Problem Set 3 Q18.5 Does the average length of a quantum

Problem Set 3 Q18.5 Does the average length of a quantum harmonic oscillator depend on its energy? Answer this question by referring to the harmonic potential function shown in Fig. 18.7. The average length is the midpoint of the horizontal line connecting the two parts of V(x). No, because the potential function is symmetric about x = 0, the average length is independent of energy. Q18.6 Does the bond length of a real molecule depend on its energy? Answer this question by referring to Fig. 18.7. The bond length is the midpoint of the horizontal line connecting the two parts of V(x). Yes, in a realistic anharmonic potential, the bond length increases with the quantum number. P18.3 In discussing molecular rotation, the quantum number J is used rather than l. Using the Boltzmann distribution calculate nJ/n0 for 1H35Cl for J = 0, 5, 10 and 20 at T = 1025 K. Does nJ/n0 go through a maximum as J increases? If so, what can you say about the value of J corresponding to the maximum? ⎡ −(E J − E 0 ) ⎤
⎡ −E J ⎤
nJ
= (2J +1)exp ⎢
⎥ = (2J +1)exp ⎢
⎥
n0
kB T
⎣
⎦
⎣ kB T ⎦ I = µr02 =
€
E=
€
1.0078amu × 34.6698amu
× 1.66 × 10 −27 kg /amu × (127.5x10 −12 m) 2 = 2.64 × 10 −47 kg m2 1.0078amu + 34.6698amu
2
(1.055 × 10 −34 J s) 2
J(J +1)
J(J +1) = 2.104 × 10 −22 J(J +1)
2I
2 × 2.64 × 10 −47 kg m2
n0
=1
n0
€
€
⎡ −2.104 x10 −22 J(30) ⎤
n5
= (2 × 5 +1)exp ⎢
⎥ = 7.04
−23
−1
n0
⎣1.38x10 JK x1025K ⎦
€
⎡ −2.104 x10 −22 J(110) ⎤
n10
= (2 × 10 +1)exp ⎢
⎥ = 4.09
−23
−1
n0
⎣1.38x10 JK x1025K ⎦
€
⎡ −2.104 x10 −22 J(420) ⎤
n 20
= (2 × 20 +1)exp ⎢
⎥ = 0.0795
−23
−1
n0
⎣1.38x10 JK x1025K ⎦
nJ/n0 goes through a maximum for J between 1 and 5. €
P18.5 1H19F has a force constant of 966 N m-‐1 and a bond length of 91.68 pm. Calculate the frequency of the light corresponding to the lowest energy pure vibrational and pure rotational transitions. In what regions of the electromagnetic spectrum do the transitions lie? 1
ν=
2π
k
1
=
µ 2π
966 kg s-2
= 1.24 × 1014 s−1 −27
1.008 amu × 18.9984 amu 1.661 × 10 kg
×
1.008 amu +18.9984 amu
amu
This frequency lies in the infrared region of the electromagnetic spectrum. €
The lowest energy transition is J = 0  J = 1. I = µr02 =
1.0078amu × 18.9984amu
× 1.66 × 10 −27 kg /amu × (91.68x10 −12 m) 2 = 1.34 × 10 −47 kg m2 1.0078amu +18.9984amu
ΔE rot = 2(J +1)
€
ν=
€
€
2
2 µr02
=
2
µr02
=
(1.055 × 10 −34 J s) 2
= 8.33 × 10 −22 J 1.34 × 10 −47 kg m2
ΔE rot
= 1.25 × 1012 s−1
h
This frequency lies in the microwave region of the electromagnetic spectrum. Q19.7 Nitrogen and oxygen do not absorb infrared radiation and are therefore not greenhouse gases. Why is this the case? They have no permanent dipole and therefore cannot have a transient dipole moment. Q19.12 What is the difference between a permanent and a dynamic dipole moment? The permanent dipole moment arises from a difference in electronegativity between the bonded atoms. The dynamic dipole moment is the variation in the dipole moment as the molecule vibrates. P19.2 The infrared spectrum of 7Li19F has an intense line at 910.57 cm-‐1. Calculate the force constant and period of vibration of this molecule. k = 4 π 2ν 2 µ
= 4 π 2 (2.998 × 1010 cm s-1 × 910.57 cm-1 ) 2 ×
7.016 amu × 18.9984 amu 1.661 × 10 −27 kg
×
= 250 N m−1
7.016 amu +18.9984 amu
amu
T = 1/ν = 3.66 x 10-‐14 s. €
P19.14 The bond length of 7Li1H is 159.49 pm. Calculate the value of B and the spacing between lines in the pure rotational spectrum of this molecule in units of s-‐1. k=
€
h
8π µcr02
2
6.626 × 10 −34 J s
=
= 7.523 cm−1
−27
7.016 amu × 18.998 amu 1.661 × 10 kg
8π 2
×
(2.998 × 1010 cm s-1 )(159.5 × 10 -10 cm) 2
7.016 amu +18.998 amu
amu
The spacing between adjacent lines is
Δν = 2cB = 2(2.998 x 1010 cm s-‐1)(7.523 cm-‐1) = 4.511 x 1011 s-‐1. P19.15 Calculating the motion of individual atoms in the vibrational modes of molecules (called normal modes) is an advanced topic. Given the normal modes for water and CO2 shown in the book, decide which if the modes have nonzero dynamical dipole moment and are therefore infrared active. All three vibrational modes of water (symmetric stretch, asymmetric stretch and bend) will lead to a change in the dipole moment and are therefore infrared active. The symmetric stretch of carbon dioxide will not lead to a change in the dipole moment and is infrared active. The other two modes will lead to a change in the dipole moment are infrared active.

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