Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
9th Edition
ISBN: 9781305372337
Author: Raymond A. Serway | John W. Jewett
Publisher: Cengage Learning
Question
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Chapter 43, Problem 21P

(a)

To determine

The moment of inertia of the hydrogen molecule about an axis through its center of mass and perpendicular to H-H bond.

(a)

Expert Solution
Check Mark

Answer to Problem 21P

The moment of inertia of the hydrogen molecule about an axis through its center of mass and perpendicular to H-H bond is 4.60×1048kgm2.

Explanation of Solution

A hydrogen molecule makes a transition from ground level to v=1 and J=1 after absorbing a photon of wavelength 2.2112μm. The molecule emit a photon of wavelength 2.4054μm after dropping to level v=0 and J=2.

Write the formula for energy levels.

    EvJ=(v+12)hf+22IJ(J+1)                                                                                 (I)

Here, EvJ is the energy, v is the vibrational quantum number, h is the Planck’s constant, f is the frequency, J is the rotational quantum number, and I is the moment of inertia.

Refer equation (I) and find energy of v=0 and J=0.

    E00=12hf

Here, E00 is the energy of level  v=0 and J=0.

Refer equation (I) and find energy of v=1 and J=1.

    E11=32hf+2I

Here, E11 is the energy of level v=1 and J=1.

Refer equation (I) and find energy of v=0 and J=2.

    E02=12hf+32I

Here, E02 is the energy of level v=0 and J=2.

Write the formula for the energy difference between v=0, J=0 level to v=1, J=1.

    ΔE1=E11E00=hf+2I=hcλ1                                                                                          (II)

Here, ΔE1 is the energy difference between v=0, J=0 level to v=1, J=1, λ1 is the wavelength of the photon required for the transition from energy difference between v=0, J=0 level to v=1, J=1.

Write the formula for the energy difference between v=1, J=1 level to v=0, J=2.

    ΔE2=E11E02=hf22I=hcλ2                                                                                    (III)

Here, ΔE2 is the energy difference between v=1, J=1 level to v=0, J=2, λ2 is the wavelength required from the transition between v=1, J=1 level to v=0, J=2.

Subtract equation (III) from (II).

    ΔE1ΔE2=(hf+2I)(hf22I)=hcλ1hcλ2

Re-write the above equation.

    32I=hc(1λ11λ2)3h4π2I=c(1λ11λ2)

Re-write the above equation to obtain I.

    I=3h4π2c(1λ11λ2)1

Conclusion:

Substitute 6.626075×1034Js for h, 2.2112μm for λ1, 2.4054μm for λ2, 2.997925×108m/s for c.

    I=3(6.626075×1034Js)4π2(2.997925×108m/s)(12.2112μm12.4054μm)1=3(6.626075×1034Js)4π2(2.997925×108m/s)(12.2112×106m12.4054×106m)1=4.60×1048kgm2

The moment of inertia of the hydrogen molecule about an axis through its center of mass and perpendicular to H-H bond is 4.60×1048kgm2.

(b)

To determine

The vibrational frequency of the hydrogen molecule.

(b)

Expert Solution
Check Mark

Answer to Problem 21P

The vibrational frequency of the hydrogen molecule is 1.32×1014Hz.

Explanation of Solution

Refer section (a) and write the formula for the energy difference between v=0, J=0 level to v=1, J=1.

    ΔE1=E11E00=hf+2I

Here, ΔE1 is the energy difference between v=0, J=0 level to v=1, J=1, J=0 level to v=1, J=1, E11 is the energy of level v=1 and J=1, E00 is the energy of level  v=0 and J=0, h is the Planck’s constant, f is the frequency, I is the moment of inertia.

Re-write the above equation to get an expression for f.

    hf=ΔE12If=ΔE1hh22πI                                                                                               (III)

Write the formula for ΔE1.

    ΔE1=hcλ1                                                                                                         (IV)

Here, λ1 is the wavelength of the photon required for the transition from energy difference between v=0, and c is the speed of light in vacuum.

Conclusion:

Substitute 6.626075×1034Js for h, 2.2112μm for λ1, 2.997925×108m/s for c in equation (IV).

    ΔE1=(6.626075×1034Js)(2.997925×108m/s)2.2112μm=(6.626075×1034Js)(2.997925×108m/s)2.2112×106m=8.983573×1020J

Substitute 8.983573×1020J for ΔE1, 6.626075×1034Js for h, 4.60×1048kgm2 for I in equation (III).

    f=8.983 573×1020J6.626 075×1034Js(6.626 075×1034Js)22π(4.600 060×1048kgm2)=1.32×1014Hz

The vibrational frequency of the hydrogen molecule is 1.32×1014Hz.

(c)

To determine

The equilibrium separation distance for the molecule.

(c)

Expert Solution
Check Mark

Answer to Problem 21P

The equilibrium separation distance for the molecule is 0.0741nm.

Explanation of Solution

Write the formula for the moment of inertia of the molecule.

    I=μr2

Here, I is the moment of inertia, μ is the reduced mass, and r is the equilibrium separation distance.

Reduced mass of hydrogen molecule is half of the mass of it.

    I=m2r2

Here, m is the mass of the molecule.

Re-write the above equation to get an expression for r.

    r=2Im

Conclusion:

Substitute 4.60×1048kgm2 for I, 1.007825u for m.

    r=2(4.60×1048kgm2)1.007825u=2(4.60×1048kgm2)1.007825u(1.67×1027kg)=0.0741nm

The equilibrium separation distance for the molecule is 0.0741nm.

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Chapter 43 Solutions

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University

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