(a) Interpretation: The eigenvalues of total angular momentum is to be evaluated using the complete forms of given wavefunction Ψ 3 , − 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 = a Ψ The total angular momentum does not depend on the mass of the particle, radius of the rotation and also the magnetic quantum number.

Question

Want to see more full solutions like this?

Answer 1

Question

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

Answer to Problem 11.54E

The eigenvalues of total angular momentum is $12ℏ2$ .

Explanation of Solution

Explanation:

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

$Ψl,ml=12π eimϕ⋅θl,ml$

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

$Ψ3,−2=10542πsin2θcosθe−2iϕ$

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

$L^2Ψ=−ℏ2(∂2∂θ2+cotθ∂∂θ+1sin2θ∂2∂ϕ2)Ψ$

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

$∂∂θ(10542πsin2θcosθe−2iϕ)$

$=10542π(2sinθcos2θ−sin3θ)e−2iϕ$ …(1)

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

$∂2∂θ2=∂∂θ(10542π(2sinθcos2θ−sin3θ)e−2iϕ)$

$=10542π(2cos3θ−4sin2θcosθ−3sin2θcosθ)e−2iϕ$ …(2)

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

$∂2∂ϕ2(10542πsin2θcosθe−2iϕ)=(−2)2i2×10542πsin2θcosθe−2iϕ=−4×10542πsin2θcosθe−2iϕ$

$=−1052πsin2θcosθe−2iϕ$ …(3)

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

$L^2Ψ3,−2=−ℏ2(10542π(2cos3θ−4sin2θcosθ−3sin2θcosθ)e−2iϕ+cotθ10542π(2sinθcos2θ−sin3θ)e−2iϕ−1sin2θ1052πsin2θcosθe−2iϕ)$

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

$L^2Ψ3,−2=−ℏ210542πe−2iϕ((2cos3θ−7sin2θcosθ)+cotθ(2sinθcos2θ−sin3θ)e−2iϕ−4cosθe−2iϕ)$

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

$L^2Ψ3,−2=−ℏ210542πe−2iϕ((2cos3θ−7sin2θcosθ)+cosθsinθ(2sinθcos2θ−sin3θ)e−2iϕ−4cosθe−2iϕ)L^2Ψ3,−2=−ℏ210542πe−2iϕ((2cos3θ−7sin2θcosθ)+(2cos3θ−cosθsin2θ)−4cosθ)L^2Ψ3,−2=−ℏ210542πe−2iϕ(4cos3θ−8sin2θcosθ−4cosθ)$

Substitute $cos2θ=1−sin2θ$ in the given equation.

$L^2Ψ3,−2=−ℏ210542πe−2iϕ(4cosθ(1−sin2θ)−8sin2θcosθ−4cosθ)L^2Ψ3,−2=−ℏ210542πe−2iϕ(4cosθ−4sin2θcosθ−8sin2θcosθ−4cosθ)L^2Ψ3,−2=−ℏ210542πe−2iϕ(−12sin2θcosθ)L^2Ψ3,−2=12ℏ210542πe−2iϕ(sin2θcosθ)$

Thus, the total angular momentum is represented as,

$L^2Ψ3,−2=12ℏ2Ψ3,−2$

The eigenvalues of total angular momentum is $12ℏ2$ .

Conclusion

The eigenvalues of total angular momentum is $12ℏ2$ .

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

Answer to Problem 11.54E

Explanation of Solution

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

$Ψl,ml=12π eimϕ⋅θl,ml$

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

$Ψ3,−2=10542πsin2θcosθe−2iϕ$

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

$L^zΨ3,−2=−iℏ∂∂ϕΨ3,−2$

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

$∂∂ϕ(10542πsin2θcosθe−2iϕ)$

$=(−2)i×10542πsin2θcosθe−2iϕ$ …(4)

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

$L^zΨ3,−2=−iℏ((−2)i×10542πsin2θcosθe−2iϕ)L^zΨ3,−2=−2ℏ(10542πsin2θcosθe−2iϕ)L^zΨ3,−2=−2ℏΨ3,−2$

The eigenvalues of z-component of angular momentum is $−2ℏ$ .

Conclusion

The eigenvalues of z-component of angular momentum is $−2ℏ$ .

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

Answer to Problem 11.54E

The eigenvalue of energy for the given wavefunctionis $6ℏ2I$ .

