Physical Chemistry
Physical Chemistry
2nd Edition
ISBN: 9781133958437
Author: Ball, David W. (david Warren), BAER, Tomas
Publisher: Wadsworth Cengage Learning,
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Chapter 11, Problem 11.54E
Interpretation Introduction

(a)

Interpretation:

The eigenvalues of total angular momentum is to be evaluated using the complete forms of given wavefunction Ψ3,2 and operators.

Concept introduction:

The eigenvalues of the wavefunction that are obtained when an operator is applied are the only possible values of observables. The expression for the eigenvalue is given by,

A^Ψ=aΨ

The total angular momentum does not depend on the mass of the particle, radius of the rotation and also the magnetic quantum number.

Expert Solution
Check Mark

Answer to Problem 11.54E

The eigenvalues of total angular momentum is 122.

Explanation of Solution

Explanation:

The general equation for the wavefunction in the 3-dimensional rotation is,

Ψl,ml=12πeimϕθl,ml

The complete form of Ψ3,2 using Table 11.3 is,

Ψ3,2=10542πsin2θcosθe2iϕ

The total angular momentum using the complete forms of operators is,

L^2Ψ=2(2θ2+cotθθ+1sin2θ2ϕ2)Ψ

The first derivative of the given wavefunction with respect to θ is,

θ(10542πsin2θcosθe2iϕ)

=10542π(2sinθcos2θsin3θ)e2iϕ…(1)

The second derivative of the given wavefunction with respect to θ is,

2θ2=θ(10542π(2sinθcos2θsin3θ)e2iϕ)

   =10542π(2cos3θ4sin2θcosθ3sin2θcosθ)e2iϕ…(2)

The second derivative of the given wavefunction with respect to ϕ is,

2ϕ2(10542πsin2θcosθe2iϕ)=(2)2i2×10542πsin2θcosθe2iϕ=4×10542πsin2θcosθe2iϕ

=1052πsin2θcosθe2iϕ…(3)

Substitute equation (1), (2) and (3) in the equation of total angular momentum as shown below.

L^2Ψ3,2=2(10542π(2cos3θ4sin2θcosθ3sin2θcosθ)e2iϕ+cotθ10542π(2sinθcos2θsin3θ)e2iϕ1sin2θ1052πsin2θcosθe2iϕ)

Take common terms together and rearrange the given equation as shown below.

L^2Ψ3,2=210542πe2iϕ((2cos3θ7sin2θcosθ)+cotθ(2sinθcos2θsin3θ)e2iϕ4cosθe2iϕ)

Substitute the value of cotθ=cosθsinθ in the given equation.

L^2Ψ3,2=210542πe2iϕ((2cos3θ7sin2θcosθ)+cosθsinθ(2sinθcos2θsin3θ)e2iϕ4cosθe2iϕ)L^2Ψ3,2=210542πe2iϕ((2cos3θ7sin2θcosθ)+(2cos3θcosθsin2θ)4cosθ)L^2Ψ3,2=210542πe2iϕ(4cos3θ8sin2θcosθ4cosθ)

Substitute cos2θ=1sin2θ in the given equation.

L^2Ψ3,2=210542πe2iϕ(4cosθ(1sin2θ)8sin2θcosθ4cosθ)L^2Ψ3,2=210542πe2iϕ(4cosθ4sin2θcosθ8sin2θcosθ4cosθ)L^2Ψ3,2=210542πe2iϕ(12sin2θcosθ)L^2Ψ3,2=12210542πe2iϕ(sin2θcosθ)

Thus, the total angular momentum is represented as,

L^2Ψ3,2=122Ψ3,2

The eigenvalues of total angular momentum is 122.

Conclusion

The eigenvalues of total angular momentum is 122.

Interpretation Introduction

(b)

Interpretation:

The eigenvalues of z-component of angular momentum is to be evaluated using the complete forms of given wavefunction Ψ3,2 and operators.

Concept introduction:

The eigenvalues of the wavefunction that are obtained when an operator is applied are the only possible values of observables. The expression for the eigenvalue is given by,

A^Ψ=aΨ

The z-component of the three dimensional angular momentum that has components in x, y and z direction is quantized.

Expert Solution
Check Mark

Answer to Problem 11.54E

Explanation of Solution

The general equation for the wavefunction in the 3-dimensional rotation is,

Ψl,ml=12πeimϕθl,ml

The complete form of Ψ3,2 using Table 11.3 is,

Ψ3,2=10542πsin2θcosθe2iϕ

The z-component of angular momentum using the complete forms of operators is,

L^zΨ3,2=iϕΨ3,2

The first derivative of the given wavefunction with respect to ϕ is,

ϕ(10542πsin2θcosθe2iϕ)

=(2)i×10542πsin2θcosθe2iϕ…(4)

Substitute equation (4) in the equation of z-component of angular momentum as shown below.

L^zΨ3,2=i((2)i×10542πsin2θcosθe2iϕ)L^zΨ3,2=2(10542πsin2θcosθe2iϕ)L^zΨ3,2=2Ψ3,2

The eigenvalues of z-component of angular momentum is 2.

Conclusion

The eigenvalues of z-component of angular momentum is 2.

Interpretation Introduction

(c)

Interpretation:

The eigenvalue of energy is to be evaluated using the complete forms of given wavefunction Ψ3,2 and operators.

Concept introduction:

The eigenvalues of the wavefunction that are obtained when an operator is applied are the only possible values of observables. The expression for the eigenvalue is given by,

A^Ψ=aΨ

The energy of the particle depends on the moment of inertia, quantum number and Planck’s constant. The total energy is quantized.

Expert Solution
Check Mark

Answer to Problem 11.54E

The eigenvalue of energy for the given wavefunctionis 62I.

Explanation of Solution

The general equation for the wavefunction in the 3-dimensional rotation is,

Ψl,ml=12πeimϕθl,ml

The complete form of Ψ3,2 using Table 11.3 is,

Ψ3,2=10542πsin2θcosθe2iϕ

The eigen equation for the Hamiltonian operator is,

H^Ψ3,2=EΨ3,2

The Hamiltonian operator for energy applied on the given wavefunction is also represented in the form of total angular momentum.

H^Ψ3,2=L^2Ψ3,22I

The value of total angular momentum is 122.

H^Ψ3,2=122Ψ3,22IH^Ψ3,2=62Ψ3,2I

The eigenvalue of energy E for the given wavefunction is 62I.

Conclusion

The eigenvalue of energy E for the given wavefunction is 62I.

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