Chem. 133 – 3/17 Lecture

Announcements

• Pass back HW + Quiz• Lab

– Term Project Proposal due Thursday– Next Lab Report also due Thursday

• Today’s Lecture– Chapter 17: (Basic Spectroscopic Theory)

• Beer’s Law– Chapter 18: Spectrometer Instrumentation

• Light Sources• Wavelength Discrimination

Beer’s Law– Best Region for Absorption Measurements

• Determine the Best Region for Most Precise Quantitative Absorption Measurements if Uncertainty in Transmittance is constant

A

% uncertainty

0 2

High A values - Poor precision due to little light reaching detector

Low A values – poor precision due to small change in light

Beer’s Law– Deviations to Beer’s Law

A. Real Deviations- Occur at higher C - Solute – solute interactions become important- Also absorption = f(refractive index)

Beer’s Law– Deviations to Beer’s Law

B. Apparent Deviations1. More than one chemical species

Example: indicator (HIn)HIn ↔ H+ + In-

Beer’s law applies for HIn and In- species individually: AHIn = ε(HIn)b[HIn] & AIn- = ε(In-)b[In-]

But if ε(HIn) ≠ ε(In-), no “Net” Beer’s law applies Ameas ≠ ε(HIn)totalb[HIn]total

Standard prepared from dilution of HIn will have [In-]/[HIn] depend on [HIn]total

0

0.05

0.10.15

0.2

0.25

0.3

0.350.4

0.45

0.5

0 0.005 0.01 0.015

Total HIn Conc.A

bso

rban

ce

In example, ε(In-) = 300 M-1 cm-1

ε(HIn) = 20 M-1 cm-1; pKa = 4.0

Beer’s Law– Deviations to Beer’s Law

More than one chemical species:Solutions to non-linearity problem1) Buffer solution so that [In-]/[HIn] =

const.2) Choose λ so ε(In-) = ε(HIn)

Beer’s Law– Deviations to Beer’s Law

B. Apparent Deviations

2. More than one wavelength

ε(λ1) ≠ ε(λ2)

0

0.5

1

1.5

2

2.5

3

3.5

4

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035

Conc.

Ab

sorb

ance

λ1λ2

Example where ε(λ1) = 3*ε(λ2)

line shows expectation where ε(λ1) = ε(λ2) = average value

Deviations are largest for large A

0

0.5

1

1.5

2

2.5

3

3.5

0 10 20 30 40 50

% deviation

Ab

so

rba

nc

e

A

λ

Beer’s Law– Deviations to Beer’s Law

More than one wavelength - continuedWhen is it a problem?

a) When polychromatic (white) light is usedb) When dε/dλ is large (best to use absoprtion maxima) and Δλ is not small (Δλ is the range of wavelengths passed to sample)c) When monochromator emits stray lightd) More serious at high A values

Luminescence SpectroscopyAdvantages to Luminescence

Spectroscopy1. Greater Selectivity (most compounds do not efficiently fluoresce)2. Greater Sensitivity – does not depend on difference in signal; with sensitive light detectors, low level light detection possibleAbsorption of light

95% transparent

(equiv. to A = 0.022)Weak light in black background

Emission of light

Chapter 19 - SpectrometersMain Components:1) Light Source (produces light in right wavelength range)2) Wavelength Descriminator (allows determination of signal

at each wavelength)3) Sample (in sample container)4) Light Transducer (converts light intensity to electrical

signal)5 )Electronics (Data processing, storage and display)Example: Simple Absorption Spectrophotometer

Light Source

(e.g tungsten lamp)

Monochromator

Sample

detector (e.g. photodiode)

Electronics

single l out

SpectrometersSome times you have to think creatively to get all the

components.Example NMR spectrometer:Light source = antenna (for exciting sample, and sample re-

emission)Light transducer = antenna Electronics = A/D board (plus many other components)Wavelength descriminator =Fourier Transformation

Radio Frequency Signal Generator

Antenna

A/D Board

Fourier Transformed Data

Spectrometers – Fluorescence/Phosphorescence

• Fluorescence Spectrometers– Need two wavelength

descriminators– Emission light usually at

90 deg. from excitation light

– Can pulse light to discriminate against various emissions (based on different decay times for different processes)

– Normally more intense light and more sensitive detector than absorption measurements since these improve sensitivity

lampExcitationmonochromator

sample

Emission monochromator

Light detector

Absorption Spectrometers

A. Sensitivity based on differentiation of light levels (P vs P0) so stable (or compensated) sources and detectors are more important