Explanation of Solution

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

$Ψl,ml=12π eimϕ⋅θl,ml$

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

$Ψ3,−2=10542πsin2θcosθe−2iϕ$

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 $12ℏ2$ .

$H^Ψ3,−2=12ℏ2Ψ3,−22IH^Ψ3,−2=6ℏ2Ψ3,−2I$

The eigenvalue of energy $E$ for the given wavefunction is $6ℏ2I$ .

Conclusion

The eigenvalue of energy $E$ for the given wavefunction is $6ℏ2I$ .

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!

Students have asked these similar questions

What are the IUPAC Names of all the compounds in the picture?

1) a) Give the dominant Intermolecular Force (IMF) in a sample of each of the following compounds. Please show your work. (8) SF2, CH,OH, C₂H₂ b) Based on your answers given above, list the compounds in order of their Boiling Point from low to high. (8)

19.78 Write the products of the following sequences of reactions. Refer to your reaction road- maps to see how the combined reactions allow you to "navigate" between the different functional groups. Note that you will need your old Chapters 6-11 and Chapters 15-18 roadmaps along with your new Chapter 19 roadmap for these. (a) 1. BHS 2. H₂O₂ 3. H₂CrO4 4. SOCI₂ (b) 1. Cl₂/hv 2. KOLBU 3. H₂O, catalytic H₂SO4 4. H₂CrO4 Reaction Roadmap An alkene 5. EtOH 6.0.5 Equiv. NaOEt/EtOH 7. Mild H₂O An alkane 1.0 2. (CH3)₂S 3. H₂CrO (d) (c) 4. Excess EtOH, catalytic H₂SO OH 4. Mild H₂O* 5.0.5 Equiv. NaOEt/EtOH An alkene 6. Mild H₂O* A carboxylic acid 7. Mild H₂O* 1. SOC₁₂ 2. EtOH 3.0.5 Equiv. NaOEt/E:OH 5.1.0 Equiv. NaOEt 6. NH₂ (e) 1. 0.5 Equiv. NaOEt/EtOH 2. Mild H₂O* Br (f) i H An aldehyde 1. Catalytic NaOE/EtOH 2. H₂O*, heat 3. (CH,CH₂)₂Culi 4. Mild H₂O* 5.1.0 Equiv. LDA Br An ester 4. NaOH, H₂O 5. Mild H₂O* 6. Heat 7. MgBr 8. Mild H₂O* 7. Mild H₂O+

Answer 2

Question

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

Answer to Problem 11.54E

The eigenvalues of total angular momentum is $12ℏ2$ .

Explanation of Solution

Explanation:

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

$Ψl,ml=12π eimϕ⋅θl,ml$

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

$Ψ3,−2=10542πsin2θcosθe−2iϕ$

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

$L^2Ψ=−ℏ2(∂2∂θ2+cotθ∂∂θ+1sin2θ∂2∂ϕ2)Ψ$

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

$∂∂θ(10542πsin2θcosθe−2iϕ)$

$=10542π(2sinθcos2θ−sin3θ)e−2iϕ$ …(1)

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

$∂2∂θ2=∂∂θ(10542π(2sinθcos2θ−sin3θ)e−2iϕ)$

$=10542π(2cos3θ−4sin2θcosθ−3sin2θcosθ)e−2iϕ$ …(2)

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

$∂2∂ϕ2(10542πsin2θcosθe−2iϕ)=(−2)2i2×10542πsin2θcosθe−2iϕ=−4×10542πsin2θcosθe−2iϕ$

$=−1052πsin2θcosθe−2iϕ$ …(3)

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

$L^2Ψ3,−2=−ℏ2(10542π(2cos3θ−4sin2θcosθ−3sin2θcosθ)e−2iϕ+cotθ10542π(2sinθcos2θ−sin3θ)e−2iϕ−1sin2θ1052πsin2θcosθe−2iϕ)$

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

$L^2Ψ3,−2=−ℏ210542πe−2iϕ((2cos3θ−7sin2θcosθ)+cotθ(2sinθcos2θ−sin3θ)e−2iϕ−4cosθe−2iϕ)$

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

$L^2Ψ3,−2=−ℏ210542πe−2iϕ((2cos3θ−7sin2θcosθ)+cosθsinθ(2sinθcos2θ−sin3θ)e−2iϕ−4cosθe−2iϕ)L^2Ψ3,−2=−ℏ210542πe−2iϕ((2cos3θ−7sin2θcosθ)+(2cos3θ−cosθsin2θ)−4cosθ)L^2Ψ3,−2=−ℏ210542πe−2iϕ(4cos3θ−8sin2θcosθ−4cosθ)$

Substitute $cos2θ=1−sin2θ$ in the given equation.

