This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Solution Key for PSE#2 [1]+[2] February 28, 2007 1a] We model the 2 delocalized πelectrons in anthracene as in a box E n = h 2 n 2 8 mL 2 (1) Since 2 electrons can fit into each energy level, the HOMO = 1, and the LUMO = 2. Thus Δ E = E 2 E 1 = (2 2 1 2 ) h 2 8 mL 2 = 5 . 02 * 10 19 J = hc λ (2) λ = 3 . 96 * 10 7 m = 396 nm (3) 1b] From this wavefunction, we see that there are 3 nodes, indicating n = 4 . Thus plug n = 4 into equation 1 where L = 0 . 867 nm . E 4 = 4 2 h 2 8 mL 2 = 3 . 21 * 10 19 J (4) 1c] The wavefunction is given as Ψ( θ ) = A * e i √ 2 IE ~ θ (5) Applying the Schrodinger equation to equation 5 gives us ~ 2 2 I d dθ A * e i √ 2 IE ~ θ = ~ 2 2 I ( i √ 2 IE ~ ) 2 * A * e i √ 2 IE ~ θ = E Ψ( θ ) (6) 1d] In order for the wavefunction to be singlevalued (Born condition), the wavefunction must have the same value for a given value of θ as it does when it rotates around one full revolution Ψ( θ ) = Ψ( θ + 2 π ) (7) 1 1e] The boundary condition in equation 7 means that e i √ 2 IE ~ θ = e i √ 2 IE ~ ( θ +2 π ) = e i √ 2 IE ~ θ e i √ 2 IE ~ 2 π (8) 1 = e i √ 2 IE ~ 2 π (9) Equation 9 is only satisfied if √ 2 IE ~ = 0 , ± 1 , ± 2 , ..., ± n . Consequently, we see that the energies for the particle on a ring must be E n = n 2 ~ 2 2 I (10) where n = 0 , ± 1 , ± 2 , ... 1f] We can see that there is one energy level for n = 0 . There are two degenerate energy levels for n = ± 1 . Since there are six electrons, we can easily see that the highest occupied level is the n = ± 1 energy level. Thus the HOMO → LUMO transition is from n = 1 → 2 . Keeping in mind that I = ma 2 , we can see that Δ E = E 2 E 1 = (2 2 1 1 ) ~ 2 2 ma 2 1 . 83 * 10 18 J = hc λ (11) λ = 1 . 09 * 10 7 m = 109 nm (12) 1g] For n = 0 , the wavefunction has no nodes. Because the Born condition must be satisfied, we will never see an odd number of nodes. Consequently, for n = 1 , we’ll see two nodes, and thus we can see that # of nodes = 2 n . Thus from the diagram, n = 2 , and thus E 2 = 2 2 ~ 2 2 ma 2 = 2 . 44 * 10 18 J 2a] In reduced units, ˆ a = i √ 2 ( ˆ P i ˆ x ) , and ˆ a ‡ = 1 i √ 2 ( ˆ P + i ˆ x ) , and so we can calculate ˆ a ˆ a ‡ , remembering that ˆ x and ˆ P don’t commute: ˆ a ˆ a ‡ = 1 2 ( ˆ P i ˆ x )( ˆ P + i ˆ x ) (13) ˆ a ˆ a ‡ = 1 2 ( ˆ P 2 i ˆ x ˆ P + i ˆ P ˆ x + ˆ x 2 ) (14) ˆ a ˆ a ‡ = 1 2 ( ˆ P 2 + ˆ x 2 ) i 2 [ˆ x, ˆ P ] (15) ˆ a ˆ a ‡ = ˆ H + 1 2 ˆ 1 (16) where ˆ 1 is the identity operator....
View
Full Document
 Winter '07
 Lin
 Electron, Quantum Chemistry, Uncertainty Principle

Click to edit the document details