Physical Chemistry
Physical Chemistry
2nd Edition
ISBN: 9781133958437
Author: Ball, David W. (david Warren), BAER, Tomas
Publisher: Wadsworth Cengage Learning,
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Concept explainers

Question
Chapter 12, Problem 12.56E
Interpretation Introduction

Interpretation:

The value of ϕH2+,1 and ϕH2+,2 as well as E1 and E2 for R=1.00, 1.15, 1.45, and 1.60A is to be determined. These values are to be combined with the determinations from Example 12.14 and a simple potential energy diagram for this system is to be constructed.

Concept introduction:

The wavefunction contains all the information about the state of the system. The wavefunction is the function of the coordinates of particles and time. The wavefunction ϕH2+,1 and ϕH2+,2 are given as follows:

ϕH2+,1=12+2S12(ΨH(1)+ΨH(2))ϕH2+,2=122S12(ΨH(1)ΨH(2))

Expert Solution & Answer
Check Mark

Answer to Problem 12.56E

The values of ϕH2+,1 and ϕH2+,2 as well as E1 and E2 for R=1.00, 1.15, 1.45, and 1.60A is given below.

R(Ao) ϕH2+,1 ϕH2+,2 E1 E2
1.00 0.558(ΨH(1)+ΨH(2)) 1.141(ΨH(1)ΨH(2)) 1.23 3.83
1.15 0.57(ΨH(1)+ΨH(2)) 1.03(ΨH(1)+ΨH(2)) 1.61 4.21
1.45 0.59(ΨH(1)+ΨH(2)) 0.91(ΨH(1)+ΨH(2)) 1.75 3.21
1.60 0.61(ΨH(1)+ΨH(2)) 0.87(ΨH(1)+ΨH(2)) 1.66 2.60

The simple potential energy diagram for this system is shown in Figure 1 and Figure 2.

Explanation of Solution

The value of S12 is calculated as follows:

S12=eR/a0(1+Ra0+R23a02)(1)

Where,

a0 is a Bohr’s radius 0.529A.

R is the distance between two nuclei.

The value of H12 is calculated as follows:

H12=K[a0S12ReR/a0(1+Ra0S122)](2)

Where,

K is the conversion factor from atomic to SI units.

The value of E1 and E2 are given by the formula,

E1=H11+H121+S12E2=H22H121S12(3)

The wavefunction ϕH2+,1 and ϕH2+,2 is given as follows:

ϕH2+,1=12+2S12(ΨH(1)+ΨH(2))ϕH2+,2=122S12(ΨH(1)ΨH(2))(4)

The value of R=1.00A, K=27.09eV and H11=0.258eV is given.

Substitute the value of R and K in the equation (1).

S12=e1.00A/0.529A(1+1.00A0.529A+(1.00A)23(0.529A)2)=0.151(1+1.89+1.194)=0.151×4.084=0.616

Thus, the value of S12 is 0.616.

Substitute the value of R, S12 and K in the equation (2).

H12=27.09eV[0.529A×0.6161.00Ae1.00A/0.529A(1+1.00A0.529A0.6162)]=27.09eV[0.3250.151(1+1.890.308)]=27.09eV[0.3250.151×2.582]=1.73eV

Thus, the value of H12 is 1.73eV.

Substitute the value of H12, S12 and H11 in the equation (3).

E1=0.258+(1.73)1+0.616=1.230eVE2=0.258(1.73)10.616=3.833eV

Thus, the value of E1 and E2 are 1.230eV and 3.833eV.

Substitute the value of S12 in the equation (4).

ϕH2+,1=12+2×0.616(ΨH(1)+ΨH(2))=0.558(ΨH(1)+ΨH(2))ϕH2+,2=122×0.616(ΨH(1)ΨH(2))=1.141(ΨH(1)ΨH(2))

Thus, the wavefunction ϕH2+,1 and ϕH2+,2 is 0.558(ΨH(1)+ΨH(2)) and 1.141(ΨH(1)ΨH(2)).

The value of R=1.15A, K=27.09eV and H11=0.258eV is given.

Substitute the value of R and K in the equation (1).

S12=e1.15A/0.529A(1+1.15A0.529A+(1.15A)23(0.529A)2)=0.113(1+2.17+1.57)=0.535

Thus, the value of S12 is 0.535.

Substitute the value of R, S12 and K in the equation (2).

H12=27.09eV[0.529A×0.5351.15Ae1.15A/0.529A(1+1.15A0.529A0.5352)]=27.09eV[0.2460.113(1+2.1730.267)]=2.22eV

Thus, the value of H12 is 2.22eV.

Substitute the value of H12, S12 and H11 in the equation (3).

E1=0.258+(2.22eV)1+0.535=1.61eVE2=0.258(2.22eV)10.535=4.21eV

Thus, the value of E1 and E2 are 1.61eV and 4.21eV.