$L^2Ψ3,−2=−ℏ210542πe−2iϕ(4cosθ(1−sin2θ)−8sin2θcosθ−4cosθ)L^2Ψ3,−2=−ℏ210542πe−2iϕ(4cosθ−4sin2θcosθ−8sin2θcosθ−4cosθ)L^2Ψ3,−2=−ℏ210542πe−2iϕ(−12sin2θcosθ)L^2Ψ3,−2=12ℏ210542πe−2iϕ(sin2θcosθ)$

Thus, the total angular momentum is represented as,

$L^2Ψ3,−2=12ℏ2Ψ3,−2$

The eigenvalues of total angular momentum is $12ℏ2$ .

Conclusion

The eigenvalues of total angular momentum is $12ℏ2$ .

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

Answer to Problem 11.54E

Explanation of Solution

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

$Ψl,ml=12π eimϕ⋅θl,ml$

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

$Ψ3,−2=10542πsin2θcosθe−2iϕ$

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

$L^zΨ3,−2=−iℏ∂∂ϕΨ3,−2$

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

$∂∂ϕ(10542πsin2θcosθe−2iϕ)$

$=(−2)i×10542πsin2θcosθe−2iϕ$ …(4)

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

$L^zΨ3,−2=−iℏ((−2)i×10542πsin2θcosθe−2iϕ)L^zΨ3,−2=−2ℏ(10542πsin2θcosθe−2iϕ)L^zΨ3,−2=−2ℏΨ3,−2$

The eigenvalues of z-component of angular momentum is $−2ℏ$ .

Conclusion

The eigenvalues of z-component of angular momentum is $−2ℏ$ .

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

Answer to Problem 11.54E

The eigenvalue of energy for the given wavefunctionis $6ℏ2I$ .

Explanation of Solution

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

$Ψl,ml=12π eimϕ⋅θl,ml$

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

$Ψ3,−2=10542πsin2θcosθe−2iϕ$

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 $12ℏ2$ .

$H^Ψ3,−2=12ℏ2Ψ3,−22IH^Ψ3,−2=6ℏ2Ψ3,−2I$

The eigenvalue of energy $E$ for the given wavefunction is $6ℏ2I$ .

Conclusion

The eigenvalue of energy $E$ for the given wavefunction is $6ℏ2I$ .

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!

Answer 3

Question

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

Answer to Problem 11.54E

The eigenvalues of total angular momentum is $12ℏ2$ .

Explanation of Solution

Explanation:

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

$Ψl,ml=12π eimϕ⋅θl,ml$

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

$Ψ3,−2=10542πsin2θcosθe−2iϕ$

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

$L^2Ψ=−ℏ2(∂2∂θ2+cotθ∂∂θ+1sin2θ∂2∂ϕ2)Ψ$

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

$∂∂θ(10542πsin2θcosθe−2iϕ)$

$=10542π(2sinθcos2θ−sin3θ)e−2iϕ$ …(1)

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

$∂2∂θ2=∂∂θ(10542π(2sinθcos2θ−sin3θ)e−2iϕ)$

$=10542π(2cos3θ−4sin2θcosθ−3sin2θcosθ)e−2iϕ$ …(2)

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

$∂2∂ϕ2(10542πsin2θcosθe−2iϕ)=(−2)2i2×10542πsin2θcosθe−2iϕ=−4×10542πsin2θcosθe−2iϕ$

$=−1052πsin2θcosθe−2iϕ$ …(3)

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

$L^2Ψ3,−2=−ℏ2(10542π(2cos3θ−4sin2θcosθ−3sin2θcosθ)e−2iϕ+cotθ10542π(2sinθcos2θ−sin3θ)e−2iϕ−1sin2θ1052πsin2θcosθe−2iϕ)$

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

$L^2Ψ3,−2=−ℏ210542πe−2iϕ((2cos3θ−7sin2θcosθ)+cotθ(2sinθcos2θ−sin3θ)e−2iϕ−4cosθe−2iϕ)$

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

$L^2Ψ3,−2=−ℏ210542πe−2iϕ((2cos3θ−7sin2θcosθ)+cosθsinθ(2sinθcos2θ−sin3θ)e−2iϕ−4cosθe−2iϕ)L^2Ψ3,−2=−ℏ210542πe−2iϕ((2cos3θ−7sin2θcosθ)+(2cos3θ−cosθsin2θ)−4cosθ)L^2Ψ3,−2=−ℏ210542πe−2iϕ(4cos3θ−8sin2θcosθ−4cosθ)$

Substitute $cos2θ=1−sin2θ$ in the given equation.

