applied physics on spectroscopy

8/16/2019 Applied Physics on Spectroscopy

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Applied PhysicsOn

UV-vis Spectrophometry

Kuwat Triyanahttp://triyana.staff.ugm.ac.id



EM Spectrum



Ultra Violet (UV)

• Three regions of UV

– UV-A:• long-wave UV, near-ultraviolet, black light, or Wood’s light

• between 320 and 400 nm

– UV-B:• medium-wave UV


– UV-C :• short-wave, far ultraviolet, or germicidal UV




Absorption Spectroscopy Introduction

A.) Absorption: electromagnetic (light) energy is transferred to atoms, ions, or molecules

in the sample. Results in a transition to a higher energy state.

- Transition can be change in electronic levels, vibrations, rotations, translation, etc.

- Concentrate on Molecular Spectrum in UV/Vis (electronic transition)

- Power (P): energy of a beam that reaches a given area per second

- Intensity (I): power per unit solid angle

- P and I related to amplitude2

Eo

E1

h Energy required of photonto give this transition:

h = DE = E1 - Eo

(excited state)

(ground state)



Absorption and Emission



Molecular Absorption and Emission

Atoms:

Electronic states only

Molecules:

Vibrational states within electronic states



B.) Terms:

1.) Beer’s Law: A = bc

The amount of light absorbed (A) by a sample is dependent on the path

length (b), concentration of the sample (c) and a proportionality constant (e

– molar absorptivity)

Amount of light absorbed is dependent on frequency ( )

c

Absorbance is directly proportional to concentration Fe+2

Increasing Fe+2 concentration



B.) Terms:

1.) Beer’s Law: A = bc

Transmittance (T) = I/Io %Transmittance = %T = 100T

Absorbance (A) = log10

Io/I

No light absorbed-

% transmittance is 100% absorbance is 0

All light absorbed-

% transmittance is 0% absorbance is infinite



Relationship Described in Terms of Beer’s Law

A = Absorbance = ebc = -log(%T/100)

e = molar absorptivity: constant for a compound at a given frequency ()

or wavelength (l)

units of L mol-1 cm-1

b = path length: cell distance in cm

c = concentration: sample concentration in moles per liter.

Therefore, by measuring absorbance or percent transmittance at a given

frequency can get information related to the amount of sample (c) present

with an identified e and l.

Note: law does not hold at high

concentrations, when A > 1



C.) Components of an Instrument for UV/Vis Absorbance Measurements:

1.) Basic Design:

Hitachi Instruments U-3010

Light Source, l selector, Sample cell holder, Detector (amplifier, recorder)



a) Desired Properties of Components of UV/Vis:

Light Source Selector

Creates Proper l Narrow Bandpass:

Stable: Selects Desired l

Constant P Large Light Throughput:

Good Precision Increase P

Intense: Increase P

Easier to See Absorbance

Samp le Cell Holder Detector Fixed Geometry: Stable

Constant b Sensitive to l of Interest

Transmits l of Interest:

Increase P



Molecular Spectra

400 500 600 700

l (nm)

Absorption Emission

I n t e n s i t y

Bands vs. Lines (atoms)



Absorbed vs. Observed

Color absorbed → Complementary color observed

R

BG

B

YG BV

VY

RVO

G



Spectroscopy Terms Describing

Absorption (Beer’s Law)

• Consider a beam of lightwith an (initial) radiantintensity Io

• The light passes through asolution of concentration (c)

• The thickness of thesolution is “b” cm.

• The intensity of the lightafter passage through thesolution (where absorptionoccurs) is I

I0 hv I

b

C o n c e n t r a t

i o n

( c )



We Define

• Transm ittance (T) = I/I0 (units = %)

• Absorbance (A ) (units = none)

– A = log (I0/I)

– A = -log (T) = log (1/T)

– A = abc (or εbc) <--- Beer’s Law • a = absorptivity (L/g cm)

• b = path length (cm)

• c = concentration (g/L)

• ε = molar absorptivity (L/mol cm) – Used when concentration is in molar units



Other expression of Lambert’s Law of Absorption

xe

I

I

I x

I

=

=

0

d

d

• The intensity I 0 if a beam of light decreases exponentially as it passes though a uniform

absorbing medium with the linear decay constant α .

