narrow excitation line), then from the preceding equations we can derive the following;
A = ( –
/
)
1 LC LA . In measurement 2 all the photoluminescence comes from the reabsorbed light whereas in measurement 3 it originates from the direct absorption of light
and also the secondary reabsorption effect. This means that the total integrated spectrum
(signal) for measurement 3: the excitation and emission comprises these two contributions.
X and Y. Thus, LC PC X Y
+
= + .
The unabsorbed part is determined by measurement 2 and hence Y = ( ) +
– (
)
1 A LB PB .
The directly absorbed light is re-emitted by the sample and hence X = ηALA, where η is
called the quantum efficiency. Thus, LC PC
+
=
+ ( ) +
– (
)
ηALA
A LB PB
1
. Therefore the
photoluminescence quantum efficiency, η, can be expressed as shown in Eq. (5.15):
η =
− −
PC
A PB
ALA
(
)
1
(5.15)
where A
LC
LB
= −
1
.
187
Integrate two regions of the spectrum:
•
The excitation spectrum from the background (zero level) before the peak to the same
•
background after the peak to give the total number of photons per second in the wavelength range of the peak, and therefore the number of excitation photons that are not
absorbed by the sample. L3
The photoluminescence emission spectrum from the background (zero level) before
•
the peak to the same background after the peak to give the total number of photons
per second in the wavelength range of the peak, and therefore the number of emission
photons that are emitted by the sample. P3
Store the spectrum and the two integrated photons per second signal rates,
•
L3 and P3.
When the light hits the sample in measurement 3, some light is absorbed, and A and a
fraction 1 – A are reflected or transmitted. The light that is not absorbed is reflected by the
integrating sphere inner surface and a fraction of this unabsorbed light is reabsorbed by the
sample, μ. It is not possible to know exactly the behavior of the sample in terms of reflectance, transmission, and refractive index when it is illuminated so we can only evaluate the
effect of secondary absorption in terms of the photons absorbed and photoluminescence
photons emitted. This is the reason two measurements are taken, that is, to simulate the secondary absorption and assume that the samples and sphere have the same manner regardless of where the excitation light comes from. Therefore, measurements 2 and 3 need to be
compared to determine the ratio of reabsorbed photons, μ. At the same time, some light,
1 – μ, is not reabsorbed and this excitation light leaves the sphere in an identical manner to
measurement 1. Thus, LB LA
µ
=
−
(
)
1
.
The absorption ratio, A, is determined from measurement 3. LA photons are in the excitation and A LA
×
are absorbed, while (
)
1− ×
A LA are reflected and μ of these reflected
photons are reabsorbed by the sample. Thus, LC LA
A
µ
=
−
−
(
)(
)
1
1
.
Assuming that absorption is independent of excitation wavelength (assuming a narrow excitation line), then from the preceding equations we can derive the following;
A = ( –
/
)
1 LC LA . In measurement 2 all the photoluminescence comes from the reabsorbed light whereas in measurement 3 it originates from the direct absorption of light
and also the secondary reabsorption effect. This means that the total integrated spectrum
(signal) for measurement 3: the excitation and emission comprises these two contributions.
X and Y. Thus, LC PC X Y
+
= + .
The unabsorbed part is determined by measurement 2 and hence Y = ( ) +
– (
)
1 A LB PB .
The directly absorbed light is re-emitted by the sample and hence X = ηALA, where η is
called the quantum efficiency. Thus, LC PC
+
=
+ ( ) +
– (
)
ηALA
A LB PB
1
. Therefore the
photoluminescence quantum efficiency, η, can be expressed as shown in Eq. (5.15):
η =
− −
PC
A PB
ALA
(
)
1
(5.15)
where A
LC
LB
= −
1
.