This set of Physical Chemistry Multiple Choice Questions & Answers (MCQs) focuses on “Quantum Theory – Rigid Rotor”.

1. What is the moment of inertia I of a body mass m rotating about a radius r?

a) I = mr^{2}

b) I = mr

c) I = m^{2} r

d) I = \(\frac{m}{r^2}\)

View Answer

Explanation: Moment of inertia is the tendency of a body to resist angular acceleration for the rotation of a classical particle about an axis. In the expression for kinetic rotational energy, moment of inertia plays the same role that mass plays in linear kinetic energy.

2. What is the mathematical definition of angular momentum L for a classical particle in terms of moment of inertia I and angular frequency ω?

a) L = I^{2}ω

b) L = Iω^{2}

c) L = \(\frac{I}{\omega}\)

d) L = Iω

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Explanation: Angular momentum is the rotational equivalent of linear momentum. While linear momentum is the product of mass and velocity, angular momentum is the product of moment of inertia and angular frequency. This relationship makes sense because moment of inertia and angular frequency are the rotational equivalents of mass and velocity respectively.

3. A particle on a ring exists in a superposition state: φ(Φ) = \(\sqrt{\frac{3}{4\pi}}\) e^{iΦ} – \(\frac{i}{\sqrt {4\pi}}\) e^{-2iΦ}. Normalized eigenfunctions for a particle on a ring are given by φ(Φ) = \(\frac{1}{\sqrt {2\pi}}\) e^{imΦ}; m = 0, 1, 2… What should this wavefunction be divided by for normalization?

a) 2

b) √2

c) \(\frac{1}{2\pi}\)

d) \(\frac{1}{\sqrt 2}\)

View Answer

Explanation: ∑|C

_{n}|

^{2}= \(\frac{3}{2} + \frac{1}{2}\) = 2 —→ divide by √2 to normalize. Normalized wavefunction given by φ(Φ) = \(\sqrt{\frac{3}{8\pi}}\) e

^{iΦ}– \(\frac{i}{\sqrt {8\pi}}\) e

^{-2iΦ}

4. A particle on a ring exists in a superposition state: φ(Φ) = \(\sqrt{\frac{3}{4\pi}}\) e^{iΦ} – \(\frac{i}{\sqrt {4\pi}}\) e^{-2iΦ}. Normalized eigenfunctions for a particle on a ring are given by φ(Φ) = \(\frac{1}{\sqrt {2\pi}}\) e^{imΦ}; m = 0, 1, 2… What is the expectation value of energy for this system?

a) < E > = \(\frac{\hbar^2}{2\mu r^2}\)

b) < E > = \(\frac{7\hbar^2}{8\mu r^2}\)

c) < E > = 0

d) < E > = \(\frac{\hbar^2}{4\pi\mu r^2}\)

View Answer

Explanation: Normalized wavefunction given by φ(Φ) = \(\sqrt{\frac{3}{8\pi}}\) e

^{iΦ}– \(\frac{i}{\sqrt {8\pi}}\) e

^{-2iΦ}

< E > = ∑|C

_{n}|

^{2}E

_{n}= \(\frac{3\hbar^2}{4.2\mu r^2} + \frac{1\hbar^2.4}{4.2\mu r^2} = \frac{7\hbar^2}{8\mu r^2}\)

5. A particle on a ring exists in a superposition state: φ(Φ) = \(\sqrt{\frac{3}{4\pi}}\) e^{iΦ} – \(\frac{i}{\sqrt {4\pi}}\) e^{-2iΦ}. Is this wavefunction an eigenfunction of the angular momentum operator L_{z} = -iη\(\frac{\partial}{\partial\varphi}\)?

a) Yes, with eigenvalue \(\frac{4\pi}{3}\)

b) Yes, with eigenvalue \(\frac{2\pi}{3}\)

c) Yes, with eigenvalue \(\sqrt{4\pi/3}\)

d) No, it is not an eigenfunction of the given operator

View Answer

Explanation: The given wavefunction is a linear superposition of two wavefunctions with different eigenvalues. It is mathematically not possible for the given wavefunction to be an eigenfunction of this angular momentum operator.

