Atkins' Physical Chemistry
Atkins' Physical Chemistry
11th Edition
ISBN: 9780198769866
Author: ATKINS, P. W. (peter William), De Paula, Julio, Keeler, JAMES
Publisher: Oxford University Press
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 11, Problem 11C.7P

(a)

Interpretation Introduction

Interpretation:

The fundamental vibrational wavenumber, force constant, moment of inertia and rotational constant has to be stated based on harmonic oscillator and rigid-rotor approximation for He2 complex.

Concept introduction:

Rotational spectroscopy measures the electronic energy between the rotational states.  The rotational constant is inversely proportional to the moment of inertia.  The rotational constant is given by an expression shown below.

  B=h8π2IBc

(a)

Expert Solution
Check Mark

Answer to Problem 11C.7P

The value of the fundamental vibrational wavenumber is 1.01×101m1_.

The value of the force constant is 1.20×1010kgs1_.

The moment of inertia is 2.93×1046kgm2_.

The value of the rotational constant is 95.6m1_.

Explanation of Solution

The fundamental vibrational wavenumber is given by the expression as shown below.

  v˜=D˜o

Where, v˜ is the fundamental vibrational wavenumber and D˜o is the depth of the potential energy minimum.

The given value for hcD˜o is shown below.

  hcD˜o=2×1026J        (1)

Where,

  • h is the Planck’s constant
  • c is the velocity of light.

The value of h is 6.626×1034Js and the value of c is 2.998×108ms1

Substitute the values of corresponding variables in equation (1) to calculate the fundamental vibrational wavenumber as shown below.

  hcD˜o=2×1026JD˜o=2×1026J(6.626×1034Js)×(2.998×108ms1)v˜=1.01×101m1_

Therefore, the value of the fundamental vibrational wavenumber is 1.01×101m1_.

The fundamental vibrational wavenumber in terms of force constant is expressed as shown below.

  v˜=12πc(kfu)1/2kf=(2πcv˜)2u        (2)

Where, kf is the force constant and u is the reduced mass.

The mass of one He atom is denoted by mHe.  The value of mHe is 4.0026mu where mu is the atomic mass unit.

  1mu=1.66054×1027kg

Therefore, the mass of He in kg is given below.

  mHe=4.0026mu=4.0026×1.66054×1027kg=6.646×1027kg

The reduced mass of the He2 is calculated as shown below.

  μ=mHe×mHemHe+mHe=6.646×1027kg×6.646×1027kg6.646×1027kg+6.646×1027kg=3.323×1027kg

Substitute the value of the reduced mass and other corresponding variables in equation (2) as shown below.

  kf=(2πcv˜)2u=(2×3.14×2.998×108ms1×1.01×101m1)2×3.323×1027kg=1.20×1010kgs1_

Therefore, the value of the force constant is 1.20×1010kgs1_.

The moment of inertia is calculated by the formula shown below.

  I=uRe2        (3)

Where

  • I is the moment of inertia.
  • u is the reduced mass.
  • Re is the internuclear distance.

The value of Re is 297pm and the value of u is 3.323×1027kg.  Substitute the values of corresponding variables in equation (3) to calculate the moment of inertia as shown below.

  I=uRe2=(3.323×1027kg)×(297pm×1012m1pm)2=(3.323×1027kg)×(2.97×1010m)2=2.93×1046kgm2_

Therefore, the moment of inertia is 2.93×1046kgm2_.

The rotational constant is given by an expression shown below.

  B=h8π2Ic        (4)

Where,

  • I is the moment of inertia.
  • h is the planks constant.
  • B is the rotational constant.
  • c is the velocity of light.

Substitute the values of corresponding variables in equation (4) to calculate the moment of inertia as shown below.

B=h8π2Ic=6.626 × 1034 m2kg s18×(3.14)2×(2.93×1046kgm2)×(2.998×108ms-1)=95.6m1_

Therefore, the value of the rotational constant is 95.6m1_.

(b)

Interpretation Introduction

Interpretation:

The vibrational wavenumber and anharmonicity constant has to be calculated on the basis of Morse potential energy.

Concept introduction:

Rotational spectroscopy measures the electronic energy between the rotational states.  The rotational constant is inversely proportional to the moment of inertia.  The Morse potential energy function is shown below.

  V=hcD˜e{1ea(RRe)}2

(b)

Expert Solution
Check Mark

Answer to Problem 11C.7P

The value the vibrational wavenumber is 76.0m1_.

The anharmonicity constant (xe) is 0.25_.

Explanation of Solution

The Morse potential energy is given by the expression as shown below.

  V=hcD˜e{1ea(RRe)}2

Where, h is the Planck’s constant, c is the velocity of light, a is a constant, D˜e is the depth of the potential minimum and R represents internuclear distance.