Restatement: In a uniform absorbing medium, the intensity of a beam of light decreases

by the same proportion for equal path lengths traveled.

• The linear decay constant α is a characteristic of the medium. It has units of reciprocal

length. α is the path length over which the intensity is attenuated to 1/e.

α

I 0

I ( x )

x

Lambert described how intensity changes with

distance in an absorbing medium.

xe I x I = 0)(

l e I I = 0

I

l

xe I I

x I I

=

=

0

dd

The distance traveled through the medium is

called the path length.



Lambert’s Law of Absorption (base 10)

xk x

e I

I ==100

Typically base 10 is used in photometry.

xk x I e I I == 1000

10ln =k

k is the path length over which the intensity is attenuated to 1/10.

xk

I

I =100



Lambert’s Law Example If one slab of absorbing material of thickness l reduces the intensity of a beam of light

to half.

α I 0

2

110

0

== l k

I

I I

l

And three slabs will reduce the intensity of a beam of light to one eight .

Then two slabs of the same absorbing material will then reduce the

intensity of a beam of light to one quarter .

4

1

2

110

2

2

0

=

== l k

I

I α I 0 I

l

α

l

8

1

2

110

3

3

0

=

== l k

I

I α I

l

α

l

α I 0

l



Beer’s Law Beer found that Lambert’s linear decay constant k for a solutionof an absorbing substance is linearly related to its

concentration c by a constant, the absorptivity ε , acharacteristic of the absorbing substance.

Restatement: The linear decay constant k is linear in concentrationc with a constant of proportionality ε .

(August Beer, 1825-1863)

A colored absorber has an absorptivity that is dependent on wavelength of thelight ε ( λ).

The absorptivity is the fundamental property of a substance. This is the property

that contains the observable spectroscopic information that can be linked to

quantum mechanics (also see absorption cross section.)

ck e =

Typical units are: k cm−1; c M (moles/liter); ε M−1cm−1



Photometric Quantities

Transmittance (T)

Absorbance (A) (AKA optical density, O.D.)

0 I

I T = usually given in percent

T I

I A loglog

0

=

= by convention, base 10 logs are used

In photometry we measure the intensity of light and characterize its

change by and object or substance. This change is typicallyexpresses as percent transmittance or absorbance.

Frequently when your primaryinterest is the light beam

Used almost exclusively when your interest

concerns the properties of the material



Beer-Lambert LawLambert’s and Beer’s Laws are combined to describe the

attenuation of light by a solution. It is easy to see how the twostandard photometric quantities can be written in terms of this law.

xc I I e = 100

xc A

T

I

I A

e =

=

= loglog

0

xcT

I

I T

e

=

=

10

0

Transmittance Absorbance



Cross-Sections and Absorptivitythe connection to single particles and molecules

The absorption of light by particles (and single molecules) ischaracterized by an absorption cross section C . In this model theparticle is replaced by a perfectly absorbing sphere with a crosssectional area C . This cross section is a property of the particleand is not related to its geometric cross sectional area. Theconcentration of particles per unit volume is N .

=

=

=

3

3

cm

liter 1010ln

10ln

C N

NC k

NC

Ae

typical units are: C cm2; N cm−3

The cross section can be directly related to the

molar absorptivity. N A is Avagadro’s number.

units are: C cm2; N cm−3; N A mole−1; ε

M−1cm−1



Efficiency

The absorption efficiency Q of a particle is the ratio of its absorption

cross section C to its geometric cross section C geo.

Absorption efficiency is dimensionless.

geoC

C Q =



Extension to Scattering and

Extinction

Attenuation of light by absorption and scattering both obey

Lambert’s Law. Thus we can extend our treatment of absorption to

scattering and extinction. (Recall that extinction is the effect of

absorption + scattering.)

cxcx A

QQQ

C C C

scaabsext

scaabsext

scaabsext

scaabsext

e e e

e e e

==

=

=

= The scattering efficiency can be much larger than unity.