6. A particle on a ring exists in a superposition state: φ(Φ) = \(\sqrt{\frac{3}{4\pi}}\) e^{iΦ} – \(\frac{i}{\sqrt {4\pi}}\) e^{-2iΦ}. What is the expectation value of angular momentum in the z-direction for this state?

a) < L_{z} > = \(\frac{\hbar}{4}\)

b) < L_{z} > = \(\frac{\hbar}{4\pi^2}\)

c) < L_{z} > = \(\frac{7\hbar}{8}\)

d) < L_{z} > = \(\frac{\hbar^2}{4}\)

View Answer

Explanation: ∑|C

_{m}|

^{2}mħ = \(\frac{3}{4}\)ħ + \(\frac{1}{4}\)ħ(-2) = \(\frac{\hbar}{4}\). m is the magnetic quantum number of each superposition state, which is in the exponential part of each function.

7. A particle on a ring exists in a superposition state: φ(Φ) = \(\sqrt{\frac{1}{8\pi}}\) e^{iΦ} – \(\frac{i}{2\sqrt{\pi}}\) e^{-3iΦ} + \(\frac{1}{2\sqrt {2\pi}}\) e^{i5Φ}. Normalized eigenfunctions for a particle on a ring are given by φ(Φ) = \(\frac{1}{\sqrt {2\pi}}\) e^{imΦ}; m = 0, 1, 2… What should this function be divided by for normalization?

a) 1

b) √2

c) 2√2

d) √3

View Answer

Explanation: The function is already normalized: ∑|C

_{m}|

^{2}= (1/2)

^{2}+ \((\frac{-i}{\sqrt 2})^2 + (\frac{1}{2})^2 = \frac{1}{4} + \frac{1}{2} + \frac{1}{4}\) = 1

8. A particle on a ring exists in a superposition state: φ(Φ) = \(\sqrt{\frac{1}{8\pi}}\) e^{iΦ} – \(\frac{i}{2\sqrt \pi}\) e^{-3iΦ} + \(\frac{1}{2\sqrt {2\pi}}\) e^{i5Φ}. What is the expectation of energy in this superposition state?

a) < E > = \(\frac{\hbar^2}{2\mu r^2}\)

b) < E > = \(\frac{\hbar^2}{\mu r^2}\)

c) < E > = \(\frac{5\hbar^2}{2\mu r^2}\)

d) < E > = \(\frac{11\hbar^2}{2\mu r^2}\)

View Answer

Explanation: < E > = ∑|C

_{n}|

^{2}E

_{n}= \(\frac{\hbar^2}{4.2\mu r^2} + \frac{1.9\hbar^2}{2.2\mu r^2} + \frac{1.25\hbar^2}{4.2\mu r^2} = \frac{\hbar^2}{2\mu r^2} (\frac{1}{4} + \frac{18}{4} + \frac{25}{4}) = \frac{11\hbar^2}{2\mu r^2}\)

9. A particle on a ring exists in a superposition state: φ(Φ) = \(\sqrt{\frac{1}{8\pi}}\) e^{iΦ} – \(\frac{i}{2\sqrt \pi}\) e^{-3iΦ} + \(\frac{1}{2\sqrt {2\pi}}\) e^{i5Φ}. Is this wavefunction an eigenfunction of the angular momentum operator L_{z} = -iη\(\frac{\partial}{\partial\varphi}\)?

a) Yes, with eigenvalue \(\frac{4\pi}{3}\)

b) Yes, with eigenvalue \(\frac{2\pi}{3}\)

c) Yes, with eigenvalue \(\sqrt{4\pi/3}\)

d) No, it is not an eigenfunction of the given operator

View Answer

Explanation: The given wavefunction is a linear superposition of two wavefunctions with different eigenvalues. It is mathematically not possible for the given wavefunction to be an eigenfunction of this angular momentum operator.

