Solution Exercise 12

Solution Exercise 12 Problem 1: The Stark effect in the hydrogen atom a) Since n = 2, the quantum numbers l can take the values, 1 and m = -1,, 1.We obtain the following basis: n, l, m = 2,,, 2, 1, 1, 2, 1,, and 2, 1,. b) In exercise 11 we used the correspondence principle to obtain the Hamiltonian operator that describes the interaction: Ĥ = ee z ẑ The first-order pertubation matrix is given by: Ĥ = ee z 2,, ẑ 2,, 2,, ẑ 2, 1, 1 2,, ẑ 2, 1, 2,, ẑ 2, 1, 2, 1, 1 ẑ 2,, 2, 1, 1 ẑ 2, 1, 1 2, 1, 1 ẑ 2, 1, 2, 1, 1 ẑ 2, 1, 2, 1, ẑ 2,, 2, 1, ẑ 2, 1, 1 2, 1, ẑ 2, 1, 2, 1, ẑ 2, 1, 2, 1, ẑ 2,, 2, 1, ẑ 2, 1, 1 2, 1, ẑ 2, 1, 2, 1, ẑ 2, 1, Ĥ = ee z (1.1) c) Using the result that the diagonal elements are zero for parity reasons, one obtains the perturbation matrix 2,, ẑ 2, 1, 2, 1, ẑ 2,,. (1.2) Note that 2,, ẑ 2, 1, = 2, 1, ẑ 2,, since the wave functions are real. To determine the first-order energy corrections we need to calculate the eigenvalues of the perturbation matrix Ĥ. We can determine the eigenvalues for each block separately, and find for the four eigenvalues λ 1 = ee z 2, 1, ẑ 2,,, (1.3a) λ 2 = ee z 2, 1, ẑ 2,,, λ 3 =, λ 4 =. d) The explicit expression for the matrix element 2, 1, ẑ 2,, is 2, 1, ẑ 2,, = 2, 1, r cos(θ) 2,, = R 21(r) r R 2 (r)r 2 dr 2π π (1.3b) (1.3c) (1.3d) Y 1(θ, ϕ) cosθ Y (θ, ϕ) sinθdθ dϕ. (1.4) Using the expressions for the radial wave functions R n,l (r) and the spherical harmonics Y l,m (θ, ϕ) given on the exercise sheet, we find 2, 1, ẑ 2,, = 1 4 3 = 1 4 3 ( 36a) ( ) r 4 a r5 4 2a 5 3 4π 4π 3 { exp r } dr a 2π π 3 4π cos2 θ sinθdθdϕ = 3a. (1.5) page 1 of 6

We therefore obtain λ 1 = ee z 2, 1, ẑ 2,, = 3aeE z, (1.6a) λ 2 = ee z 2, 1, ẑ 2,, = 3aeE z, (1.6b) λ 3 =, λ 4 =. (1.6c) (1.6d) We see that the degeneracy of the n = 2 level is partially lifted and that the energy shifts are linear in E z. e) A schematic drawing of the results is depicted in figure 1-1. Figure 1-1: The linear Stark effect of the n = 2 states of the hydrogen atom. f) The force F acting on a hydrogen atom in the given electric field is F = ( V (1) s ) where only the z-component F z = 3ae E has a non-vanishing value. The hydrogen atoms have a kinetic energy of E kin = 1m 2 Hv 2 = 1.338e 22 J. In order to stop the beam of hydrogen, this kinetic energy has to be removed from the atoms and we find d stop F z = E kin which yields d stop = E kinµ. (1.7) 3m e a e E With the given parameters for hydrogen in n = 2 one calculates d stop =.526 m, so the beam would be stopped after travelling roughly half a meter into the electric field gradient. Let us consider the electric potential required to produce such a gradient: if we set U(z = ) = V then U(d stop ) = d stop E(z)dz = 1 E 2 d2 stop 1.4 MV. Applying such high voltages is experimentally very challenging and thus this is not a feasible approach to stop hydrogen atoms in the n = 2 state. The n = 2 state has a very short lifetime of a few nanoseconds which also makes this approach not feasible experimentally. g) We only have to slightly modify Eq. (1.7) to find E kin µ d stop = 3 nkm. (1.8) 2 ea e E which yields for n = 3, k = 29 a stopping distance d stop = 1.21 mm. Applying the maximum voltage U(d stop ) on the order of ten volt is trivial and thus this is a very well suited method of producing a sample of slow stopped hydrogen atoms. page 2 of 6