Extinction paradox: Qext = 2 (Qabs = 1; Qsca = 1)

for an perfectly absorbing particle very large compared

to the wavelength of light.

Note:

• All of these quantities are in general wavelength dependent.

•Our discussion has not included the mechanism (cause) of absorption and scattering.

•There are many different mechanisms that cause of absorption and scattering.



Instrumentation

• Spectrometer: measures I vs λ.Simply measures the spectrum of the light (e.g. emission spectroscopy).

• Spectrophotometer: measures I /I 0 vs λ.Measures how the sample changes the spectrum of the light (e.g.

transmission, reflection, scattering, fluorescence).All spectrophotometers contain a spectrometer.

• -meter: the detector is electronic

• -graph: light intensity recorded on film

• photometer: measures I /I 0 without λ selection.



OU NanoLab/NSFNUE/Bumm & Johnson

The Spectrophotometer

Measures absorbance as a function of wavelengthComponents: light source, monochromator, sample cell, detector,

optical system.

monochromator sample cell

detector

light source

s l i t

d

i f f r a c t i o n

g r a t i n g



balance the

forces:

Computer controlled

acquisitionof absorption spectra

Cary 50 UV-Vis Spectrophotometer

sample

detector

lightsource

monochromator

Can you find thediffraction grating and theslit?

www.varianinc.co



Making a Measurement with the Cary 50

• First, measure the baseline using a blank sample. This is raw I 0.The blank sample is the cuvette with deionized water(everything but your nanoparticles). This corrects for anyabsorption due to the cuvette, water, and variations of the lightintensity of the light source, monochromator, etc.

• Second, measure the zero by inserting the beam block . Thiscorrects the instrument for the detector background.

• Third, measure your sample. This is the raw I . The Cary 50automatically calculates the corrected intensities (I and I 0) bysubtracting the zero from each of the raw intensities.

• Subsequent measurements do not require re-measuring the

blank and zero, simply repeat step 3.

T A

zero I raw

zero I raw

I

I T

log

00

=

==



Transmittance

T => transmittance

IT = -----

Io

Io I

b



I0 = 10,000 I = 5,000

5.010000

5000

0

=== P

P T

-b-

Example

A = -log T = -log (0.5) = 0.3010



Beer’s Law

A = abc = ebc

where T => transmittance

%T => percentage transmittance

I => transmitted Iower of radiation

Io => incident power of radiation

A => absorbance

a => absorptivity

b => path length

c => concentration

e => molar absorptivity, extinction coefficient



Beer’s Law

A = abc = ebc

A

c



Beer’s Law A = ebc

Path Length Dependence, b

Readout

Absorbance

0.82

Source

Detector





Readout

Absorbance

0.62

Source

Detector

b

Sample





Readout

Absorbance

0.42

Source

Detector Samples





Readout

Absorbance

0.22

Source

Detector Samples




Wavelength Dependence, a

Readout

Absorbance

0.80

Source

Detector

b





Readout

Absorbance

0.82

Source

Detector





Readout

Absorbance

0.30

Source

Detector

b





Readout

Absorbance

0.80

Source

Detector

b



Non-Absorption Losses

"Reflection andscattering losses."



Limitations to Beer’s Law

• Real – At high concentrations charge distribution effects

occur causing electrostatic interactions betweenabsorbing species

• Chemical – Analyte dissociates/associates or reacts with solvent

• Instrumental

– ε = f(λ); most light sources are polychromatic notmonochromatic (small effect)

– Stray light – comes from reflected radiation in themonochromator reaching the exit slit.



Chemical Limitations

A reaction is occurring as you record

Absorbance measurements

Cr 2O72- + H2O 2H+ + CrO4

2-

A550 A446

concentration concentrationwavelength

400 500300

CrO42-

Cr 2O72-



Instrumental Limitations - ε = f(λ)

• How/Why does εvary with λ?