Substitute the value of S12 in the equation (4).

ϕH2+,1=12+2×0.535(ΨH(1)+ΨH(2))=0.57(ΨH(1)+ΨH(2))ϕH2+,2=122×0.535(ΨH(1)ΨH(2))=1.03(ΨH(1)ΨH(2))

Thus, the wavefunction ϕH2+,1 and ϕH2+,2 is 0.57(ΨH(1)+ΨH(2)) and 1.03(ΨH(1)ΨH(2)).

The value of R=1.45A, K=27.09eV and H11=0.258eV is given.

Substitute the value of R and K in the equation (1).

S12=e1.45A/0.529A(1+1.45A0.529A+(1.45A)23(0.529A)2)=0.064(1+2.74+2.50)=0.399

Thus, the value of S12 is 0.399.

Substitute the value of R, S12 and K in the equation (2).

H12=27.09eV[0.529A×0.3991.45Ae1.45A/0.529A(1+1.45A0.529A0.3992)]=27.09eV[0.1450.064(1+2.7410.1995)]=2.19eV

Thus, the value of H12 is 2.19eV.

Substitute the value of H12, S12 and H11 in the equation (3).

E1=0.258+(2.19)1+0.399=1.75eVE2=0.258(2.19)10.399=3.21eV

Thus, the value of E1 and E2 are 1.75eV and 3.21eV.

Substitute the value of S12 in the equation (4).

ϕH2+,1=12+2×0.399(ΨH(1)+ΨH(2))=0.59(ΨH(1)+ΨH(2))ϕH2+,2=122×0.399(ΨH(1)ΨH(2))=0.91(ΨH(1)ΨH(2))

Thus, the wavefunction ϕH2+,1 and ϕH2+,2 is 0.59(ΨH(1)+ΨH(2)) and 0.91(ΨH(1)ΨH(2)).

The value of R=1.60A, K=27.09eV and H11=0.258eV is given.

Substitute the value of R and K in the equation (1).

S12=e1.60A/0.529A(1+1.60A0.529A+(1.60A)23(0.529A)2)=0.048(1+3.02+3.05)=0.339

Thus, the value of S12 is 0.339.

Substitute the value of R, S12 and K in the equation (2).

H12=27.09eV[0.529A×0.3391.60Ae1.60A/0.529A(1+1.60A0.529A0.3392)]=27.09eV[0.1120.048(1+3.020.16)]=1.977eV

Thus, the value of H12 is 1.977eV.

Substitute the value of H12, S12 and H11 in the equation (3).

E1=0.258+(1.977eV)1+0.339=1.66eVE2=0.258(1.977eV)10.339=2.60eV

Thus, the value of E1 and E2 are 1.66eV and 2.60eV.

Substitute the value of S12 in the equation (4).

ϕH2+,1=12+2×0.339(ΨH(1)+ΨH(2))=0.61(ΨH(1)+ΨH(2))ϕH2+,2=122×0.339(ΨH(1)ΨH(2))=0.87(ΨH(1)ΨH(2))

Thus, the wavefunction ϕH2+,1 and ϕH2+,2 is 0.61(ΨH(1)+ΨH(2)) and 0.87(ΨH(1)ΨH(2)).

The values of potential energy and R is given below.

R(Ao) Energy (eV)
1.00 3.83
1.15 4.21
1.32 16.23
1.45 3.21
1.60 2.60

The plot between the potential energy with R is shown below.

Physical Chemistry, Chapter 12, Problem 12.56E , additional homework tip  1

Figure 1

The values of potential energy and R is given below.

R(Ao) Energy (eV)
1.00 1.23
1.15 1.61
1.32 6.37
1.45 1.75
1.60 1.66

The plot between the potential energy with R is shown below.

Physical Chemistry, Chapter 12, Problem 12.56E , additional homework tip  2

Figure 2

Conclusion

The values of ϕH2+,1 and ϕH2+,2 as well as E1 and E2 for R=1.00, 1.15, 1.45, and 1.60A is given below.

R(Ao) ϕH2+,1 ϕH2+,2 E1 E2
1.00 0.558(ΨH(1)+ΨH(2)) 1.141(ΨH(1)ΨH(2)) 1.23 3.83
1.15 0.57(ΨH(1)+ΨH(2)) 1.03(ΨH(1)+ΨH(2)) 1.61 4.21
1.45 0.59(ΨH(1)+ΨH(2)) 0.91(ΨH(1)+ΨH(2)) 1.75 3.21
1.60 0.61(ΨH(1)+ΨH(2)) 0.87(ΨH(1)+ΨH(2)) 1.66 2.60

The simple potential energy diagram for this system is shown in Figure 1 and Figure 2.

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