h) We note that the radial wave functions (2.3a) to (2.3c) on the exercise sheet are given in the so-called mass normalised coordinate r me where a = a a µ = me 4πɛ 2 µ m ee. In positronium, the 2 reduced mass is µ = meme m e+m e = me and thus a 2 Ps = 2a. We then find V s (1) = 2 3a ee z and Eq. (1.7) becomes d stop = E kin. (1.9) 6a e E 2E The given kinetic energy of 42 mev corresponds to a velocity of v = kin 2m e = 9.1 1 4 m s, the stopping distance would be d stop = 15 m. Problem 2: Hückel model (Exam question) a) The Hückel matrices for the allyl radical, H 1, and cyclopropenyl radical, H 2, are given by: α β H 1 = β α β (2.1) β α α β β H 2 = β α β (2.2) β β α b) The eigenvalues of the Hückel matrix for the allyl radical are given by: α ɛ β det β α ɛ β = (2.3) β α ɛ (α ɛ) 3 (α ɛ)β 2 β 2 (α ɛ) = (2.4) (α ɛ)((α ɛ) 2 2β 2 ) = (2.5) ɛ 1 = α, ɛ 2 = α + 2β, ɛ 3 = α 2β (2.6) A schematic drawing of the energy level diagram for the conjugated π-electron system is depicted in figure 2-1. Note that β < Figure 2-1: Energy level diagram for the conjugated π-electron system for allyl radical. c) A schematic drawing of the ground state configuration for the electron system for the conjugated π-electron system is depicted in figure 2-2. page 3 of 6

Figure 2-2 d) A schematic drawing of the possible transitions are depicted in figure 2-3. We have N A = 6.2214 1 23 mol, β = 25 kjmol β/n A = 4.1513 1 9 J (2.7) E 1 = E 2 = 1 /2 E 3 = 2β (2.8) λ 1 = λ 2 = hc / E 1 = hc / 2β = 338.36 nm (2.9) λ 3 = λ 2 /2 = 169.18 nm (2.1) ν 1 = ν 2 = 1 /λ 2 = 29554.4 cm (2.11) The calculated wave number is not the same as determined experimentally (24485 cm ). The difference can be explained as lack of electron-electron interactions in the model and imperfect value for β. Figure 2-3 e) A schematic drawing of the ground state configuration for the electron system of cyclopropenyl for the conjugated π-electron system is depicted in figure 2-4. page 4 of 6

Figure 2-4 f) The ground state energies as sums of orbital energies of the electrons are: c C 3 H 3 : E = 2(α + 2β) + 2(α β) = 4α + 2β (2.12) c C 3 H 3 : E = 2(α + 2β) + α β = 3α + 3β (2.13) c C 3 H + 3 : E = 2(α + 2β) = 2α + 4β (2.14) C 3 H 5 : E = 2(α + 2β) + 2α = 4α + 2 2β (2.15) C 3 H 5 : E = 2(α + 2β) + α = 3α + 2 2β (2.16) C 3 H + 5 : E = 2(α + 2β) = 2α + 2 2β (2.17) g) It is only c-c 3 H 3 that is a cyclic system with c-c 3 H 3 being antiaromatic (with 4 electrons) and c-c 3 H + 3 being aromatic (with 2 electrons). An electron in one double bond has the energy E = α+β which we can compare to energies for the c-c 3 H 3 and c-c 3 H + 3 systems: c C 3 H 3 : E = 4α + 2β 4(α + β) = 2β (2.18) c C 3 H + 3 : E = 2α + 4β 2(α + β) = 2β (2.19) As β <, the antiaromatic c-c 3 H 3 is destabilized by -2β and the aromatic c-c 3 H + 3 is stabilized by 2β. h) The eigenvector for ɛ 1 =α+2β is determined by: α (α + 2β) β β x 1 β α (α + 2β) β x 2 = (2.2) β β α (α + 2β) x 3 2x 1 + x 2 + x 3 = (2.21) x 1 2x 2 + x 3 = (2.22) x 1 + x 2 2x 3 = (2.23) x 1 = x 2 = x 3 (2.24) The eigenvector for ɛ 1 =α+2β is (normalized) 1 1 1 (2.25) 3 1 page 5 of 6

The eigenvector for ɛ 2 =ɛ 3 =α-β is determined by: α (α β) β β x 1 β α (α β) β x 2 = (2.26) β β α (α β) x 3 As ɛ 2 and ɛ 2 are degenerated, a linear combination of the corresponding eigenvectors for ɛ 2 and ɛ 3 will also be eigenvector to the eigenvalues. One have to find two eigenvectors that fulfill x 1 + x 2 + x 3 = and are orthogonal. Lets assume 1 Ψ 1 = 1 and Ψ 2 = (2.27) Ψ 1 and Ψ 2 are not orthogonal (Note that Ψ 1 = Ψ 2 ) but the following are: 1 Ψ 1 + Ψ 2 = and Ψ 1 Ψ 2 = 2 (2.28) These are eigenvectors (not normalized) of the Hückel matrix. i) A schematic drawing of the molecular orbitals are depicted in figure 2-5.The coefficients for the eigenvectors determine the size of the orbitals. Figure 2-5 page 6 of 6