ATKINS' PHYSICAL CHEMISTRY-ACCESS
ATKINS' PHYSICAL CHEMISTRY-ACCESS
11th Edition
ISBN: 9780198834700
Author: ATKINS
Publisher: OXF
bartleby

Concept explainers

Question
Book Icon
Chapter 14, Problem 14.2IA

(a)

Interpretation Introduction

Interpretation:

The Lennard-Jones parameters r0 and ε has to be calculated for He2.  The plot of Lennard-Jones potential for HeHe has to be drawn.

Concept introduction:

Two neutral atoms or molecule first experience an attractive force when they come in contact with each other.  After a certain distance these particles experience repulsion force.  This type of interaction can be represented by Lennard-Jones potential equation.  The Lennard-Jones potential energy (V) is given by the expression as shown below.

    V=4ε{(r0r)12(r0r)6}

(a)

Expert Solution
Check Mark

Answer to Problem 14.2IA

The Lennard-Jones parameters r0 and ε for He2 are 264.5969 pm_ and 1.51×1023 J_ respectively.

The plot of Lennard-Jones potential for HeHe is shown below.

ATKINS' PHYSICAL CHEMISTRY-ACCESS, Chapter 14, Problem 14.2IA , additional homework tip  1

Explanation of Solution

The Lennard-Jones potential energy (V) is given by the expression as shown below.

    V=4ε{(r0r)12(r0r)6}        (1)

Where,

  • ε is the dept of the well.
  • r is the distance between the distance between the atoms.
  • r0 is the distance between the atoms at V=0.

The value of hcD˜e is 1.51×1023 J.

The value of hcD˜0 is 2×1026 J.

The value of distance between the atoms at well minima (Re) is 297 pm.

The dept of the well is give by the expression as shown below.

    ε=hcD˜e

Where,

  • h is the Plank’s constant.
  • c is the speed of light.
  • D˜e is the wavelength corresponding to the dissection of molecule.

Substitute the value of hcD˜e in the above equation.

    ε=1.51×1023 J_

The distance between the atoms at which potential energy is zero is given by the expression as shown below.

    r0=21/6Re

Substitute the value of Re in the above equation.

    r0=(297 pm)21/6=264.5969 pm_

Therefore, the Lennard-Jones parameters r0 and ε for He2 are 264.5969 pm_ and 1.51×1023 J_ respectively.

The Lennard-Jones potential equation for He2 is written as shown below.

    V=4(1.51×1023 J){(264.5969 pmr)12(264.5969 pmr)6}        (2)

Substitute the value of r=0 pm in the above equation.

    V=4(1.51×1023 J){(264.5969 pm0 pm)12(264.5969 pm0 pm)6}=

Similarly, the Lennard-Jones potential energy is from r=0 pm to r=600 pm by the use of equation (2). The calculated values of Lennard-Jones potential energy is shown below in the table.

r(pm)Lennard-Jones potential energy (J)
0 pm J
100 pm7.09×1018 J
200 pm1.41×1021 J
300 pm1.5×1023 J
400 pm4.6×1024 J
500 pm1.3×1024 J
600 pm4.4×1025 J

The plot of Lennard-Jones potential for HeHe is shown below.

ATKINS' PHYSICAL CHEMISTRY-ACCESS, Chapter 14, Problem 14.2IA , additional homework tip  2

Figure 1

(b)

Interpretation Introduction

Interpretation:

The plot of Morse potential for for HeHe is to be drawn.

Concept introduction:

The curve that represents the interatomic interaction model corresponding to the diatomic molecule’s potential energy is known as Morse potential curve.  The curve gives the approximation for the vibrational structure of a diatomic molecule.

(b)

Expert Solution
Check Mark

Answer to Problem 14.2IA

The plot of Morse potential for HeHe is shown below.

ATKINS' PHYSICAL CHEMISTRY-ACCESS, Chapter 14, Problem 14.2IA , additional homework tip  3

Explanation of Solution

The expression to calculate the Morse potential energy is mentioned as follows.

    V(R)=hcDe˜{1ea(RRe)}2        (3)

Where,

  • De˜ is the molecular partition function.
  • h is the Planck’s constant.
  • c is the speed of light.
  • Re is the equilibrium bond length.
  • R is the bond length.

The value of hcD˜e is 1.51×1023 J.

The value of hcD˜0 is 2×1026 J.

The value of distance between the atoms at well minima (Re) is 297 pm.

The value of a is 5.79×1010 m1.

The conversion of a in cm1 is shown below.

    a=(5.79×1010 m1)(1 m1012 pm)=5.79×102 pm1

Substitute the values of hcD˜e, Re, and a in the above equation.

    V(R)=(1.51×1023 J){1e(5.79×102 pm1)(R297 pm)}2        (4)

Substitute the value of r=0 pm in the above equation.

    V(R)=(1.51×1023 J){1e(5.79×102 pm1)(0 pm297 pm)}2=(1.51×1023 J){129393966.79}2=(1.51×1023 J){29393965.79}2=1.3046×108 J

Similarly, the Morse potential energy is from r=100 pm to r=600 pm by the use of equation (4). The calculated values of Morse potential energy is shown below in the table.

r(pm)Morse potential energy (J)
100 pm1.21998×1013 J
200 pm1.13258×1018 J
300 pm3.83912×1025 J
400 pm1.50225×1023 J
500 pm1.50998×1023 J
600 pm1.51×1023 J

The plot of Morse potential for HeHe is shown below.

