Study Guide 5: Light Absorption by π Electrons in Biological Molecules

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1 Study Guide 5: Light Absorption by π Electrons in Biological Molecules Text: Chapter 4, sections 5 (from end of example 4.5) 9. Upcoming quizzes: For quiz 3 (final day, Friday, Feb 29, 2008) you should know: 1. Material from SG Experiment 4: Sketch oscilloscope patterns for 1- and 2-slit interference, and for a circular aperture. From these patterns, calculate the wavelength of the laser light.) 3. Experiment 6: Block diagrams for absorption and fluorescent spectrophotometers. Qualitative relation between λ max of absorption and fluorescent spectra - SG 6! quiz 4 (final day, Friday, March 14, 2008) you should know: Material from SG 5 (current SG) and SG 6. 1

2 light E = hf Light Absorption by π - electrons in Molecules molecule in excited state Nature of Excited State depends on the energy of the photon absorbed. In SG5 we will study the absorption of visible and ultraviolet light causing transitions between different π - electron energy levels. We will study other types of transitions in SG6. π - electron Energy Levels From SG4: π - electron wave function: Ψ n (x) = 2/L sin(nπx/l) 1. How is quantum number, n, chosen? 2. How is n related to electron energy? 2

3 We saw in SG4 that n = number of ½ wavelengths inside box of length L containing conjugated sequence. Therefore: L = nλ e /2 λ e = 2L/n Electron Energy: sub - in E = p 2 /2m e and p = h/λ e E = h 2 2 2meλ e sub-in λ e we get E = n 2 2 n h 8m L e 2 E E n Energy of electron with quantum number n So we have answered question#2. We have related n to the electron energy. How is n chosen? 3

4 Electron Structure and the Pauli Exclusion Principle The molecules we study generally have many π electrons. (Number = number of C, O and N atoms in chain). How are quantum numbers n assigned to all these electrons? Unlike nucleons (protons, neutrons) which are composed of quarks, electrons CANNOT be broken into pieces consisting of more fundamental particles. Electrons are elementary particles. Electrons have three fundamental characteristics: 1) mass, m e = kg 2) charge, e = C 3) spin, s = h/2 What is spin? 4

5 Spin Analagous to rotation of a planet or baseball about it axis. s However, electron has no apparent size. It is a true point particle Characterized mechanically by Spin Angular Momentum, S. S causes electron to have a magnetic dipole moment (as is classically produced by a circulating electric charge) and is responsible for most magnetic properties of materials. The direction of the spin angular momentum, S, can be either UP or DOWN with respect to any arbitrary reference direction. This is represented by arrows: for UP and for DOWN. An electron may be in any energy state, E n with spin either UP or DOWN. How do we decide which electrons go where? 5

6 PAULI EXCLUSION PRINCIPLE (P.E.P.) Two electrons with the same spin cannot exist in the same energy state at the same time. a) Suppose we have 2 electrons in a linear molecule with the same quantum number n (i.e. they are in the same energy state, E n ). By P.E.P., these electrons must have opposite spins b) Cannot have 3 or more electrons with same n. An energy state, En, can be occupied by at most 2 electrons and these MUST have opposite spins. Electrons tend to occupy the lowest energy state available. To occupy higher energy states requires more energy. When each electron is in the lowest energy state allowed by the PEP, the molecule is in the GROUND STATE. We can classify the π-electronic states of a molecule by the occupation of the different energy levels, n = 1, 2, 3,. 6

7 Solved Problem I - Octene Determine the wavelength for the primary transition energy for π-electrons in octene. H H H H H H C 1 C 2 C 3 C 4 C 6 C 7 C 8 C 5 H H H H Eight C atoms 8 π electrons. Ground state occupy energy levels so that electrons have the lowest possible energy. n = 5 ENERGY LEVEL n = 4 DIAGRAM FOR n = 3 GROUND STATE n = 2 OF OCTENE n = 1 The 1 st two electrons can occupy n = 1. Lowest available state for next 2 e - is n = 2 etc n = 5 state present but unoccupied for octene 7

8 NB: configurations of the following types are disallowed: n = 4 n = 4 n = 3 n = 3 n = 2 n = 2 n = 1 n = 1 2 electrons, more than 2 same n, same spin electrons, same n 1 st Excited state Next lowest energy state above ground state. ground state 1 st excited state Octene n = 5 n = 4 n = 3 n = 2 n = 1 n = 4 is highest filled π-electron level in ground state. n = 5 is the lowest empty level in ground state. These spins could be reversed (see SG6) 8

9 Aside: (from Self Test I, #3). What is the lowest completely empty level in the 1 st excited state of octene? NB: other excited states are possible, e.g. ground state Octene n = 5 n = 4 n = 3 n = 2 n = 1 excited state but they have higher energy than 1 st excited state (the electron in the n = 4 level can drop to n = 3) A transition between any two states ( ground to 1 st excited, etc.) involves energy. A photon with E = hf = (energy difference between states) can induce a transition. Maximum (light) Absorption occurs for the primary transition, which is between the ground state and 9

