ATKINS' PHYSICAL CHEMISTRY-ACCESS
ATKINS' PHYSICAL CHEMISTRY-ACCESS
11th Edition
ISBN: 9780198834700
Author: ATKINS
Publisher: OXF
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Chapter 7, Problem 7E.5P

(a)

Interpretation Introduction

Interpretation:

The wavenumber of the radiation corresponding to photons, ν˜ is given by the relation ν˜=ω/2πc has to be shown.

Concept Introduction:

A harmonic oscillator consists of a particle of mass m that undergoes a restoring force which is proportional to its amplitude x.  The restoring force is represented by kf which is known as the force constant.  The vibrational frequency is ω.  It is directly proportional to the force constant.

(a)

Expert Solution
Check Mark

Answer to Problem 7E.5P

The wavenumber of the radiation corresponding to photons, ν˜ is given by the relation ν˜=ω/2πc.

Explanation of Solution

The energy of the photons is given by the relation given below.

    E=ω        (1)

Where,

is a constant.

ω is the vibrational frequency.

The value of is given below.

    =h2π        (2)

Where,

h is Planck’s constant.

The energy of photons is also given by the relation below.

    E=hcλ        (3)

Where,

 h is Planck’s constant.

 c is the speed of light.

 λ is the wavelength.

Substitute the equations (3) and (2) in (1).

    hcλ=h2πωc=ωλ2πλ=2πcω        (4)

The relation wavelength (λ) and wavenumber (ν˜) is given below.

    1λ=ν˜        (5)

Substitute equation (5) in (4).

    1ν˜=2πcων˜=ω2πc

Therefore, the wavenumber of the radiation corresponding to photons, ν˜ is given by the relation ν˜=ω/2πc.

(b)

Interpretation Introduction

Interpretation:

The ν˜ of the molecule, 1H35Cl with ω=5.63×1014s1 has to be calculated.

Concept Introduction:

Refer part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 7E.5P

The ν˜ of the molecule, 1H35Cl with ω=5.63×1014s1 is 0.299×106m1_.

Explanation of Solution

The formula to calculate ν˜ is given by the expression below.

    ν˜=ω2πc        (6)

Where,

 c is the speed of light.

 ω is the vibrational frequency.

The speed of light (c) is 3×108ms1.

The vibrational frequency (ω) is 5.63×1014s1

Substitute the value of c and ω in equation (6).

    ν˜=5.63×1014s12×3.14×3×108ms1=0.299×106m1_

Therefore, the ν˜ of the molecule, 1H35Cl with ω=5.63×1014s1 is 0.299×106m1_.

(c)

Interpretation Introduction

Interpretation:

An expression for the force constant, kf has to derived in terms of ν˜.

Concept introduction:

Refer part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 7E.5P

The expression for the force constant, kf is derived in terms of ν˜ as shown below.

    ν˜=12πckfμ

Explanation of Solution

The formula to calculate ν˜ is given by the expression below.

    ν˜=ω2πc        (6)

Where,

 c is the speed of light.

 ω is the vibrational frequency.

The force constant, kf is given by the expression below.

    ω=kfμ

Where,

μ is the reduced mass of the molecule.

Substitute the equation (7) in equation (6).

    ν˜=12πckfμ

(d)

Interpretation Introduction

Interpretation:

The force constant and the wavenumber corresponding to the transition v=01 in 12C16O has to be calculated for the molecule 14C16O.

Concept introduction:

Refer part (a).

(d)

Expert Solution
Check Mark

Answer to Problem 7E.5P

The force constant of the 14C16O is 1902.1Nm-1_.  The wavenumber at which the transition takes place from transition from v=01 for 14C16O is 2.07×105m1_.

Explanation of Solution

The formula to calculate the wavenumber of the transition from v=01 is given below.

    ν˜=12πckfμ        (7)

Where,

kf is force constant.

μ is the reduced mass of the molecule.

c is the speed of light.

The reduced mass of 12C16O is calculated by the formula shown below.

    μ=m12Cm16Om12C+m16O        (8)

The mass of 12C (m12C) is 12amu.