ATKINS' PHYSICAL CHEMISTRY-ACCESS, Chapter 14, Problem 14.2IA , additional homework tip  4

Figure 2

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 14 Solutions

ATKINS' PHYSICAL CHEMISTRY-ACCESS

Ch. 14 - Prob. 14A.3DQCh. 14 - Prob. 14A.1AECh. 14 - Prob. 14A.1BECh. 14 - Prob. 14A.2AECh. 14 - Prob. 14A.2BECh. 14 - Prob. 14A.3AECh. 14 - Prob. 14A.3BECh. 14 - Prob. 14A.4AECh. 14 - Prob. 14A.4BECh. 14 - Prob. 14A.5AECh. 14 - Prob. 14A.5BECh. 14 - Prob. 14A.6AECh. 14 - Prob. 14A.6BECh. 14 - Prob. 14A.7AECh. 14 - Prob. 14A.7BECh. 14 - Prob. 14A.8AECh. 14 - Prob. 14A.8BECh. 14 - Prob. 14A.9AECh. 14 - Prob. 14A.9BECh. 14 - Prob. 14A.1PCh. 14 - Prob. 14A.2PCh. 14 - Prob. 14A.3PCh. 14 - Prob. 14A.4PCh. 14 - Prob. 14A.5PCh. 14 - Prob. 14A.6PCh. 14 - Prob. 14A.7PCh. 14 - Prob. 14A.8PCh. 14 - Prob. 14A.10PCh. 14 - Prob. 14A.12PCh. 14 - Prob. 14A.13PCh. 14 - Prob. 14B.1DQCh. 14 - Prob. 14B.2DQCh. 14 - Prob. 14B.3DQCh. 14 - Prob. 14B.4DQCh. 14 - Prob. 14B.5DQCh. 14 - Prob. 14B.1AECh. 14 - Prob. 14B.1BECh. 14 - Prob. 14B.2AECh. 14 - Prob. 14B.2BECh. 14 - Prob. 14B.3AECh. 14 - Prob. 14B.3BECh. 14 - Prob. 14B.4AECh. 14 - Prob. 14B.4BECh. 14 - Prob. 14B.5AECh. 14 - Prob. 14B.5BECh. 14 - Prob. 14B.6AECh. 14 - Prob. 14B.6BECh. 14 - Prob. 14B.1PCh. 14 - Prob. 14B.2PCh. 14 - Prob. 14B.3PCh. 14 - Prob. 14B.4PCh. 14 - Prob. 14B.5PCh. 14 - Prob. 14B.6PCh. 14 - Prob. 14B.7PCh. 14 - Prob. 14B.8PCh. 14 - Prob. 14B.10PCh. 14 - Prob. 14C.1DQCh. 14 - Prob. 14C.2DQCh. 14 - Prob. 14C.1AECh. 14 - Prob. 14C.1BECh. 14 - Prob. 14C.2AECh. 14 - Prob. 14C.2BECh. 14 - Prob. 14C.3AECh. 14 - Prob. 14C.3BECh. 14 - Prob. 14C.4AECh. 14 - Prob. 14C.4BECh. 14 - Prob. 14C.1PCh. 14 - Prob. 14C.2PCh. 14 - Prob. 14D.1DQCh. 14 - Prob. 14D.2DQCh. 14 - Prob. 14D.3DQCh. 14 - Prob. 14D.4DQCh. 14 - Prob. 14D.5DQCh. 14 - Prob. 14D.1AECh. 14 - Prob. 14D.1BECh. 14 - Prob. 14D.2AECh. 14 - Prob. 14D.2BECh. 14 - Prob. 14D.3AECh. 14 - Prob. 14D.3BECh. 14 - Prob. 14D.4AECh. 14 - Prob. 14D.4BECh. 14 - Prob. 14D.5AECh. 14 - Prob. 14D.5BECh. 14 - Prob. 14D.6AECh. 14 - Prob. 14D.6BECh. 14 - Prob. 14D.8AECh. 14 - Prob. 14D.8BECh. 14 - Prob. 14D.9AECh. 14 - Prob. 14D.9BECh. 14 - Prob. 14D.2PCh. 14 - Prob. 14D.3PCh. 14 - Prob. 14D.4PCh. 14 - Prob. 14D.6PCh. 14 - Prob. 14D.7PCh. 14 - Prob. 14D.8PCh. 14 - Prob. 14D.9PCh. 14 - Prob. 14D.10PCh. 14 - Prob. 14E.1DQCh. 14 - Prob. 14E.2DQCh. 14 - Prob. 14E.3DQCh. 14 - Prob. 14E.4DQCh. 14 - Prob. 14E.5DQCh. 14 - Prob. 14E.1AECh. 14 - Prob. 14E.1BECh. 14 - Prob. 14E.1PCh. 14 - Prob. 14E.3PCh. 14 - Prob. 14.1IACh. 14 - Prob. 14.2IACh. 14 - Prob. 14.6IACh. 14 - Prob. 14.8IA
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