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Exam2_key

# Exam2_key - Chem 5210 Exam 2 Answer Key Question 1 From...

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Chem 5210 Exam 2 Answer Key Question 1 From Homework S4.3(a) = ~ 2 2 I ± 1 sin θ ∂θ ² sin θ ∂ψ ∂θ ³ + 1 sin 2 θ 2 ψ ∂φ 2 ´ ψ = A e 2 sin 2 θ The θ partial derivatives we will need: ∂ψ ∂θ = 2 A e 2 sin θ cos θ ∂θ ² sin θ ∂ψ ∂θ ³ = 2 A e 2 ∂θ ( sin 2 θ cos θ ) = 2 A e 2 µ - sin 3 θ + 2cos 2 θ sin θ 1 sin θ ∂θ ² sin θ ∂ψ ∂θ ³ = 2 A e 2 µ - sin 2 θ + 2cos 2 θ = 2 A e 2 µ - sin 2 θ + 2 - 2sin 2 θ = - 6 A e 2 sin 2 θ + 4 A e 2 = - 6 ψ + 4 A e 2 The φ partial derivatives we will need: ∂ψ ∂φ = 2 iA e 2 sin 2 θ 2 ψ ∂φ 2 = (2 i ) 2 A e 2 sin 2 θ = - 4 A e 2 sin 2 θ 1 sin 2 θ 2 ψ ∂φ 2 = - 4 A e 2 1

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Chem 5210 Exam 2 Answer Key So we have: = - ~ 2 2 I ±( - 6 ψ + 4 A e 2 ) - 4 A e 2 ² = +6 ~ 2 2 I ψ = E = 6 ~ 2 2 I = 3 ~ 2 I Question 2 (a) Based on Homework S4.8. First get the moment of inertia. r (H - COM) = 2 . 9637 ˚ A r (C1 - COM) = 1 . 8997 ˚ A r (C2 - COM) = 0 . 6905 ˚ A r (C3 - COM) = 0 . 6905 ˚ A r (N - COM) = 1 . 8542 ˚ A I = X m i r 2 i = (1 × 2 . 9637 2 ) + (12 × 1 . 8997 2 ) + (12 × 0 . 6905 2 ) + (12 × 0 . 6905 2 ) + (14 × 1 . 8542 2 ) = 111 . 67amu ˚ A 2 = 1 . 8537 × 10 - 45 kgm 2 B (cm) - 1 = h 8 π 2 Ic = 6 . 63 × 10 - 34 kgm 2 s - 1 8 π 2 × 1 . 8537 × 10 - 45 kgm 2 × 3 × 10 10 cms - 1 = 0 . 151cm - 1 2
Chem 5210 Exam 2 Answer Key To convert to Hz (s - 1 ) we multiply by the speed of light to give B = 4529.86 MHz. In units of either wavenumbers or Hz:

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Exam2_key - Chem 5210 Exam 2 Answer Key Question 1 From...

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