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The quantum theory of atoms and molecules Spectroscopy in a box Dr Grant Ritchie UV absorption spectrum of cyanine dyes + − (−CH = CH −) − CH = N (CH ) (CH 3 ) 2 N k 3 2 + (CH ) (CH 3 ) 2 N = CH − (−CH = CH −) k − N 3 2 Dye I: k = 1 Dye II: k = 2 Dye III: k = 3 (Taken from www.jce.divched.org ) Aims To use the particle in a box model to describe the electronic structure of these dyes. To calculate selection rules which will allow us to predict whether a transition between two energy levels will occur. To explain why the intensity and position of the absorption spectrum varies as a function of the length of the molecule. Electronic structure of cyanine dyes Have both σ and π electrons ⇒ σ electrons have largest probabilities in the plane of the molecules while π electrons are most like to be found above and below plane of molecule. Decouple π electrons from σ framework ⇒ treat π electrons as being delocalised over the length of the molecule between the N atoms. UV and visible light can then be absorbed and energy used to cause transition of π electrons from one energy level to another. NB: Each carbon contributes one electron to the π system while the two N atoms contribute 3 electrons. Is the particle in a box model justified for this problem? Length of the box, L = bβ +2δ, where b is the number of bonds. Remember, remember……. For a particle in a 1d box: Eigenvalues: n2 h2 En = 8mL2 Eigenfunctions: ψ n ( x) = n = 1, 2, 3,.... 2 nπ x sin L L n = 1, 2,.... ψ that describe electrons in atoms and molecules are called orbitals ⇒ an orbital is a wavefunction for a single electron. If we say that the electron is in orbital n then it is described by the wavefunction ψn and has the corresponding energy En . Resonance condition Absorption of a photon occurs when the energy of the photon (= hν) matches the difference in the energy between the two states involved in the transition (∆E): E photon h2 2 2 = hν = ∆E = E f − Ei = n − n ( f i ) 8mL2 where ni and nf are the quantum numbers associated with the initial and final states respectively. All that is required now is to know which values of ni and nf has to be used…. This depends on the number of π electrons present and the Pauli Exclusion Principle. Pauli exclusion principle – maximum of 2 electrons per orbital. Electron pairs must have opposite spins. See later in the course. Electronic configuration For Dye I, k =1 and we have a total of (3+3) = 6 π electrons. To create the lowest energy electron configuration we must pair the electrons in levels n = 1, 2 and 3. Therefore the highest occupied molecular orbital (HOMO) has n = 3, while the lowest unoccupied molecular orbital (LUMO) has n = 4. Hence the lowest energy transition involves promotion of a π electron from n = 3 → n = 4. Example: If the length of the box L is 8.5Å, what is the peak absorption wavelength for dye I? Are all transitions possible? – Selection rules Must always obey Pauli exclusion principle. Transitions are electric dipole transitions – the oscillating electric field component of the radiation interacts with electrical charges, i.e. the positive nuclei and negative electrons that comprise an atom or molecule, and cause the transitions observed in uv-visible absorption and emission spectroscopies. The interaction energy, U, between a system of charged particles and an electric field, E, is given by: U = − µ .E The dipole moment is defined for a collection of charges by: µ = ∑ qi ri i where ri is the position vector of charged particle i. ( See electrostatics lectures in Michaelmas term). Interaction energy The expectation value of the interaction energy is thus: U = ψ n* ( − µˆ .Eˆ )ψ n dτ ∫ µˆ and Eˆ are vectors and are the same as the classical quantities µ and E. Usually, the wavelength of the light used in electronic spectroscopy is long compared with the size of the molecule e.g. 500 nm v 1nm say. Thus, if we assume that the magnitude of the electric field is constant over the length of the molecule (and that ψ is finite only over the length of the molecule) we can write: U = − µ .E ∫ where µ = ψ n* µˆψ n dτ i.e. the strength of interaction between a distribution of charges and an electric field depends on the dipole moment of the charge distribution. The transition dipole moment In order to obtain the strength of interaction that causes a transition between two states, the transition dipole moment is used rather than the dipole moment. For a transition between and initial state, ψ i , to a final state ψ f , the transition dipole moment integral is. µ fi = ∫ψ *f µˆψ i dτ Just like the probability density is given by ψ*ψ, so