Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)
Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)
9th Edition
ISBN: 9781305266292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 43, Problem 19P

(a)

To determine

The longest wavelength in the rotational spectrum of the HCl molecule.

(a)

Expert Solution
Check Mark

Answer to Problem 19P

The longest wavelength in the rotational spectrum of the HCl molecule is 472μm.

Explanation of Solution

The longest wavelength for a rotational spectrum of a molecule will be obtained if the transition occurs between J=0 and J=1 state.

Write expression for the energy of rotational levels.

  EJ=22IJ(J+1)                                                                                                      (I)

Here, EJ is the energy of rotational levels, is the reduced Planck’s constant, J is the quantum number of rotational energy levels and I is the moment of inertia of molecule.

Write the expression for the energy difference between J=0 and J=1 states.

  ΔEmin=E1E0                                                                                                         (II)

Here, ΔEmin is the energy of the minimum energy transition, E1 is the energy of J=1 level and E0 is the energy of J=0 state.

Write the expression for the change in energy levels in terms of wavelength of photon emitted.

  ΔE=hcλ                                                                                                                  (III)

Here, ΔE is the change in energy levels, c is the speed of light in vacuum and λ is the wavelength of photon emitted.

Rearrange above two equation to get λ.

  λ=hcΔE                                                                                                                  (IV)

Write the expression for the reduced mass of HCl molecule.

  μ=mHmClmH+mCl                                                                                                           (V)

Here, μ is the reduced mass HCl molecule, mH is the mass of Hydrogen atom and mCl is the mass of Chlorine atom.

Write the expression for the moment of inertia of HCl molecule.

  I=μr2                                                                                                                    (VI)

Here, r is the equilibrium separation of the H and Cl atom.

Conclusion:

The mass of H atom is 1.007825u, mass of 35Cl atom is 34.968853u and equilibrium separation of atoms is 0.12746nm.

Substitute 0 for J in equation (I) to get E0.

  E0=22I0(0+1)=0

Substitute 1 for J in equation (I) to get E1.

  E1=22I1(1+1)=2I

Substitute 0 for E0 and 2I for E1 in (I) to get ΔEmin.

  ΔEmin=2I0=2I

Substitute 2I for ΔE in (IV) to get λ.

  λ=hc(h2π)2I=4π2Ich                                                                                                           (VII)

Substitute 1.007825u for mH and 34.968853u for mCl equation (V) to get reduced mass of HCL with Cl atom to be isotope of 35Cl.

  μ35=(1.007825u)(34.968853u)(1.007825u)+(34.968853u)×(1.660540×1027kgu)=1.626653×1027kg

Here, μ35 is the reduced mass of HCL with Cl atom to be isotope of 35Cl.

Substitute 1.626653×1027kg for μ and 0.12746nm for r in equation (VI) to get I.

  I=(1.626653×1027kg)(0.12746nm1 m109 nm)2=2.6426688×1047kgm2

Substitute 6.626075×1034Js for h, 3×108m/s2 for c and 2.6426688×1047kgm2 for I in equation (VII) to get λ.

  λ=4π2(2.6426688×1047kgm2)(3×108m/s2)6.626075×1034Js=4.718749601×104m×1m106μm=4.72×104m×1m106μm=472μm

Therefore, the longest wavelength in the rotational spectrum of the HCl molecule is 472μm.

(b)

To determine

The longest wavelength in the rotational spectrum of the HCl molecule in which Cl atom is an isotope 37Cl.

(b)

Expert Solution
Check Mark

Answer to Problem 19P

The longest wavelength in the rotational spectrum of the HCl molecule in which Cl atom is an isotope 37Cl is 473μm.

Explanation of Solution

Use equation (V) to get reduced mass of HCl, equation (VI) to get moment of inertia of HCl molecule and (VII) to get λ.

Conclusion:

Substitute 1.007825u for mH and 36.965903u for mCl equation (V) to get reduced mass of HCL with Cl atom to be isotope of 35Cl.

  μ35=(1.007825u)(36.965903u)(1.007825u)+(36.965903u)×(1.660540×1027kgu)=1.629118×1027kg

Here, μ37 is the reduced mass of HCL with Cl atom to be isotope of 37Cl.

Substitute 1.629118×1027kg for μ and 0.12746nm for r in equation (VI) to get I.

  I=(1.629118×1027kg)(0.12746nm×1m109nm)2=2.6466735×1047kgm2

Substitute 6.626×1034Js for h, 3×108m/s2 for c and 2.6466735×1047kgm2 for I in equation (VII) to get λ.

  λ=4π2(2.6466735×1047kgm2)(3×108m/s2)6.626×1034Js=4.726063702×104m×1m106μm=4.73×104m×1m106μm=473μm

Therefore, the longest wavelength in the rotational spectrum of the HCl molecule in which Cl atom is an isotope 37Cl is 473μm.

(c)

To determine

The separation in wavelength between the doublet lines for the longest wavelength.

(c)

Expert Solution
Check Mark

Answer to Problem 19P

The separation in wavelength between the doublet lines for the longest wavelength is 0.731μm.

Explanation of Solution

Write the expression for the separation of wavelength.

  Δλ=λ37λ35

Here, λ35 is the λ is the wavelength of photon emitted in H35Cl and λ37 is the wavelength of photon emitted in H37Cl.

Conclusion:

Substitute 4.726063702×104 m for λ37 and 4.718749601×104 m for λ35 in above equation to get Δλ.

  Δλ=4.726063702×104 m4.718749601×104 m=7.31×107m×106μm1m=0.731μm

Therefore, the separation in wavelength between the doublet lines for the longest wavelength is 0.731μm.

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Chapter 43 Solutions

Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)

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