$L^2Ψ3,−2=−ℏ210542πe−2iϕ(4cosθ(1−sin2θ)−8sin2θcosθ−4cosθ)L^2Ψ3,−2=−ℏ210542πe−2iϕ(4cosθ−4sin2θcosθ−8sin2θcosθ−4cosθ)L^2Ψ3,−2=−ℏ210542πe−2iϕ(−12sin2θcosθ)L^2Ψ3,−2=12ℏ210542πe−2iϕ(sin2θcosθ)$

Thus, the total angular momentum is represented as,

$L^2Ψ3,−2=12ℏ2Ψ3,−2$

The eigenvalues of total angular momentum is $12ℏ2$ .

Conclusion

The eigenvalues of total angular momentum is $12ℏ2$ .

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

Answer to Problem 11.54E

Explanation of Solution

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

$Ψl,ml=12π eimϕ⋅θl,ml$

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

$Ψ3,−2=10542πsin2θcosθe−2iϕ$

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

$L^zΨ3,−2=−iℏ∂∂ϕΨ3,−2$

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

$∂∂ϕ(10542πsin2θcosθe−2iϕ)$

$=(−2)i×10542πsin2θcosθe−2iϕ$ …(4)

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

$L^zΨ3,−2=−iℏ((−2)i×10542πsin2θcosθe−2iϕ)L^zΨ3,−2=−2ℏ(10542πsin2θcosθe−2iϕ)L^zΨ3,−2=−2ℏΨ3,−2$

The eigenvalues of z-component of angular momentum is $−2ℏ$ .

Conclusion

The eigenvalues of z-component of angular momentum is $−2ℏ$ .

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

Answer to Problem 11.54E

The eigenvalue of energy for the given wavefunctionis $6ℏ2I$ .

Explanation of Solution

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

$Ψl,ml=12π eimϕ⋅θl,ml$

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

$Ψ3,−2=10542πsin2θcosθe−2iϕ$

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 $12ℏ2$ .

$H^Ψ3,−2=12ℏ2Ψ3,−22IH^Ψ3,−2=6ℏ2Ψ3,−2I$

The eigenvalue of energy $E$ for the given wavefunction is $6ℏ2I$ .

Conclusion

The eigenvalue of energy $E$ for the given wavefunction is $6ℏ2I$ .

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!

Answer 4

Question

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

Answer to Problem 11.54E

The eigenvalues of total angular momentum is $12ℏ2$ .

Explanation of Solution

Explanation:

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

$Ψl,ml=12π eimϕ⋅θl,ml$

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

$Ψ3,−2=10542πsin2θcosθe−2iϕ$

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

$L^2Ψ=−ℏ2(∂2∂θ2+cotθ∂∂θ+1sin2θ∂2∂ϕ2)Ψ$

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

$∂∂θ(10542πsin2θcosθe−2iϕ)$

$=10542π(2sinθcos2θ−sin3θ)e−2iϕ$ …(1)

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

$∂2∂θ2=∂∂θ(10542π(2sinθcos2θ−sin3θ)e−2iϕ)$

$=10542π(2cos3θ−4sin2θcosθ−3sin2θcosθ)e−2iϕ$ …(2)

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

$∂2∂ϕ2(10542πsin2θcosθe−2iϕ)=(−2)2i2×10542πsin2θcosθe−2iϕ=−4×10542πsin2θcosθe−2iϕ$

$=−1052πsin2θcosθe−2iϕ$ …(3)

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

$L^2Ψ3,−2=−ℏ2(10542π(2cos3θ−4sin2θcosθ−3sin2θcosθ)e−2iϕ+cotθ10542π(2sinθcos2θ−sin3θ)e−2iϕ−1sin2θ1052πsin2θcosθe−2iϕ)$

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

$L^2Ψ3,−2=−ℏ210542πe−2iϕ((2cos3θ−7sin2θcosθ)+cotθ(2sinθcos2θ−sin3θ)e−2iϕ−4cosθe−2iϕ)$

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

$L^2Ψ3,−2=−ℏ210542πe−2iϕ((2cos3θ−7sin2θcosθ)+cosθsinθ(2sinθcos2θ−sin3θ)e−2iϕ−4cosθe−2iϕ)L^2Ψ3,−2=−ℏ210542πe−2iϕ((2cos3θ−7sin2θcosθ)+(2cos3θ−cosθsin2θ)−4cosθ)L^2Ψ3,−2=−ℏ210542πe−2iϕ(4cos3θ−8sin2θcosθ−4cosθ)$

Substitute $cos2θ=1−sin2θ$ in the given equation.

