classes_winter07_113AID181_PracticeSelfEvaluation_2_Chem113A_W07_s0

Classes_winter07_113AID181_PracticeSelfEvaluation_2_Chem113A_W07_s0

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Unformatted text preview: Solution Key for PSE#2 [1]+[2] February 28, 2007 1a] We model the 2 delocalized -electrons 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 E = E 2- E 1 = (2 2- 1 2 ) h 2 8 mL 2 = 5 . 02 * 10- 19 J = hc (2) = 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 ( ) = A * e i 2 IE ~ (5) Applying the Schrodinger equation to equation 5 gives us- ~ 2 2 I d d A * e i 2 IE ~ =- ~ 2 2 I ( i 2 IE ~ ) 2 * A * e i 2 IE ~ = E ( ) (6) 1d] In order for the wavefunction to be single-valued (Born condition), the wavefunction must have the same value for a given value of as it does when it rotates around one full revolution ( ) = ( + 2 ) (7) 1 1e] The boundary condition in equation 7 means that e i 2 IE ~ = e i 2 IE ~ ( +2 ) = e i 2 IE ~ e i 2 IE ~ 2 (8) 1 = e i 2 IE ~ 2 (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 E = E 2- E 1 = (2 2- 1 1 ) ~ 2 2 ma 2 1 . 83 * 10- 18 J = hc (11) = 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 , well 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, a = i 2 ( P- i x ) , and a = 1 i 2 ( P + i x ) , and so we can calculate a a , remembering that x and P dont commute: a a = 1 2 ( P- i x )( P + i x ) (13) a a = 1 2 ( P 2- i x P + i P x + x 2 ) (14) a a = 1 2 ( P 2 + x 2 )- i 2 [ x, P ] (15) a a = H + 1 2 1 (16) where 1 is the identity operator....
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Classes_winter07_113AID181_PracticeSelfEvaluation_2_Chem113A_W07_s0

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