The vibrational wavenumber (v˜) is given by the expression as shown below.

  v˜=D˜e

The given value for hcD˜e is shown below.

  hcD˜e=1.51×1023J        (1)

Substitute the values of corresponding variables in equation (1) to calculate the fundamental vibrational wavenumber as shown below.

  hcD˜e=1.51×1023JD˜e=1.51×1023J(6.626×1034Js)×(2.998×108ms1)v˜=76.0m1_

Therefore, the value the vibrational wavenumber is 76.0m1_.

The anharmonicity constant (xe) is given by the expression as shown below.

  xe=v˜4D˜e        (2)

Substitute the values of corresponding variables in equation (2) to calculate the anharmonicity constant (xe) as shown below.

  xe=v˜4D˜e=v˜4v˜=14=0.25_

Therefore, the anharmonicity constant (xe) is 0.25_.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 11 Solutions

Atkins' Physical Chemistry

Ch. 11 - Prob. 11E.2STCh. 11 - Prob. 11F.2STCh. 11 - Prob. 11A.1DQCh. 11 - Prob. 11A.2DQCh. 11 - Prob. 11A.3DQCh. 11 - Prob. 11A.1AECh. 11 - Prob. 11A.1BECh. 11 - Prob. 11A.2AECh. 11 - Prob. 11A.2BECh. 11 - Prob. 11A.3AECh. 11 - Prob. 11A.3BECh. 11 - Prob. 11A.4AECh. 11 - Prob. 11A.4BECh. 11 - Prob. 11A.5AECh. 11 - Prob. 11A.5BECh. 11 - Prob. 11A.6AECh. 11 - Prob. 11A.6BECh. 11 - Prob. 11A.7AECh. 11 - Prob. 11A.7BECh. 11 - Prob. 11A.8AECh. 11 - Prob. 11A.8BECh. 11 - Prob. 11A.9AECh. 11 - Prob. 11A.9BECh. 11 - Prob. 11A.10AECh. 11 - Prob. 11A.10BECh. 11 - Prob. 11A.11AECh. 11 - Prob. 11A.11BECh. 11 - Prob. 11A.12AECh. 11 - Prob. 11A.12BECh. 11 - Prob. 11A.1PCh. 11 - Prob. 11A.2PCh. 11 - Prob. 11A.3PCh. 11 - Prob. 11A.4PCh. 11 - Prob. 11A.5PCh. 11 - Prob. 11A.7PCh. 11 - Prob. 11A.8PCh. 11 - Prob. 11A.9PCh. 11 - Prob. 11A.11PCh. 11 - Prob. 11B.1DQCh. 11 - Prob. 11B.2DQCh. 11 - Prob. 11B.3DQCh. 11 - Prob. 11B.4DQCh. 11 - Prob. 11B.5DQCh. 11 - Prob. 11B.6DQCh. 11 - Prob. 11B.7DQCh. 11 - Prob. 11B.8DQCh. 11 - Prob. 11B.1AECh. 11 - Prob. 11B.1BECh. 11 - Prob. 11B.3AECh. 11 - Prob. 11B.3BECh. 11 - Prob. 11B.4BECh. 11 - Prob. 11B.5AECh. 11 - Prob. 11B.5BECh. 11 - Prob. 11B.6AECh. 11 - Prob. 11B.6BECh. 11 - Prob. 11B.7AECh. 11 - Prob. 11B.7BECh. 11 - Prob. 11B.8AECh. 11 - Prob. 11B.8BECh. 11 - Prob. 11B.9AECh. 11 - Prob. 11B.9BECh. 11 - Prob. 11B.10AECh. 11 - Prob. 11B.10BECh. 11 - Prob. 11B.11AECh. 11 - Prob. 11B.11BECh. 11 - Prob. 11B.12AECh. 11 - Prob. 11B.12BECh. 11 - Prob. 11B.13AECh. 11 - Prob. 11B.13BECh. 11 - Prob. 11B.14AECh. 11 - Prob. 11B.14BECh. 11 - Prob. 11B.1PCh. 11 - Prob. 11B.2PCh. 11 - Prob. 11B.3PCh. 11 - Prob. 11B.4PCh. 11 - Prob. 11B.5PCh. 11 - Prob. 11B.6PCh. 11 - Prob. 11B.7PCh. 11 - Prob. 11B.8PCh. 11 - Prob. 11B.9PCh. 11 - Prob. 11B.10PCh. 11 - Prob. 11B.11PCh. 11 - Prob. 11B.12PCh. 11 - Prob. 11B.13PCh. 11 - Prob. 11B.14PCh. 11 - Prob. 11C.1DQCh. 11 - Prob. 11C.2DQCh. 11 - Prob. 11C.3DQCh. 11 - Prob. 11C.4DQCh. 11 - Prob. 11C.1AECh. 11 - Prob. 11C.1BECh. 11 - Prob. 