10. A particle on a ring exists in a superposition state: φ(Φ) = \(\sqrt{\frac{1}{8\pi}}\) e^{iΦ} – \(\frac{i}{2\sqrt \pi}\) e^{-3iΦ} + \(\frac{1}{2\sqrt {2\pi}}\) e^{i5Φ}. What is the expectation of energy in this superposition state?

a) < L_{z} > = \(\frac{\hbar}{4}\)

b) < L_{z} > = \(\frac{\hbar}{4\pi^2}\)

c) < L_{z} > = \(\frac{7\hbar}{8}\)

d) < L_{z} > = 0

View Answer

Explanation: < L

_{z}> = ∑|C

_{m}|

^{2}mħ = \(\frac{\hbar}{4} + \frac{1}{2}\)(-3)ħ + \(\frac{\hbar.5}{4}\) = 0.

11. A wavefunction is given by φ(x) = \((\frac{4\alpha^3}{\pi})^{1/4} xe^{\frac{-\alpha x^2}{2}}\). Is this function odd or even?

a) Odd, because φ(-x) = -φ(x)

b) Odd, because φ(x) = -φ(x)

c) Even, because φ(-x) = -φ(x)

d) Even, because φ(-x) = φ(x)

View Answer

Explanation: A function is odd if φ(-x) = -φ(x). The integral over all space for such a function is always equal to zero since the function is symmetric about the origin.

12. A wavefunction is given by φ(x) = \((\frac{4\alpha^3}{\pi})^{1/4} xe^{\frac{-\alpha x^2}{2}}\). Is this function acceptable according to the Born conditions?

a) Yes, because the function and its derivatives are continuous and finite everywhere

b) No, because the function and its derivatives are continuous and finite everywhere

c) Yes, because the function and its derivatives are neither continuous nor finite everywhere

d) No, because the function and its derivatives are neither continuous nor finite everywhere

View Answer

Explanation: The given wavefunction is continuous and finite and so are its derivatives. A plot of the function is included below:

13. A wavefunction is given by φ(x) = \((\frac{4\alpha^3}{\pi})\)^{1/4} \(xe^{\frac{-\alpha x^2}{2}}\). Which stationary state does this function describe?

a) Ground state

b) First excited state

c) Second excited state

d) Third excited state

View Answer

Explanation: This function has only one node, i.e. it intersects the x-axis at only one point (x = 0). This represents the first energy level or excited state of the given wavefunction. It is the first excited state of the quantum mechanical harmonic oscillator.

14. What is an element of differential volume dv in spherical coordinates?

a) dv = r^{2} sin θ drdθ

b) dv = drdθdΦ

c) dv = r^{2} drdθdΦ

d) dv = r^{2} sinθ drdθdΦ

What is an element of differential volume dv in spherical coordinates?

View Answer

Explanation: Spherical coordinates contains three variables: r (radius), radial angle, and azimuthal angle. The addition of the r

^{2}sinθ term is from the conversion from Cartesian to spherical coordinates.

15. By what factor does the energy of a rigid rotor change if the moment of inertia of a particle on a sphere is increased by a factor of 2?

a) Decreases by a factor of 2

b) Increases by a factor of 2

c) Stays the same

d) Increases by a factor of 4

What is an element of differential volume dv in spherical coordinates?

View Answer

Explanation: Energies of a rigid rotor are given by E = \(\frac{l(l+1)\hbar^2}{2I}\); l is the angular quantum number, and I is the moment of inertia Increasing I by a factor of two will subsequently decrease overall energy by a factor of 2 by the given equation.

**Sanfoundry Global Education & Learning Series – Physical Chemistry.**

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