$L^2Ψ3,−2=−ℏ210542πe−2iϕ(4cosθ(1−sin2θ)−8sin2θcosθ−4cosθ)L^2Ψ3,−2=−ℏ210542πe−2iϕ(4cosθ−4sin2θcosθ−8sin2θcosθ−4cosθ)L^2Ψ3,−2=−ℏ210542πe−2iϕ(−12sin2θcosθ)L^2Ψ3,−2=12ℏ210542πe−2iϕ(sin2θcosθ)$

Thus, the total angular momentum is represented as,

$L^2Ψ3,−2=12ℏ2Ψ3,−2$

The eigenvalues of total angular momentum is $12ℏ2$ .

Conclusion

The eigenvalues of total angular momentum is $12ℏ2$ .

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

Answer to Problem 11.54E

Explanation of Solution

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

$Ψl,ml=12π eimϕ⋅θl,ml$

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

$Ψ3,−2=10542πsin2θcosθe−2iϕ$

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

$L^zΨ3,−2=−iℏ∂∂ϕΨ3,−2$

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

$∂∂ϕ(10542πsin2θcosθe−2iϕ)$

$=(−2)i×10542πsin2θcosθe−2iϕ$ …(4)

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

$L^zΨ3,−2=−iℏ((−2)i×10542πsin2θcosθe−2iϕ)L^zΨ3,−2=−2ℏ(10542πsin2θcosθe−2iϕ)L^zΨ3,−2=−2ℏΨ3,−2$

The eigenvalues of z-component of angular momentum is $−2ℏ$ .

Conclusion

The eigenvalues of z-component of angular momentum is $−2ℏ$ .

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

Answer to Problem 11.54E

The eigenvalue of energy for the given wavefunctionis $6ℏ2I$ .

Explanation of Solution

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

$Ψl,ml=12π eimϕ⋅θl,ml$

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

$Ψ3,−2=10542πsin2θcosθe−2iϕ$

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 $12ℏ2$ .

$H^Ψ3,−2=12ℏ2Ψ3,−22IH^Ψ3,−2=6ℏ2Ψ3,−2I$

The eigenvalue of energy $E$ for the given wavefunction is $6ℏ2I$ .

Conclusion

The eigenvalue of energy $E$ for the given wavefunction is $6ℏ2I$ .

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!

Answer 5

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

Answer to Problem 11.54E

The eigenvalues of total angular momentum is $12ℏ2$ .

Explanation of Solution

Explanation:

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

$Ψl,ml=12π eimϕ⋅θl,ml$

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

$Ψ3,−2=10542πsin2θcosθe−2iϕ$

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

$L^2Ψ=−ℏ2(∂2∂θ2+cotθ∂∂θ+1sin2θ∂2∂ϕ2)Ψ$

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

$∂∂θ(10542πsin2θcosθe−2iϕ)$

$=10542π(2sinθcos2θ−sin3θ)e−2iϕ$ …(1)

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

$∂2∂θ2=∂∂θ(10542π(2sinθcos2θ−sin3θ)e−2iϕ)$

$=10542π(2cos3θ−4sin2θcosθ−3sin2θcosθ)e−2iϕ$ …(2)

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

$∂2∂ϕ2(10542πsin2θcosθe−2iϕ)=(−2)2i2×10542πsin2θcosθe−2iϕ=−4×10542πsin2θcosθe−2iϕ$

$=−1052πsin2θcosθe−2iϕ$ …(3)

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

$L^2Ψ3,−2=−ℏ2(10542π(2cos3θ−4sin2θcosθ−3sin2θcosθ)e−2iϕ+cotθ10542π(2sinθcos2θ−sin3θ)e−2iϕ−1sin2θ1052πsin2θcosθe−2iϕ)$