• Consider awavelength scan fora molecular

compound at twodifferent wavelengthbands

• In reality, a

monochromator cannot isolate a singlewavelength, butrather a smallwavelength band

Larger the Bandwidth – larger deviation



Instrumental Limitations – Stray Light

• How does stray light effect Absorbanceand Beer’s Law?

• A = -log I/Io = log Io/I

• When stray light (Ps) is present, theabsorbance observed (Aapparent) is the sum

of the real (Areal) and stray absorbance

(Astray)




• Aapp = Areal + Astray =

• As the analyte concentration increases([analyte]↑), the intensity of light exiting the

absorbance cell decreases (I↓)

• Eventually, I < Is

o s

s

I Ilog

I I




• Result – non-linearabsorption (Analyte

vs. Conc.) as a

function of analyteconcentration

– Similar to

polychromatic light

limitations



Solution too Concentrated

• Refractive index changes with larger

concentrations

A e e n n

n2 + 2



Allura Red

Formula: C18H14N2Na2O8S2

Molar mass: 496.42 g/mol

2Na+



During Absorption…

Starting e- arrangement: After photon absorption:



Absorption is random

•Photons collide with molecules•Certain probability of absorption



Transmittance, T

I0 I1 TII

0

1 =

T: ratio of light “in” vs. “out”

= fraction of light passing through

Depends on:

# of molecules b & c

molecules’ identity e

Pathlength, b

Concentration,c



Transmittance and Pathlength

1

2

0

1

II

II =

For cell with same pathlength: constant T

1

2

II

I0 I1

0

1

II

I2



Transmittance and Pathlength

1

2

I

I

0

1

I

I

2

0

1

1

2

0

1

0

2

I

I

I

I

I

I

I

I

==I0

For cells with same pathlength: constant T

Double pathlength: square T

I2

#1 #2

2

121total TTTT ==



T and Pathlength

1- b

b

I

I

0

1

I

I

1- b

b

0

1

b

0

1

0

b

I

I

I

I

I

I

I

I

=

=

I0

For b cells (b pathlengths = 1cm)

Tb

Ib

#1 #b

b1

b

1total TTTT ==

…

b in power



Concentration

• # photons absorbed depends on #molecules in path

I0 I1

Transmittance and



Transmittance and

Concentration

1

2

I

I

b

0

cm1

I

IT

=

0

1

I

I

I0 I1 I2

1.0 M reference solution gives

2.0 M: Double concentration,c→“Double pathlength, b”…

bc

0

1Mcm,1 bc

1I

ITT

==

…

bc in powerpathlength and

concentration



Molar Extinction Coefficient, e

e

Probability of absorption

– Specific to molecule

– Function of l

lmax Most efficient absorption

– Peak of absorption curve

–“Best l” for experiment

T: concentration c & pathlength b What about: molecular identity?



b, c, and e

εcb- bcε-

bc

0

1Mcm,11010

I

IT ==

=

I

Ilogε

0

1Mcm,1

=

e in powermolar extinction coeff.

ε

0

1Mcm,110

I

I =

Therefore:

bc

0

1Mcm,1 bc

1I

ITT

==

bc in powerpathlength and concentration



Absorbance

Beer’s Law : Defines absorbance, A

εbc A

A

T A

εbc

=

=

=10log

log

A : Directly proportional to concentration

high A ≡ low T

Part 1 Spectral Profile of Allura



Part 1 Spectral Profile of Allura

Red, lmax

•Use stock solution (record conc.) cstock

•Record %T, 400 – 700 nm %T

– 10 nm intervals near lmax

– 20 nm intervals elsewhere

•Record cell width, b = pathlength b

•Calculate A A

•Plot A vs. l

•Determine lmax l

max



Part 1: lmax

400 500 600 700l (nm)

Absorption Transmittance

I n t e n s i t y

Find l where

%T is lowest

A is highest

This is not

necessarily

Allura Red

(but would

appear red)