The mass of 16O (m16O) is 16amu.

Substitute the value of m12C and m16O in equation (8).

    μ=12amu×16amu12amu+16amu=6.857amu

The conversion of 1amu to kg is shown below.

    1amu=1.66×1027kg

Therefore, the conversion of 6.857amu to kg is shown below.

    6.857amu=6.857×1.66×1027kg=1.138×1026kg

The value of ν˜ is 2170cm1.

The conversion of 1cm1 to m1 is shown below.

    1cm1=102m1

Therefore, the conversion of 2170cm1 to m1 is shown below.

    2170cm1=2170×102m1

The speed of light (c) is 3×108ms1.

Substitute the value of ν˜, c and μ in equation (7).

    2170×102m1=12×3.14×3×108ms1kf1.138×1026kgkf=4×(3.14)2×(3×108ms1)2×(2170×102m1)2×1.138×1026kg=1902.1kgm2s2m2×1Nm-1kgm2s2m2=1902.1Nm-1

The force constant of 12C16O is equal to force constant of 14C16O.

Therefore, the force constant of the 12C16O is 1902.1Nm-1_.

The reduced mass of 14C16O is calculated by the formula shown below.

    μ=m14Cm16Om14C+m16O        (8)

The mass of 14C (m12C) is 14amu.

The mass of 16O (m16O) is 16amu.

Substitute the value of m14C and m16O in equation (8).

    μ=14amu×16amu14amu+16amu=7.47amu

The conversion of 1amu to kg is shown below.

    1amu=1.66×1027kg

Therefore, the conversion of 7.47amu to kg is shown below.

    7.47amu=7.47×1.66×1027kg=1.239×1026kg

The force constant (kf) is 1902.1Nm-1.

Substitute the value of c, kf and μ in equation (1).

    ν˜=12×3.14×3×108ms11902.1Nm-11.239×1026kg=16.28×3×108ms11902.1Nm-11.239×1026kg×kgm2s21N=16.28×3×108ms1×39.18×1013s1=2.07×105m1_

Therefore, the wavenumber at which the transition takes place from transition from v=01 is 2.07×105m1_.

