tue 3nov15 beer-lambert law (a = cx )€¦ · beer-lambert law (a = εcx ) suppose a = 2 what...

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1 Beer-Lambert Law (A = εcx ) suppose A = 2 What fraction of light is transmitted? 10 -2 = 0.01 = 1 % Now, double the concentration. What fraction of light is transmitted? 10 -4 = 0.0001 = 0.01 % Now, double the path length using this concentration. What fraction of light is transmitted? 10 -8 = 10 -6 % Now, change the wavelength until ε is doubled. What fraction of light is transmitted at this wavelength? 10 -16 = 10 -14 % 361 Lec 37 Tue 3nov15 f= Transmittance = T = 10 -εcx = 10 -A =

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Page 1: Tue 3nov15 Beer-Lambert Law (A = cx )€¦ · Beer-Lambert Law (A = εcx ) suppose A = 2 What fraction of light is transmitted? 10-2 = 0.01 = 1 % . Now, double the concentration

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Beer-Lambert Law (A = εcx ) suppose A = 2 What fraction of light is transmitted?

10-2 = 0.01 = 1 %

Now, double the concentration. What fraction of light is transmitted?

10-4 = 0.0001 = 0.01 % Now, double the path length using this concentration. What fraction of light is transmitted? 10-8 = 10-6 % Now, change the wavelength until ε is doubled. What fraction of light is transmitted at this wavelength?

10-16 = 10-14 %

361 Lec 37 Tue 3nov15

f= Transmittance = T = 10-εcx = 10-A =

Page 2: Tue 3nov15 Beer-Lambert Law (A = cx )€¦ · Beer-Lambert Law (A = εcx ) suppose A = 2 What fraction of light is transmitted? 10-2 = 0.01 = 1 % . Now, double the concentration

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UV absorption of Amino Acids

Page 3: Tue 3nov15 Beer-Lambert Law (A = cx )€¦ · Beer-Lambert Law (A = εcx ) suppose A = 2 What fraction of light is transmitted? 10-2 = 0.01 = 1 % . Now, double the concentration

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Tyr -

UV absorption of Amino Acids

Note the Log scale

What is absorbance of a 0.01 M solution of phenylalanine if path is 1 cm A = εcx A = 100 x .01 x 1= 1 Fraction transmitted = ? 0.1

What is absorbance of a 0.01 M solution of tyrosine is path if 1 cm A = 1000 x .01 x 1= 10 Fraction transmitted = ? 10-10

Page 4: Tue 3nov15 Beer-Lambert Law (A = cx )€¦ · Beer-Lambert Law (A = εcx ) suppose A = 2 What fraction of light is transmitted? 10-2 = 0.01 = 1 % . Now, double the concentration

Using the Beer-Lambert Law to determine concentrations of a mixture of two absorbing species

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What if εM1 = εN

1???

Then A1 = ε[M] + ε[N] = ε ([M] + [N]) In other words at isosbestic point you get the TOTAL concentration

Page 5: Tue 3nov15 Beer-Lambert Law (A = cx )€¦ · Beer-Lambert Law (A = εcx ) suppose A = 2 What fraction of light is transmitted? 10-2 = 0.01 = 1 % . Now, double the concentration

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Green is the unknown mixure of Trp and Tyr Red is pure Tyr - Blue is pure Trp

Purple is 50-50 Trp + Tyr - Example 13.1

Page 6: Tue 3nov15 Beer-Lambert Law (A = cx )€¦ · Beer-Lambert Law (A = εcx ) suppose A = 2 What fraction of light is transmitted? 10-2 = 0.01 = 1 % . Now, double the concentration

Highest Occupied Molecular Orbital (a linear combination of atomic p orbitals)

Ground State

LUMO

What happens during absorption of light by Tryptophan?

LUMO +1 , etc,

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Page 7: Tue 3nov15 Beer-Lambert Law (A = cx )€¦ · Beer-Lambert Law (A = εcx ) suppose A = 2 What fraction of light is transmitted? 10-2 = 0.01 = 1 % . Now, double the concentration

Lowest Unoccupied Molecular Orbital (electron excited)

Excited State (fluorescing state)

LUMO

LUMO +1 , etc,

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Page 8: Tue 3nov15 Beer-Lambert Law (A = cx )€¦ · Beer-Lambert Law (A = εcx ) suppose A = 2 What fraction of light is transmitted? 10-2 = 0.01 = 1 % . Now, double the concentration

Fluorescence lifetime ~5 ns (exponential decay)

Similar to Fig. 13.5 of our Textbook.

vibrational relaxation ~1 ps

internal conversion~1 ps

∆E = hν

HOT!!!

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Page 9: Tue 3nov15 Beer-Lambert Law (A = cx )€¦ · Beer-Lambert Law (A = εcx ) suppose A = 2 What fraction of light is transmitted? 10-2 = 0.01 = 1 % . Now, double the concentration

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What is fluorescence lifetime? d(Intensity)/dt = -k (Intensity) Fluor. intensity at time t = (Fluor. Intensity at time 0) x e-kt

= (Fluor. Intensity at time 0 x) e-t/τ

τ = “lifetime = 1/k τ = inverse of 1st order rate constant

Fluorescence “Quenching” A reaction with another molecule that competes with the rate of fluorescence

Page 10: Tue 3nov15 Beer-Lambert Law (A = cx )€¦ · Beer-Lambert Law (A = εcx ) suppose A = 2 What fraction of light is transmitted? 10-2 = 0.01 = 1 % . Now, double the concentration

Highest Occupied Molecular Orbital (a linear combination of atomic p orbitals)

Ground State

LUMO

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Page 11: Tue 3nov15 Beer-Lambert Law (A = cx )€¦ · Beer-Lambert Law (A = εcx ) suppose A = 2 What fraction of light is transmitted? 10-2 = 0.01 = 1 % . Now, double the concentration

Lowest Unoccupied Molecular Orbital (electron excited)

Excited State (fluorescing state)

LUMO

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Page 12: Tue 3nov15 Beer-Lambert Law (A = cx )€¦ · Beer-Lambert Law (A = εcx ) suppose A = 2 What fraction of light is transmitted? 10-2 = 0.01 = 1 % . Now, double the concentration

Iodide

Iodide ion collides; (has higher HOMO) will quench fluorescence

LUMO

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Page 13: Tue 3nov15 Beer-Lambert Law (A = cx )€¦ · Beer-Lambert Law (A = εcx ) suppose A = 2 What fraction of light is transmitted? 10-2 = 0.01 = 1 % . Now, double the concentration

Electron transferred from iodide to vacancy in HOMO of ring i.e., QUENCHING

Radical Pair (can’t fluoresce)

Ring Iodine

Electron transfer from I- to indole makes a radical pair that cannot fluoresce. (would violate Pauli exclusion)

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Page 14: Tue 3nov15 Beer-Lambert Law (A = cx )€¦ · Beer-Lambert Law (A = εcx ) suppose A = 2 What fraction of light is transmitted? 10-2 = 0.01 = 1 % . Now, double the concentration

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Page 15: Tue 3nov15 Beer-Lambert Law (A = cx )€¦ · Beer-Lambert Law (A = εcx ) suppose A = 2 What fraction of light is transmitted? 10-2 = 0.01 = 1 % . Now, double the concentration

13.22

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