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

$L^2Ψ3,−2=−ℏ210542πe−2iϕ((2cos3θ−7sin2θcosθ)+cotθ(2sinθcos2θ−sin3θ)e−2iϕ−4cosθe−2iϕ)$

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

$L^2Ψ3,−2=−ℏ210542πe−2iϕ((2cos3θ−7sin2θcosθ)+cosθsinθ(2sinθcos2θ−sin3θ)e−2iϕ−4cosθe−2iϕ)L^2Ψ3,−2=−ℏ210542πe−2iϕ((2cos3θ−7sin2θcosθ)+(2cos3θ−cosθsin2θ)−4cosθ)L^2Ψ3,−2=−ℏ210542πe−2iϕ(4cos3θ−8sin2θcosθ−4cosθ)$

Substitute $cos2θ=1−sin2θ$ in the given equation.

$L^2Ψ3,−2=−ℏ210542πe−2iϕ(4cosθ(1−sin2θ)−8sin2θcosθ−4cosθ)L^2Ψ3,−2=−ℏ210542πe−2iϕ(4cosθ−4sin2θcosθ−8sin2θcosθ−4cosθ)L^2Ψ3,−2=−ℏ210542πe−2iϕ(−12sin2θcosθ)L^2Ψ3,−2=12ℏ210542πe−2iϕ(sin2θcosθ)$

Thus, the total angular momentum is represented as,

$L^2Ψ3,−2=12ℏ2Ψ3,−2$

The eigenvalues of total angular momentum is $12ℏ2$ .

Conclusion

The eigenvalues of total angular momentum is $12ℏ2$ .

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

Answer to Problem 11.54E

Explanation of Solution

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

$Ψl,ml=12π eimϕ⋅θl,ml$

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

$Ψ3,−2=10542πsin2θcosθe−2iϕ$

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

$L^zΨ3,−2=−iℏ∂∂ϕΨ3,−2$

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

$∂∂ϕ(10542πsin2θcosθe−2iϕ)$

$=(−2)i×10542πsin2θcosθe−2iϕ$ …(4)

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

$L^zΨ3,−2=−iℏ((−2)i×10542πsin2θcosθe−2iϕ)L^zΨ3,−2=−2ℏ(10542πsin2θcosθe−2iϕ)L^zΨ3,−2=−2ℏΨ3,−2$

The eigenvalues of z-component of angular momentum is $−2ℏ$ .

Conclusion

The eigenvalues of z-component of angular momentum is $−2ℏ$ .

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

Answer to Problem 11.54E

The eigenvalue of energy for the given wavefunctionis $6ℏ2I$ .

Explanation of Solution

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

$Ψl,ml=12π eimϕ⋅θl,ml$

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

$Ψ3,−2=10542πsin2θcosθe−2iϕ$

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 $12ℏ2$ .

$H^Ψ3,−2=12ℏ2Ψ3,−22IH^Ψ3,−2=6ℏ2Ψ3,−2I$

The eigenvalue of energy $E$ for the given wavefunction is $6ℏ2I$ .

Conclusion

The eigenvalue of energy $E$ for the given wavefunction is $6ℏ2I$ .

Answer 6

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

Answer to Problem 11.54E

The eigenvalues of total angular momentum is $12ℏ2$ .

Explanation of Solution

Explanation:

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

$Ψl,ml=12π eimϕ⋅θl,ml$

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

$Ψ3,−2=10542πsin2θcosθe−2iϕ$

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

$L^2Ψ=−ℏ2(∂2∂θ2+cotθ∂∂θ+1sin2θ∂2∂ϕ2)Ψ$

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

$∂∂θ(10542πsin2θcosθe−2iϕ)$

$=10542π(2sinθcos2θ−sin3θ)e−2iϕ$ …(1)

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

$∂2∂θ2=∂∂θ(10542π(2sinθcos2θ−sin3θ)e−2iϕ)$

$=10542π(2cos3θ−4sin2θcosθ−3sin2θcosθ)e−2iϕ$ …(2)

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

$∂2∂ϕ2(10542πsin2θcosθe−2iϕ)=(−2)2i2×10542πsin2θcosθe−2iϕ=−4×10542πsin2θcosθe−2iϕ$

$=−1052πsin2θcosθe−2iϕ$ …(3)

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

$L^2Ψ3,−2=−ℏ2(10542π(2cos3θ−4sin2θcosθ−3sin2θcosθ)e−2iϕ+cotθ10542π(2sinθcos2θ−sin3θ)e−2iϕ−1sin2θ1052πsin2θcosθe−2iϕ)$