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Chapter 7 Solutions

ATKINS' PHYSICAL CHEMISTRY-ACCESS

Ch. 7 - Prob. 7D.1STCh. 7 - Prob. 7E.1STCh. 7 - Prob. 7E.2STCh. 7 - Prob. 7F.1STCh. 7 - Prob. 7A.1DQCh. 7 - Prob. 7A.2DQCh. 7 - Prob. 7A.3DQCh. 7 - Prob. 7A.4DQCh. 7 - Prob. 7A.1AECh. 7 - Prob. 7A.1BECh. 7 - Prob. 7A.2AECh. 7 - Prob. 7A.2BECh. 7 - Prob. 7A.3AECh. 7 - Prob. 7A.3BECh. 7 - Prob. 7A.4AECh. 7 - Prob. 7A.4BECh. 7 - Prob. 7A.5AECh. 7 - Prob. 7A.5BECh. 7 - Prob. 7A.6AECh. 7 - Prob. 7A.6BECh. 7 - Prob. 7A.7AECh. 7 - Prob. 7A.7BECh. 7 - Prob. 7A.8AECh. 7 - Prob. 7A.8BECh. 7 - Prob. 7A.9AECh. 7 - Prob. 7A.9BECh. 7 - Prob. 7A.10AECh. 7 - Prob. 7A.10BECh. 7 - Prob. 7A.11AECh. 7 - Prob. 7A.11BECh. 7 - Prob. 7A.12AECh. 7 - Prob. 7A.12BECh. 7 - Prob. 7A.13AECh. 7 - Prob. 7A.13BECh. 7 - Prob. 7A.1PCh. 7 - Prob. 7A.2PCh. 7 - Prob. 7A.3PCh. 7 - Prob. 7A.4PCh. 7 - Prob. 7A.5PCh. 7 - Prob. 7A.6PCh. 7 - Prob. 7A.7PCh. 7 - Prob. 7A.8PCh. 7 - Prob. 7A.9PCh. 7 - Prob. 7A.10PCh. 7 - Prob. 7B.1DQCh. 7 - Prob. 7B.2DQCh. 7 - Prob. 7B.3DQCh. 7 - Prob. 7B.1AECh. 7 - Prob. 7B.1BECh. 7 - Prob. 7B.2AECh. 7 - Prob. 7B.2BECh. 7 - Prob. 7B.3AECh. 7 - Prob. 7B.3BECh. 7 - Prob. 7B.4AECh. 7 - Prob. 7B.4BECh. 7 - Prob. 7B.5AECh. 7 - Prob. 7B.5BECh. 7 - Prob. 7B.6AECh. 7 - Prob. 7B.6BECh. 7 - Prob. 7B.7AECh. 7 - Prob. 7B.7BECh. 7 - Prob. 7B.8AECh. 7 - Prob. 7B.8BECh. 7 - Prob. 7B.1PCh. 7 - Prob. 7B.2PCh. 7 - Prob. 7B.3PCh. 7 - Prob. 7B.4PCh. 7 - Prob. 7B.5PCh. 7 - Prob. 7B.7PCh. 7 - Prob. 7B.8PCh. 7 - Prob. 7B.9PCh. 7 - Prob. 7B.11PCh. 7 - Prob. 7C.1DQCh. 7 - Prob. 7C.2DQCh. 7 - Prob. 7C.3DQCh. 7 - Prob. 7C.1AECh. 7 - Prob. 7C.1BECh. 7 - Prob. 7C.2AECh. 7 - Prob. 7C.2BECh. 7 - Prob. 7C.3AECh. 7 - Prob. 7C.3BECh. 7 - Prob. 7C.4AECh. 7 - Prob. 7C.4BECh. 7 - Prob. 7C.5AECh. 7 - Prob. 7C.5BECh. 7 - Prob. 7C.6AECh. 7 - Prob. 7C.6BECh. 7 - Prob. 7C.7AECh. 7 - Prob. 7C.7BECh. 7 - Prob. 7C.8AECh. 7 - Prob. 7C.8BECh. 7 - Prob. 7C.9AECh. 7 - Prob. 7C.9BECh. 7 - Prob. 7C.10AECh. 7 - Prob. 7C.10BECh. 7 - Prob. 7C.1PCh. 7 - Prob. 7C.2PCh. 7 - Prob. 7C.3PCh. 7 - Prob. 7C.4PCh. 7 - Prob. 7C.5PCh. 7 - Prob. 7C.6PCh. 7 - Prob. 7C.7PCh. 7 - Prob. 7C.8PCh. 7 - Prob. 7C.9PCh. 7 - Prob. 7C.11PCh. 7 - Prob. 7C.12PCh. 7 - Prob. 7C.13PCh. 7 - Prob. 7C.14PCh. 7 - Prob. 7C.15PCh. 7 - Prob. 7D.1DQCh. 7 - Prob. 7D.2DQCh. 7 - Prob. 7D.3DQCh. 7 - Prob. 7D.1AECh. 7 - Prob. 7D.1BECh. 7 - Prob. 7D.2AECh. 7 - Prob. 7D.2BECh. 7 - Prob. 7D.3AECh. 7 - Prob. 7D.3BECh. 7 - Prob. 7D.4AECh. 7 - Prob. 7D.4BECh. 7 - Prob. 7D.5AECh. 7 - Prob. 7D.5BECh. 7 - Prob. 7D.6AECh. 7 - Prob. 7D.6BECh. 7 - Prob. 7D.7AECh. 7 - Prob. 7D.7BECh. 7 - Prob. 7D.8AECh. 7 - Prob. 7D.8BECh. 7 - Prob. 7D.9AECh. 7 - Prob. 7D.9BECh. 7 - Prob. 7D.10AECh. 7 - Prob. 7D.10BECh. 7 - Prob. 7D.11AECh. 7 - Prob. 7D.11BECh. 7 - Prob. 7D.12AECh. 7 - Prob. 7D.12BECh. 7 - Prob. 7D.13AECh. 7 - Prob. 7D.13BECh. 7 - Prob. 7D.14AECh. 7 - Prob. 7D.14BECh. 7 - Prob. 7D.15AECh. 7 - Prob. 7D.15BECh. 7 - Prob. 7D.1PCh. 7 - Prob. 7D.2PCh. 7 - Prob. 7D.3PCh. 7 - Prob. 7D.4PCh. 7 - Prob. 7D.5PCh. 7 - Prob. 7D.6PCh. 7 - Prob. 7D.7PCh. 7 - Prob. 7D.8PCh. 7 - Prob. 7D.9PCh. 7 - Prob. 7D.11PCh. 7 - Prob. 7D.12PCh. 7 - Prob. 7D.14PCh. 7 - Prob. 7E.1DQCh. 7 - Prob. 7E.2DQCh. 7 - Prob. 7E.3DQCh. 7 - Prob. 7E.1AECh. 7 - Prob. 7E.1BECh. 7 - Prob. 7E.2AECh. 7 - Prob. 7E.2BECh. 7 - Prob. 7E.3AECh. 7 - Prob. 7E.3BECh. 7 - Prob. 7E.4AECh. 7 - Prob. 7E.4BECh. 7 - Prob. 7E.5AECh. 7 - Prob. 7E.5BECh. 7 - Prob. 7E.6AECh. 7 - Prob. 7E.6BECh. 7 - Prob. 7E.7AECh. 7 - Prob. 7E.7BECh. 7 - Prob. 7E.8AECh. 7 - Prob. 7E.8BECh. 7 - Prob. 7E.9AECh. 7 - Prob. 7E.9BECh. 7 - Prob. 7E.1PCh. 7 - Prob. 7E.2PCh. 7 - Prob. 7E.3PCh. 7 - Prob. 7E.4PCh. 7 - Prob. 7E.5PCh. 7 - Prob. 7E.6PCh. 7 - Prob. 7E.7PCh. 7 - Prob. 7E.8PCh. 7 - Prob. 7E.9PCh. 7 - Prob. 7E.12PCh. 7 - Prob. 7E.15PCh. 7 - Prob. 7E.16PCh. 7 - Prob. 7E.17PCh. 7 - Prob. 7F.1DQCh. 7 - Prob. 7F.2DQCh. 7 - Prob. 7F.3DQCh. 7 - Prob. 7F.1AECh. 7 - Prob. 7F.1BECh. 7 - Prob. 7F.2AECh. 7 - Prob. 7F.2BECh. 7 - Prob. 7F.3AECh. 7 - Prob. 7F.3BECh. 7 - Prob. 7F.4AECh. 7 - Prob. 7F.4BECh. 7 - Prob. 7F.5AECh. 7 - Prob. 7F.5BECh. 7 - Prob. 7F.6AECh. 7 - Prob. 7F.6BECh. 7 - Prob. 7F.7AECh. 7 - Prob. 7F.7BECh. 7 - Prob. 7F.8AECh. 7 - Prob. 7F.8BECh. 7 - Prob. 7F.9AECh. 7 - Prob. 7F.9BECh. 7 - Prob. 7F.10AECh. 7 - Prob. 7F.10BECh. 7 - Prob. 7F.11AECh. 7 - Prob. 7F.11BECh. 7 - Prob. 7F.12AECh. 7 - Prob. 7F.12BECh. 7 - Prob. 7F.13AECh. 7 - Prob. 7F.13BECh. 7 - Prob. 7F.14AECh. 7 - Prob. 7F.14BECh. 7 - Prob. 7F.1PCh. 7 - Prob. 7F.4PCh. 7 - Prob. 7F.6PCh. 7 - Prob. 7F.7PCh. 7 - Prob. 7F.8PCh. 7 - Prob. 7F.9PCh. 7 - Prob. 7F.10PCh. 7 - Prob. 7F.11PCh. 7 - Prob. 7.3IACh. 7 - Prob. 7.4IACh. 7 - Prob. 7.5IACh. 7 - Prob. 7.6IA
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