11C.2AECh. 11 - Prob. 11C.2BECh. 11 - Prob. 11C.3AECh. 11 - Prob. 11C.3BECh. 11 - Prob. 11C.4AECh. 11 - Prob. 11C.4BECh. 11 - Prob. 11C.5AECh. 11 - Prob. 11C.5BECh. 11 - Prob. 11C.6AECh. 11 - Prob. 11C.6BECh. 11 - Prob. 11C.7AECh. 11 - Prob. 11C.7BECh. 11 - Prob. 11C.8AECh. 11 - Prob. 11C.8BECh. 11 - Prob. 11C.2PCh. 11 - Prob. 11C.3PCh. 11 - Prob. 11C.4PCh. 11 - Prob. 11C.5PCh. 11 - Prob. 11C.6PCh. 11 - Prob. 11C.7PCh. 11 - Prob. 11C.8PCh. 11 - Prob. 11C.9PCh. 11 - Prob. 11C.10PCh. 11 - Prob. 11C.11PCh. 11 - Prob. 11C.12PCh. 11 - Prob. 11C.13PCh. 11 - Prob. 11C.15PCh. 11 - Prob. 11C.17PCh. 11 - Prob. 11C.18PCh. 11 - Prob. 11C.19PCh. 11 - Prob. 11D.1DQCh. 11 - Prob. 11D.2DQCh. 11 - Prob. 11D.3DQCh. 11 - Prob. 11D.1AECh. 11 - Prob. 11D.1BECh. 11 - Prob. 11D.2AECh. 11 - Prob. 11D.2BECh. 11 - Prob. 11D.3AECh. 11 - Prob. 11D.3BECh. 11 - Prob. 11D.4AECh. 11 - Prob. 11D.4BECh. 11 - Prob. 11D.5AECh. 11 - Prob. 11D.5BECh. 11 - Prob. 11D.6AECh. 11 - Prob. 11D.6BECh. 11 - Prob. 11D.7AECh. 11 - Prob. 11D.7BECh. 11 - Prob. 11D.2PCh. 11 - Prob. 11E.1DQCh. 11 - Prob. 11E.1AECh. 11 - Prob. 11E.1BECh. 11 - Prob. 11E.2AECh. 11 - Prob. 11E.2BECh. 11 - Prob. 11E.3AECh. 11 - Prob. 11E.3BECh. 11 - Prob. 11E.1PCh. 11 - Prob. 11E.2PCh. 11 - Prob. 11F.1DQCh. 11 - Prob. 11F.2DQCh. 11 - Prob. 11F.3DQCh. 11 - Prob. 11F.4DQCh. 11 - Prob. 11F.5DQCh. 11 - Prob. 11F.6DQCh. 11 - Prob. 11F.1AECh. 11 - Prob. 11F.1BECh. 11 - Prob. 11F.2AECh. 11 - Prob. 11F.2BECh. 11 - Prob. 11F.3AECh. 11 - Prob. 11F.3BECh. 11 - Prob. 11F.4AECh. 11 - Prob. 11F.4BECh. 11 - Prob. 11F.5AECh. 11 - Prob. 11F.5BECh. 11 - Prob. 11F.6AECh. 11 - Prob. 11F.6BECh. 11 - Prob. 11F.7AECh. 11 - Prob. 11F.7BECh. 11 - Prob. 11F.8AECh. 11 - Prob. 11F.8BECh. 11 - Prob. 11F.9AECh. 11 - Prob. 11F.9BECh. 11 - Prob. 11F.10AECh. 11 - Prob. 11F.10BECh. 11 - Prob. 11F.11AECh. 11 - Prob. 11F.11BECh. 11 - Prob. 11F.12AECh. 11 - Prob. 11F.12BECh. 11 - Prob. 11F.13AECh. 11 - Prob. 11F.13BECh. 11 - Prob. 11F.1PCh. 11 - Prob. 11F.2PCh. 11 - Prob. 11F.3PCh. 11 - Prob. 11F.4PCh. 11 - Prob. 11F.5PCh. 11 - Prob. 11F.6PCh. 11 - Prob. 11F.7PCh. 11 - Prob. 11F.8PCh. 11 - Prob. 11F.9PCh. 11 - Prob. 11F.10PCh. 11 - Prob. 11F.11PCh. 11 - Prob. 11F.12PCh. 11 - Prob. 11G.1DQCh. 11 - Prob. 11G.2DQCh. 11 - Prob. 11G.3DQCh. 11 - Prob. 11G.4DQCh. 11 - Prob. 11G.5DQCh. 11 - Prob. 11G.1AECh. 11 - Prob. 11G.1BECh. 11 - Prob. 11G.2AECh. 11 - Prob. 11G.2BECh. 11 - Prob. 11G.1PCh. 11 - Prob. 11G.2PCh. 11 - Prob. 11G.3PCh. 11 - Prob. 11G.4PCh. 11 - Prob. 11G.5PCh. 11 - Prob. 11G.6PCh. 11 - Prob. 11.1IACh. 11 - Prob. 11.5IACh. 11 - Prob. 11.8IA
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
The Bohr Model of the atom and Atomic Emission Spectra: Atomic Structure tutorial | Crash Chemistry; Author: Crash Chemistry Academy;https://www.youtube.com/watch?v=apuWi_Fbtys;License: Standard YouTube License, CC-BY