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

$L^2Ψ3,−2=−ℏ210542πe−2iϕ((2cos3θ−7sin2θcosθ)+cotθ(2sinθcos2θ−sin3θ)e−2iϕ−4cosθe−2iϕ)$

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

$L^2Ψ3,−2=−ℏ210542πe−2iϕ((2cos3θ−7sin2θcosθ)+cosθsinθ(2sinθcos2θ−sin3θ)e−2iϕ−4cosθe−2iϕ)L^2Ψ3,−2=−ℏ210542πe−2iϕ((2cos3θ−7sin2θcosθ)+(2cos3θ−cosθsin2θ)−4cosθ)L^2Ψ3,−2=−ℏ210542πe−2iϕ(4cos3θ−8sin2θcosθ−4cosθ)$

Substitute $cos2θ=1−sin2θ$ in the given equation.

$L^2Ψ3,−2=−ℏ210542πe−2iϕ(4cosθ(1−sin2θ)−8sin2θcosθ−4cosθ)L^2Ψ3,−2=−ℏ210542πe−2iϕ(4cosθ−4sin2θcosθ−8sin2θcosθ−4cosθ)L^2Ψ3,−2=−ℏ210542πe−2iϕ(−12sin2θcosθ)L^2Ψ3,−2=12ℏ210542πe−2iϕ(sin2θcosθ)$

Thus, the total angular momentum is represented as,

$L^2Ψ3,−2=12ℏ2Ψ3,−2$

The eigenvalues of total angular momentum is $12ℏ2$ .

Conclusion

The eigenvalues of total angular momentum is $12ℏ2$ .

Answer 7

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.

Answer 8

Expert Solution

Answer to Problem 11.54E

The eigenvalues of total angular momentum is $12ℏ2$ .

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Explanation:

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

$Ψl,ml=12π eimϕ⋅θl,ml$

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

$Ψ3,−2=10542πsin2θcosθe−2iϕ$

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

$L^2Ψ=−ℏ2(∂2∂θ2+cotθ∂∂θ+1sin2θ∂2∂ϕ2)Ψ$

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

$∂∂θ(10542πsin2θcosθe−2iϕ)$

$=10542π(2sinθcos2θ−sin3θ)e−2iϕ$ …(1)

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

$∂2∂θ2=∂∂θ(10542π(2sinθcos2θ−sin3θ)e−2iϕ)$

$=10542π(2cos3θ−4sin2θcosθ−3sin2θcosθ)e−2iϕ$ …(2)

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

$∂2∂ϕ2(10542πsin2θcosθe−2iϕ)=(−2)2i2×10542πsin2θcosθe−2iϕ=−4×10542πsin2θcosθe−2iϕ$

$=−1052πsin2θcosθe−2iϕ$ …(3)

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

$L^2Ψ3,−2=−ℏ2(10542π(2cos3θ−4sin2θcosθ−3sin2θcosθ)e−2iϕ+cotθ10542π(2sinθcos2θ−sin3θ)e−2iϕ−1sin2θ1052πsin2θcosθe−2iϕ)$

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

$L^2Ψ3,−2=−ℏ210542πe−2iϕ((2cos3θ−7sin2θcosθ)+cotθ(2sinθcos2θ−sin3θ)e−2iϕ−4cosθe−2iϕ)$

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

$L^2Ψ3,−2=−ℏ210542πe−2iϕ((2cos3θ−7sin2θcosθ)+cosθsinθ(2sinθcos2θ−sin3θ)e−2iϕ−4cosθe−2iϕ)L^2Ψ3,−2=−ℏ210542πe−2iϕ((2cos3θ−7sin2θcosθ)+(2cos3θ−cosθsin2θ)−4cosθ)L^2Ψ3,−2=−ℏ210542πe−2iϕ(4cos3θ−8sin2θcosθ−4cosθ)$

Substitute $cos2θ=1−sin2θ$ in the given equation.

$L^2Ψ3,−2=−ℏ210542πe−2iϕ(4cosθ(1−sin2θ)−8sin2θcosθ−4cosθ)L^2Ψ3,−2=−ℏ210542πe−2iϕ(4cosθ−4sin2θcosθ−8sin2θcosθ−4cosθ)L^2Ψ3,−2=−ℏ210542πe−2iϕ(−12sin2θcosθ)L^2Ψ3,−2=12ℏ210542πe−2iϕ(sin2θcosθ)$

Thus, the total angular momentum is represented as,

$L^2Ψ3,−2=12ℏ2Ψ3,−2$

The eigenvalues of total angular momentum is $12ℏ2$ .

