Physics for Scientists and Engineers with Modern Physics
Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN: 9781337553292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 42, Problem 11P

(a) In an HCl molecule, take the Cl atom to be the isotope 35Cl. The equilibrium separation of the H and Cl atoms is 0.127 46 nm. The atomic mass of the H atom is 1.007 825 u and that of the 35Cl atom is 34.968 853 u. Calculate the longest wavelength in the rotational spectrum of this molecule. (b) What If? Repeat the calculation in part (a), but take the Cl atom to be the isotope 37Cl, which has atomic mass 36.965 903 u. The equilibrium separation distance is the same as in part (a). (c) Naturally occurring chlorine contains approximately three parts of 35Cl to one part of 37Cl. Because of the two different Cl masses, each line in the microwave rotational spectrum of HCl is split into a doublet as shown in Figure P42.11. Calculate the separation in wavelength between the doublet lines for the longest wavelength.

Chapter 42, Problem 11P, (a) In an HCl molecule, take the Cl atom to be the isotope 35Cl. The equilibrium separation of the H

(a)

Expert Solution
Check Mark
To determine

The longest wavelength in the rotational spectrum of HCl molecule when C35l isotope is used.

Answer to Problem 11P

The longest wavelength in the rotational spectrum of HCl molecule when C35l isotope is used is 472μm.

Explanation of Solution

Write the equation for the smallest energy difference between the rotational energy levels.

    ΔEmin=2I                                                                                                           (I)

Here, ΔEmin is the smallest energy difference between the rotational energy levels, is the reduced Plank’s constant and I is the moment of inertia.

Write the equation for the wavelength.

    λ=hcΔEmin                                                                                                       (II)

Here, h is the Plank’s constant and c is the speed of light.

Substitute equation (I) in the equation (II) and substitute h2π for to find λ.

    λ=hc(2I)=4π2Ich                                                                                                         (III)

Write the equation for the reduced mass.

    μ=mHmClmH+mCl                                                                                                  (IV)

Here, μ is the reduced mass, mH is the atomic mass of hydrogen and mCl is the atomic mass of chlorine.

Write the equation for moment of inertia.

    I=μr2                                                                                                            (V)

Here, I is the moment of inertia and r is the separation distance of the atoms between hydrogen and chlorine.

Conclusion:

Substitute equation (V) in equation (III).

    λ=4π2μr2ch

Substitute 0.12746nm for r, 2.997925×108m/s for c and 6.626075×1034Js for h to find λ in above equation.

λ=4π2μ((0.12746nm)(109m1nm))2(2.997925×108m/s)6.626075×1034Js=(2.901830×1023m/kg)μ (VI)

Substitute 1.007825u for mH and 34.968853u for mCl in equation (IV) to find μ35.

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

Here, μ35 is the reduced mass for C35l.

Substitute 1.626653×1027kg for μ in equation (VI) to find λ35.

    λ35=(2.901830×1023m/kg)(1.626653×1027kg)=(472×106m)(106μm1m)=472μm

Thus, the longest wavelength in the rotational spectrum of HCl molecule when C35l isotope is used is 472μm.

(b)

Expert Solution
Check Mark
To determine

The longest wavelength in the rotational spectrum of HCl molecule when C37l isotope is used.

Answer to Problem 11P

The longest wavelength in the rotational spectrum of HCl molecule when C37l isotope is used is 473μm.

Explanation of Solution

From the equation (IV) for the reduced mass.

    μ=mHmClmH+mCl

From the equation (VI) for wavelength.

    λ=(2.901830×1023m/kg)μ

Conclusion:

Substitute 1.007825u for mH and 36.965903u for mCl in equation (IV) to find μ37.

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

Here, μ35 is the reduced mass for C35l.

Substitute 1.629118×1027kg for μ in equation (VI) to find λ37.

    λ37=(2.901830×1023m/kg)(1.629118×1027kg)=(473×106m)(106μm1m)=473μm

Thus, the longest wavelength in the rotational spectrum of HCl molecule when C37l isotope is used is 473μm.

(c)

Expert Solution
Check Mark
To determine

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

Answer to Problem 11P

The separation in wavelength between the doublet lines for the longest wavelength is 0.715μm.

Explanation of Solution

Write the equation for the separation in wavelength between the doublet lines for the longest wavelength.

    d=λ37λ35

Here, d is the separation in wavelength between the doublet lines for the longest wavelength.

Conclusion:

Substitute 472.7424μm for λ37 and 472.0270μm for λ35 in the above equation find d.

    d=472.7424μm472.0270μm=0.715μm

Thus, the separation in wavelength between the doublet lines for the longest wavelength is 0.715μm.

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Students have asked these similar questions
(a) In an HCl molecule, take the Cl atom to be the isotope 35Cl. The equilibrium separation of the H and Cl atoms is 0.127 46 nm. The atomic mass of the H atom is 1.007 825 u and that of the 35Cl atom is 34.968 853 u. Calculate the longest wavelength in the rotational spectrum of this molecule. (b) What If? Repeat the calculation in part (a), but take the Cl atom to be the isotope 37Cl, which has atomic mass 36.965 903 u. The equilibrium separation distance is the same as in part (a). (c) Naturally occurring chlorine contains approximately three parts of 35Cl to one part of 37Cl. Because of the two different Cl masses, each line in the microwave rotational spectrum of HCl is split into a doublet as shown. Calculate the separation in wavelength between the doublet lines for the longest wavelength.
Calculate the radius of a nickel atom in cm, given that Ni has an FCC crystal structure, a density of 7.982 g/cm³, and an atomic weight of 58.69 g/mol.
Pls help ASAP. Pls show all work annd circle the final answer.
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