B. Dual beam instruments account for drifts in light intensity or detector response

Light Source

(tungsten lamp)

Monochromator

Sample

Electronics

chopper or beam splitter

Reference

detector

Spectrometers – Specific ComponentsLight Sources

A. Continuous Sources - General1) Provide light over a distribution of

wavelengths2) Needed for multi-purpose instruments that

read over range of wavelengths3) Sources are usually limited to wavelength

ranges (e.g. D2 source for UV)

Spectrometers – Light Sources

A. Continuous Sources – Specific1) For visible through infrared,

sources are “blackbody” emitters

2) For UV light, discharge lamps (e.g. deuterium) are more common (production of light through charged particle collision excitation)

3) Similar light sources (based on charged particle collisions) are used for X-rays and for higher intensity lamps used for fluorescence

4) For radio waves, light generated by putting AC signal on bare wire (antenna). Wide range of AC frequencies will produce a broad band of wavelengths.

UV Vis IR

high T

low T (max shifted to larger l)in

tens

ity

Spectrometers – Light Sources

B. Discrete Light Sources - General1. More common in “specific” instruments (e.g.

industrial process instrument that measures single constituent)

2. Light source usually is a (or the) wavelength discriminator also.

Specific Sources3. LEDs (inexpensive light sources – relatively narrow

band of wavelengths)4. Hollow cathode lamps (used in atomic absorption –

discussed later)5. Lasers (intense, coherent, unidirectional, and very

narrow wavelength distribution)

Some Questions1. Does the intensity of a light source have a large effect on

the sensitivity of a UV absorption spectrometer? What about a fluorescence spectrometer?

2. If a sample is known to fluoresce and phosphoresce, how can you discriminate against one of these processes?

3. If a sample can both fluoresce and absorb light, why would one want to use a fluorescent spectrometer?

4. What is the advantage of using a dual beam UV absorption spectrometer?

5. Would an LED light source be very useful for a general purpose light absorption spectrometer?

6. List 5 components of spectrometers.7. Why could the use of a broad band light source in the

absence of wavelength discrimination lead to poor quantification of light absorbing constituents?

Spectrometers – Wavelength Discrimination

A. Filters1. Mostly used with

specific instruments2. “Standard Filters” –

act to pass band of light or cut-off low or high wavelengths

3. Interference filters- pass a narrow band of

light- based on interference

(show on board)- used with other filters

to reduce other orders

- some “tuning” of wavelength possible by changing gap or refractive index

inte

nsity

before filter

after filter

inte

nsity

before filter

wavelength

wavelength

after filter

Spectrometers – Wavelength Discrimination

B. Monochromators1. Allows selection of a

narrow band of wavelength from “broad band” source of light

2. Most monochromators allow continuous adjustment of the selected wavelengths

3. Some monochromators also allow adjustment of the range of wavelengths passed (Dl)

inte

nsity

wavelength

after filter

before filterdesired l

Dl

Spectrometers – Monochromators

A. Components1. Entrance Slit (to

match exit slit)2. Light Collimator

(optics to make light beam parallel when falling on dispersive element)

3. Dispersing Element (to disperse light at different angles for different l values)

4. Focusing Optics (to focus light on exit slit)

5. Exit Slit (to select range of l values passed – Dl)

entrance slit

light

grating

collimating optics

l1

l2

Focusing opticsexit slit

In this example, wavelength selection occurs through rotation of the grating

Spectrometers – Monochromators

B. Dispersion of Light1. Prisms – based on

refractive index (n) = f(l)

2. Gratings – based on constructive interferencea. 2 beams hitting grating

will travel different distances

b. travel difference = a – bc. this difference must be

an integral # of l to lead to constructive interference

d. a – b = n l (n = integer)e. from geometry, nl =

d(sinq – sinf)f. Each groove acts as a

light source

extra distance traveled by beam 2 = a

12

extra distance traveled by beam 1 = b

d

qf

d = groove spacing

q = incoming light angle

f = outgoing light angle

Spectrometers – Monochromators

B. Performance of Grating1. Resolution = l/Dl = nN

where n = order (1, 2, 3...) and N = No. grooves illuminated2. To increase resolution,

a. decrease d (groove spacing)b. increase length of grating illuminated (perpendicular to

grooves)c. use higher diffraction order (n = 5 vs. n = 1)

3. Dispersion from gratings:a. Angular dispersion = Df/Dl = n/dcosfb. Linear dispersion = D = Dy/Dl = FDf/Dl

f

Exit slit y-axis

F = focal length