Spectral Profilel (nm) Absorbance

400 0.078

420 0.130

440 0.268 460 0.555

480 0.966

490 1.189

500 1.383

510 1.485

520 1.515

530 1.459

540 1.217

560 0.395

580 0.063

600 0.019

620 0.015

640 0.010

660 0.009

680 0.006

700 0.000

Spectral profile - Allura Red

0.00

0.20

0.40

0.60

0.80

1.00

1.20

1.40

1.60

400 450 500 550 600 650 700

wavelength (nm)

A b s o r b a n c

e

Part 2 Absorbance for Various



Part 2 Absorbance for Various

Concentrations

•Stock solution, 4 dilutions, and blank• n1 = n2

• M1 . V1 = M2

. V2

•

•Determine %T at lmax for each

%T

•Calculate A

A

•Plot A vs. conc (Beer’s Law plot)

•Slope = eb



Part 2 Concentrations•1 blank: pure DI water M0 = 0.00 M allura red

•1 stock: 0.002% allura red M5 = 4×10-5 M

•4 dilutions, each by ½:

– (4×10-5 M)(25.00 mL) = (M1 )(50.00 mL) M1 = 2×10-5 M – (2×10-5 M)(25.00 mL) = (M2 )(50.00 mL) M2 = 1×10-5 M

– (1×10-5 M)(25.00 mL) = (M3 )(50.00 mL) M3 = 5×10-6 M

– (5×10-6 M)(25.00 mL) = (M4 )(50.00 mL) M4 = 2×10-6 M

M solution L

solution g

red allura g

red alluramol

solution g

red allura g 5

001.0

1

42.496

1

100

002.0104%002.0 ==

diluteed concentrat dilutediluteed concentrat ed concentrat nnV M V M ==



Beer’s Law

e,b constant

xm ==

=

y

cεbA

TlogA

A === 1010

%100

%TT εbc

c varied



Beer’s Law Plot

cεbA =

x y =m

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

0.45

0.50

Molarity

A b s o r b a n c e

y = 1.9 x

yD

εb=D

D=

x

y slope

xD

B ’ L Pl l



Beer’s Law Plot example

Dilution Factor M (mol/L) T A = -logT

0 0 1.00 0.000

1/16 2.E-06 0.87 0.060

1/8 5.E-06 0.73 0.137

1/4 1.E-05 0.58 0.240

1/2 2.E-05 0.32 0.495

1 4.E-05 0.10 1.000

Stock M 4E-05 mol/L

A = e×b×c

Slope = D

A/D

c Here:

e×b = 24943

A 1-cm cell: e

24943 cm-1M-1

Absorbance vs. Concentration - Allura Red

y = 24943x

R 2 = 0.9995

0.000

0.200

0.400

0.600

0.800

1.000

1.200

0.E+00 1.E-05 2.E-05 3.E-05 4.E-05

Molarity (mol/L)

A b s o r b a n c e

Part 3 Allura Red Concentration in



Part 3 Allura Red Concentration in

Mouthwash

•Use 1:25 dilution

•

•Determine %T at lmax %T

•Calculate AA

•Measure b b

•Find concentration, c c

soεbc A =b

Ac

e

=



Part 3 Example

Mouthwash trials (1:25 dilution): T A M

dilute Mconcentrated

0.68 0.167 7E-06 1.7E-04

0.70 0.155 6E-06 1.6E-04 0.65 0.187 8E-06 1.9E-04

0.69 0.161 6E-06 1.6E-04

Mdilute= A/(b×e ) if cell length b is constant

Average Mconcentrated = 1.7×10-4 M = 2×10-4 M



Report• Abstract

• Data/Results

• Sample calculations including:

– Absorbance from transmittance

– Dilution

– Slope and extinction coefficient

– [Allura red]cuvette and [Allura red]mouthwash

– # allura red molecules in 1 mL mouthwash

• Discussion/review questions

applied physics on spectroscopy

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