Explanation Given info: The transition is between first three rotational states in the n = 1 level to any states in the n = 0 level. The rotational inertia is 4.60 × 10 − 48 kg ⋅ m 2 and classical vibrational frequency is 3 .69 × 10 14 Hz . Explanation: The allowed transitions are given by Δ n = ± 1 , Δ l = ± 1 . The allowed transitions are ( 1 , 0 ) → ( 0 , 1 ) , ( 1 , 1 ) → ( 0 , 0 ) , ( 1 , 1 ) → ( 0 , 2 ) , ( 1 , 2 ) → ( 0 , 1 ) and ( 1 , 2 ) → ( 0 , 3 ) . Write the expression for the vibrational energy difference for transition between adjacent states. Δ E vib = ( n f + 1 2 ) ℏ ω − ( n i + 1 2 ) ℏ ω Here, n i is the principal quantum number of the initial state n f is the principal quantum number of the final state ℏ is the reduced Planck’s constant ω is the angular frequency Substitute 0 for n f , 1 for n i , h / 2 π for ℏ and 2 π f for ω in the above equation to find the energy difference between adjacent vibrational levels. Δ E vib = − h f Here, h is the Planck’s constant f is the classical vibrational frequency Write the expression for the rotational energy difference for transition between adjacent states. Δ E rot = ℏ 2 2 I l f ( l f + 1 ) − ( ℏ 2 2 I l i ( l i + 1 ) ) Here, l i is the orbital quantum number of the initial state l f is the orbital quantum number of the final state I is the rotational inertia of the molecule Substitute l i = 0 and l f = 1 to find Δ E rot 0 → 1 . Δ E rot 0 → 1 = ℏ 2 2 I 1 ( 1 + 1 ) − 0 = ℏ 2 I Substitute l i = 1 and l f = 0 to find Δ E rot 1 → 0 . Δ E rot 1 → 0 = 0 − ℏ 2 2 I 1 ( 1 + 1 ) = − ℏ 2 I Substitute l i = 1 and l f = 2 to find Δ E rot 1 → 2 . Δ E rot 1 → 2 = ℏ 2 2 I 2 ( 2 + 1 ) − ℏ 2 2 I 1 ( 1 + 1 ) = 2 ℏ 2 I Substitute l i = 2 and l f = 1 to find Δ E rot 2 → 1 . Δ E rot 2 → 1 = ℏ 2 2 I 1 ( 1 + 1 ) − ℏ 2 2 I 2 ( 2 + 1 ) = − 2 ℏ 2 I Substitute l i = 2 and l f = 3 to find Δ E rot 2 → 3 . Δ E rot 2 → 3 = ℏ 2 2 I 3 ( 3 + 1 ) − ℏ 2 2 I 2 ( 2 + 1 ) = 3 ℏ 2 I Write the expression for total energy difference due to transitions. Δ E = Δ E vib + Δ E rot l i → l f (1) Substitute − h f for Δ E vib and ℏ 2 I for Δ E rot 0 → 1 in the equation (1). Δ E ( 1 , 0 ) → ( 0 , 1 ) = − h f + ℏ 2 I Substitute 4.136 × 10 − 15 eV ⋅ s for h , 3.69 × 10 14 Hz for f and 4.60 × 10 − 48 kg ⋅ m 2 for I in the above equation. Δ E ( 1 , 0 ) → ( 0 , 1 ) = − ( 4.136 × 10 − 15 eV ⋅ s ) ( 3.69 × 10 14 Hz ) + ( 4.136 × 10 − 15 eV ⋅ s ) 2 ( 1.602 × 10 − 19 J / eV ) 4 π 2 ( 4.60 × 10 − 48 kg ⋅ m 2 ) = − 1.51 eV Substitute − h f for Δ E vib and − ℏ 2 I for Δ E rot 1 → 0 in the equation (1). Δ E ( 1 , 1 ) → ( 0 , 0 ) = − h f − ℏ 2 I Substitute 4.136 × 10 − 15 eV ⋅ s for h , 3.69 × 10 14 Hz for f and 4.60 × 10 − 48 kg ⋅ m 2 for I in the above equation. Δ E ( 1 , 1 ) → ( 0 , 0 ) = − ( 4.136 × 10 − 15 eV ⋅ s ) ( 3.69 × 10 14 Hz ) − ( 4

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ISBN: 9780134988566

Author: Wolfson, Richard

Publisher: Pearson Education,

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Chapter 37, Problem 44P

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The wavelengths for excitation of H2 in all allowed transitions between first three rotational states in the n=1 level to any states in the n=0 level.

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Chapter 37, Problem 43P

Chapter 37, Problem 45P

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The wavelengths for excitation of H 2 in all allowed transitions between first three rotational states in the n = 1 level to any states in the n = 0 level.

Solution Summary: The author determines the wavelengths for excitation of H_2 in all allowed transitions between first three rotational states in the n=1 level to any states

The wavelengths for excitation of H2 in all allowed transitions between first three rotational states in the n=1 level to any states in the n=0 level.

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Chapter 37 Solutions