111: 900 Page 3 (b) Construct an energy-level diagram of 2-pentene showing the relative energy ordering of the lowest 14 molecular orbitals when temperature is 30 °C. Label the levels in terms of their quantum numbers and Eo (Eo = h²/8ma²). Be alert for degeneracies. You need to start from writing E(30) from ground state in the form of nx, ny, and n₂ as we did in the class and show all equations and calculations for the diagram. E30 Ex + Ey + E₂ = n²h²³ 2 + r²₂ h² h² 8ma² 8maz Ima² 8m. a² (* h2 n ny ・ + 8m 92 (Sa) h₁₂ (Sa) h² + + दर 0.25 a² n- 2 025a 80 (min) A 121=216. 112=218 211=120 DE: 24-16 15t 122=33E0 221=24E 8maz (nx² + 4 m² + 4 m² E. (x + 4 + 4)
111: 900 Page 3 (b) Construct an energy-level diagram of 2-pentene showing the relative energy ordering of the lowest 14 molecular orbitals when temperature is 30 °C. Label the levels in terms of their quantum numbers and Eo (Eo = h²/8ma²). Be alert for degeneracies. You need to start from writing E(30) from ground state in the form of nx, ny, and n₂ as we did in the class and show all equations and calculations for the diagram. E30 Ex + Ey + E₂ = n²h²³ 2 + r²₂ h² h² 8ma² 8maz Ima² 8m. a² (* h2 n ny ・ + 8m 92 (Sa) h₁₂ (Sa) h² + + दर 0.25 a² n- 2 025a 80 (min) A 121=216. 112=218 211=120 DE: 24-16 15t 122=33E0 221=24E 8maz (nx² + 4 m² + 4 m² E. (x + 4 + 4)
Biochemistry
9th Edition
ISBN:9781319114671
Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Chapter1: Biochemistry: An Evolving Science
Section: Chapter Questions
Problem 1P
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Transcribed Image Text:111: 900
Page 3
(b) Construct an energy-level diagram of 2-pentene showing the relative energy ordering of the
lowest 14 molecular orbitals when temperature is 30 °C. Label the levels in terms of their
quantum numbers and Eo (Eo = h²/8ma²). Be alert for degeneracies. You need to start from
writing E(30) from ground state in the form of nx, ny, and n₂ as we did in the class and show all
equations and calculations for the diagram.
E30 Ex + Ey + E₂
= n²h²³
2
+
r²₂ h²
h²
8ma²
8maz
Ima²
8m.
a²
(*
h2
n
ny
・
+
8m
92
(Sa)
h₁₂
(Sa)
h²
+
+
दर
0.25 a²
n-
2
025a
80
(min) A
121=216.
112=218
211=120
DE: 24-16 15t
122=33E0
221=24E
8maz
(nx² + 4 m² + 4 m²
E. (x + 4 + 4)
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