Spectrophotometry Part 3 Flashcards

(11 cards)

1
Q

Stray Light Mitigation

A
  • When making absorbance measurements you are limited to the instrument that is available.
  • If that instrument has an unfavorable stray light limit you will have to make do.
  • Let’s say the stray light limit is 5% like the slides from last lecture and your sample’s true absorbance is 2.00 in a 1.00 cm cell at 𝜆 max of your compound.
  • Can you think of ways to measure the absorbance of your sample on that instrument with decreased stray light effect?
  • Discuss in your groups and come back with some suggestions
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

Jablonski

A

See picture

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

Fluorescent Molecules

A
  • Compounds containing aromatic rings give the most intense and most useful molecular fluorescence emission.
  • Most unsubstituted aromatic hydrocarbons fluorescence in solution. Quantum efficiency increases with number of rings.
  • Simplest heterocycles do not exhibit molecular fluorescence, but fused-ring structures containing these rings often do.
  • Fluorene has much higher quantum yield than biphenyl due to effect of rigidity on quantum yield.
  • Likewise 8-hydroxyquinoline (aka oxine) is non-fluorescent while Zn(8-HQ) 2 fluoresces. aka Zn(oxine) 2
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

Two reasons for fluorescence’s enhanced sensitivity over absorbance

A
  • Absorbance is a ratio of incident and transmitted power, Po/P , all at the same wavelength
    A = log (Po/P)
  • Fluorescence intensity I(subf) at the emission wavelength 𝜆em is given by the equation,

where Φ is quantum yield and I ex is intensity at the excitation wavelength 𝜆em
* Fluorescence signal may be enhanced by…
* using brighter light, eg. 1000× larger I ex
* This would not help absorbance because increasing Po also increases P
* increasing the gain on the detector by 1000×
* This would not help absorbance because gain would make both Po and P 1000× larger

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

Fluorescence’s enhanced sensitivity over absorbance

A

See picture

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

Diffraction Grating

A
  • A diffraction grating is a reflective or transmissive optical component with a series of closely ruled lines that separates a beam of white light into multiple beams, called orders.
  • Most orders are further separated into wavelengths (colors).
  • The naturally transmitted beam is the zero (n = 0)-order beam and its colors are not separated.
  • All higher-order beams (n < 0 or n > 0) exhibit color separation because diffraction causes longer wavelengths to bend more strongly than shorter wavelengths.
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

Transmission Grating

A
  • Transmission grating with incident light normal to two grooves separated by distance d
  • Each groove acts as a radiation source from which transmitted light propagates at all angles.
  • When the light rays are diffracted at angle φ, the lower ray travels an increased path (d sin φ).
  • At angles φ where an integer number of wavelengths (n) is equal to the increased path length, the adjacent light rays are said to be in phase, and reinforce one
    another.
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

Czerny-Turner Reflection Grating Monochromator

A
  1. Polychromatic light (light of many wavelengths) enters through the entrance slit.
  2. The expanding beam of polychromatic light is collimated (made into a beam of parallel rays) and reflected to the grating by the
    concave collimating mirror.
  3. The grating:
    * Separates the beam into orders (only one is shown)
    * Within the selected order, diffracts different wavelengths at different angles
  4. The concave focusing mirror focuses each wavelength at a
    different point on the focal plane.
  5. The narrow range of wavelengths falling on the exit slit travel on
    to the sample. All other wavelengths are absorbed by the black paint on the monochromator walls.
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

Why do we need a monochromator?

A
  • The need for monochromator was driven home to me as I prepared a demo of Hollow Cathode Lamp (HCL)
  • HCL gives off ”orange” light emitted as Ar + of filling gas recombining with electrons, this appears at many wavelengths and overwhelms the line spectrum of the metal of interest.
  • Monochromator comes after furnace, or flame, or plasma, because each of these also emits light at many wavelengths.
  • For molecular absorption (UV-VIS) we need monochromator to get a narrow bandpass of light to obtain linear Beer’s Law plot
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

Dispersion

A

Dispersion measures the ability to separate wavelengths differing by
Δλ through the difference in angle, Δφ (radians). Dispersion increases
with decreasing groove spacing d and also depends on order n and
diffraction angle 𝜙

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

Molecular UV-VIS Spectroscopy – Mixtures

A

Select two wavelengths subject to the constraints
* Large difference in absorbance between the two compounds
* A changes slowly with wavelength
* i.e. near a 𝜆 max

How well did you know this?
1
Not at all
2
3
4
5
Perfectly