10 the 1 st excited state. Therefore the primary transition energy, ΔE, in octene is that of one electron going from n = 4 to n = 5. ΔE = E 5 E 4 = ( )h 2 /(8m e L 2 ) = (9/8) h 2 /(m e L 2 ) L = (8 1) b CC = 7 (0.15) = 1.05 nm 2 34 ( Js i ) 31 9 ( kg )( m) 9 Δ E = 8 = Then the wavelength of the photon which can induce this transition is: 19 J E photon = hc/λ = ΔE λ = hc/(δe) ( Js i )( m/ s) = = J = 400 nm (violet light) 7 m These predictions are not exact because: 10

11 1) Electrons are NOT independent of each other 2) Molecules are not linear Transitions to and between higher excited states (2 nd, 3 rd ) also occur (less probable and higher ΔE). There is also a correction due to the fact that conjugated π-bond chains are not linear. Adjacent bonds are actually at an angle of 120 o to each other. e.g., retinal (real-life version, differs from pictures in text) If b 0 is the actual bondlength (which we have taken as 0.15 nm), then the effective bondlength to be used in calculations is b = b 0 sin(60 o ) = b = (0.15) = 0.13nm 11

12 (But text sometimes uses 0.12 nm!) Use this revised b value in self-test 1, #4, and in text problems 4-15, 4-17 and Solved Problem II The π-electron energy level diagram of Molecule X is given as: E = J = J = J = J (a) If a photon of 203 nm is absorbed by the molecule, what is the resulting energy level diagram? (b) Repeat for photon of wavelength 284 nm. Ans. There is no change! 12

13 Solved Problem III Similar to killer problem 4-19 in text and SG Self-test II #2. A molecule in which there is a sequence of conjugated double bonds 4 atoms long is observed to have a maximum absorption at 650 nm. For a similar molecule in which the conjugated sequence is 6 atoms long, at what wavelength would you expect to find the maximum absorption. Note: maximum absorption occurs for transition from ground state to 1 st excited state. In this problem we do not know the bond length, b, of each bond. Ans. λ 6 = 1290 nm Conjugated Ring Molecules E.g. hemes, chlorophylls, and others. These molecules share in common the porphyrin substructure of 20 Carbon atoms in a conjugated ring shown in the following illustration: 13

14 This structure has a specific absorption spectrum: 14

15 Chlorophyll - a: The green colour of chlorophyll is due to its strong absorbence in the red and blue regions of the spectrum, shown above. Therefore, the light it reflects and transmits appears green. If a solution has only a single absorption maximum in the red, it appears blue or violet (and vice versa). Another e.g.: benzene: 15

Ring molecules are closed: model by a torus. x TORUS 0 r Measure position x along circumference of ring from some arbitrary initial point. Total length of circumference = L.

16 Ring molecules are closed: model by a torus. x TORUS 0 r Measure position x along circumference of ring from some arbitrary initial point. Total length of circumference = L. After traveling a distance, L, the electron is back where it started. Ψ(x+L) = Ψ(x) So Ψ(x) = either C sin(kx) or C cos(kx) C sin(kx + L) = C sin (kx) C cos(kx + L) = C cos (kx) These equ ns are true when kl = 2πn, n = 0, 1, 2,3 NB: k = 2π/λ e L = nλ e an integer # of wavelengths must fit on ring 16

17 Normalized wave functions are: Ψ n (x) = 2/L sin(2πnx/l) n = 1, 2, 3, OR Ψ n (x) = 2/L cos(2πnx/l) n = 0,1,2,3, n = 0 only allowed for cos functions because for sin n = 0 gives Ψ 0 (x) = 0, i.e. NO electron. Energy levels for electrons in ring molecules E = p 2 /(2m e ), p = h/λ e Therefore: E = (2m e ) -1 (h/λ e ) Substituting λ e = L/n (shown earlier) we get: E = n 2 2 n h 2m L e 2 vs. LINEAR E 2 2 nh = 8m L e 2 n = 0, 1, 2, Also L = 2πr for circular ring so (but no longer used in text or quizzes.) E 2 2 nh = 8π mr n 2 2 e 17

18 There are TWO possible states (cos and sin) for all values of n EXCEPT n = 0. Each state could be occupied by two electrons of opposite spin. E.g.: benzene six π-electrons E.g.: Chlorophyll Ground state Energy Level Diagram n = 6 n = 5 n = 4 n = 3 n = 2 n = 1 n = 0 n = 2 n = 1 n = 0 20 π-electrons Highest occupied level in ground state is only half filled. Lowest completely empty level, n = 6. Lowest completely filled? n = 5 is not because it costs more energy 18

19 1 st excited state of chlorophyll. Can raise one electron from n = 4 to n = 5. ΔE = E 5 E 4 = ( )h 2 /(2m e L 2 ) In text: circumference of porphyrin ring L = 20(0.12)*= 2.4 nm. ΔE = J Wavelength of photon absorbed: λ = hc/δe = 527 nm * Bondlengths to use (0.12 or 0.15 nm) are usually given on quizzes. 19

20 Solved Problem IV Two similar organic molecules have circular conjugated rings, one with 10 atoms and the other with 6 atoms. What is the ratio of their maximum absorption wavelengths? Ans. λ 10 /λ 6 = 5/3 =

Diffraction, EM and particle Waves

$Diffraction, EM and particle Waves$ Diffraction, EM and particle Waves RESOURCE MATERIAL Web: www.physics.uoguelph.ca/biophysics/ www.physics.uoguelph.ca/phys1070/webst.html Text: chapter, 3 pp.33-36 4 sections 1-5 andbook: study guide 4