the probability for a transition (as measured by the absorption coefficient) is proportional to µfi* µfi.. * µfi may be positive, negative or imaginary. If µfi = 0 then the interaction energy is zero and no transition occurs – the transition is said to be electric dipole forbidden. Conversely, if µfi is large, then the transition probability and absorption coefficient are large. Perturbations and transitions • If a molecule interacts with something, the interaction is represented by an operator Â. • If the wavefunction is not an eigenfunction of  it is transformed into another function. e.g. when momentum operator acts upon sin(kx). • The new function may be expressed as a mixture of eigenstates of the energy i.e. the interaction induces transitions to these states. • The extent to which the perturbation mixes states is given by an integral of the form * ψ  ψ d τ k j ∫ • Such an integral is called a matrix element. Transition rates If the matrix element is non-zero, the interaction causes transitions between the states labelled k and j. 2 * ˆ The rate of transition is proportional to kA jd ∫ψ ψ τ An important example is in spectroscopy. Interaction with light is represented by the electric dipole operator, µ, representing the interaction of the electric field of the light with the dipole of the molecule. The intensity of the transition is therefore proportional to. ˆ ψ µ ∫ ψ j dτ * k 2 Transition dipole moment integral for particle in a box Need to consider the transition dipole moment integral for one electron. The dipole moment operator for an electron in one dimension is –ex and so L µ fi = −e ∫ψ *f ( x) xψ i ( x)dx 0 So now we have to evaluate µfi for various wavefunctions to see which are allowed (µfi ≠ 0) and which are forbidden (µfi = 0). Example: Is n = 1 → n = 2 an allowed transition? General selection rules for the particle in a box L Consider the generalised µ = − 2e sin n f πx x sin niπx dx fi L 0 L L transition ni → nf : ∫ L (n f + ni )πx e (n f − ni )πx dx =− x cos − cos L0 L L ∫ If we then define ∆n = nf – ni and ntot = nf + ni then the above integral becomes: 2 eL 1 1 1 1 µ fi = − 2 (cos(∆nπ ) − 1) − 2 (cos(ntot π ) − 1) + sin( ∆nπ ) − sin( ntot π ) L π ∆n ∆n ntot ntot Looks bad!!!!! However, if ∆n is even then ntot is even and overall µfi = 0 – Forbidden! ∆n is odd is allowed If ∆n is odd then ntot is also odd and overall µfi ≠ 0 and is given by 2eL 1 1 8eL ni n f µ fi = − 2 2 − 2 = 2 2 2 2 π ntot ∆n π (n f − ni ) The general selection rule is ∆n is odd. However, we only really see a single peak in the absorption spectrum for each dye because other allowed transitions have very much smaller transition moments. Example: For dye 1, compare the values of the transition dipole moment integrals for the two transitions n = 3 → n = 4 and n = 3 → n = 6. Also note that longer molecules have larger absorption coefficients because µfi increases with the length of the molecule. Using symmetry to evaluate integrals An alternative to evaluating integrals is to use symmetry…… Firstly we consider the parity of the particle in a box wavefunctions by shifting the positions of the potential barriers from 0 and L to –L/2 to L/2. (We already did this in a previous lecture and so that the wavefunctions look the same for the two co-ordinate systems!) n=1 − L 2 0 n=2 L 2 − L 2 0 L 2 Symmetry II Consider the transition n =1 → n = 2 for which the transition dipole moment integral is: L/2 L/2 e e πx πx sin x cos dx = − f (x )dx µ 21 = − L −L / 2 L L −L / 2 2L ∫ ∫ The figure below show the product function f(x). n=1 x L 2 0 L 2 − ∫ n=2 × − L/2 −L/ 2 → × L 2 0 L 2 − f ( x) ≠ 0 L 2 Clearly ∫ f (x)dx ≠ 0 and the transition is allowed. 0 L 2 − L 2 0 L 2 Symmetry III In contrast, for the transition n = 1 → n = 3, ∫ f(x) dx = 0 and transition is forbidden. L/2 ∫ x n=1 n=3 × − L 2 0 L 2 − → × L 2 0 L 2 − f ( x) = 0 −L/ 2 L 2 0 L 2 − L 2 0 Conclusion: if the integrand is odd / antisymmetric / ungerade then ∫ f(x) dx = 0 and transition is forbidden. * Can determine if the product of two functions is u using the following rules: g*g= u ; u*u = g ; g*u = u. Then can readily extend result to products of more than two functions. L 2 Summary The particle in a box model can be used to interpret (i) the positions and (ii) relative intensities of absorption bands in the electronic absorption spectra of conjugated polyenes / dyes. The key quantity is determining whether a transition is allowed or not is the transition dipole moment integral. µ fi = ∫ψ *f µˆψ i dτ For cases in which this integral is zero the transition is forbidden and has zero intensity. Symmetry may be used to readily evaluate whether the integral is zero.