Answer 13

Conclusion

The eigenvalues of total angular momentum is $12ℏ2$ .

Answer 14

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

Answer to Problem 11.54E

Explanation of Solution

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

$Ψl,ml=12π eimϕ⋅θl,ml$

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

$Ψ3,−2=10542πsin2θcosθe−2iϕ$

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

$L^zΨ3,−2=−iℏ∂∂ϕΨ3,−2$

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

$∂∂ϕ(10542πsin2θcosθe−2iϕ)$

$=(−2)i×10542πsin2θcosθe−2iϕ$ …(4)

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

$L^zΨ3,−2=−iℏ((−2)i×10542πsin2θcosθe−2iϕ)L^zΨ3,−2=−2ℏ(10542πsin2θcosθe−2iϕ)L^zΨ3,−2=−2ℏΨ3,−2$

The eigenvalues of z-component of angular momentum is $−2ℏ$ .

Conclusion

The eigenvalues of z-component of angular momentum is $−2ℏ$ .

Answer 15

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.

Answer 16

Expert Solution

Answer to Problem 11.54E

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Expert Solution

Answer 20

Explanation of Solution

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

$Ψl,ml=12π eimϕ⋅θl,ml$

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

$Ψ3,−2=10542πsin2θcosθe−2iϕ$

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

$L^zΨ3,−2=−iℏ∂∂ϕΨ3,−2$

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

$∂∂ϕ(10542πsin2θcosθe−2iϕ)$

$=(−2)i×10542πsin2θcosθe−2iϕ$ …(4)

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

$L^zΨ3,−2=−iℏ((−2)i×10542πsin2θcosθe−2iϕ)L^zΨ3,−2=−2ℏ(10542πsin2θcosθe−2iϕ)L^zΨ3,−2=−2ℏΨ3,−2$

The eigenvalues of z-component of angular momentum is $−2ℏ$ .

Answer 21

Conclusion

The eigenvalues of z-component of angular momentum is $−2ℏ$ .

Answer 22

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

Answer to Problem 11.54E

The eigenvalue of energy for the given wavefunctionis $6ℏ2I$ .

Explanation of Solution

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

$Ψl,ml=12π eimϕ⋅θl,ml$

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

$Ψ3,−2=10542πsin2θcosθe−2iϕ$

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 $12ℏ2$ .

$H^Ψ3,−2=12ℏ2Ψ3,−22IH^Ψ3,−2=6ℏ2Ψ3,−2I$

The eigenvalue of energy $E$ for the given wavefunction is $6ℏ2I$ .

Conclusion

The eigenvalue of energy $E$ for the given wavefunction is $6ℏ2I$ .

Answer 23

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.

Answer 24

Expert Solution

Answer to Problem 11.54E

The eigenvalue of energy for the given wavefunctionis $6ℏ2I$ .

Answer 25

Expert Solution

Answer 26

Expert Solution

Answer 27

Expert Solution

Answer 28

Explanation of Solution

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

$Ψl,ml=12π eimϕ⋅θl,ml$

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

$Ψ3,−2=10542πsin2θcosθe−2iϕ$

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 $12ℏ2$ .

$H^Ψ3,−2=12ℏ2Ψ3,−22IH^Ψ3,−2=6ℏ2Ψ3,−2I$

The eigenvalue of energy $E$ for the given wavefunction is $6ℏ2I$ .

Answer 29

Conclusion

The eigenvalue of energy $E$ for the given wavefunction is $6ℏ2I$ .

Answer 30

Ch. 11 - Convert 3.558mdyn/A into units of N/m.Ch. 11 - Prob. 11.2E Ch. 11 - Prob. 11.3E Ch. 11 - Prob. 11.4E Ch. 11 - Prob. 11.5E Ch. 11 - Prob. 11.6E Ch. 11 - Prob. 11.7E Ch. 11 - Prob. 11.8E Ch. 11 - Prob. 11.9E Ch. 11 - Prob. 11.10E

Concept explainers

Answer to Problem 11.54E

Explanation of Solution

Answer to Problem 11.54E

Explanation of Solution

Answer to Problem 11.54E

Explanation of Solution

Want to see more full solutions like this?

Chapter 11 Solutions