Physical Chemistry 2 Quiz: Rotational Energy Levels

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Rotational Energy LevelsQuestion 1 of 20

For a linear rotor with B=2 cm^-1, what is the first rotational Raman Stokes shift?

12 cm^-1

4 cm^-1

8 cm^-1

20 cm^-1

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Physical Chemistry 2 Quiz

Physical Chemistry 2 Quiz: Rotational Energy Levels

Practice Rotational Energy Levels in Physical Chemistry 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Rotational Energy Levels, giving you a quick way to practice the rules, question types, and explanations that matter most for Physical Chemistry 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

For a linear rotor with B=2 cm^-1, what is the first rotational Raman Stokes shift?

12 cm^-1 (correct answer)
4 cm^-1
8 cm^-1
20 cm^-1

Explanation: Rotational Raman Stokes transitions follow Delta J = +2. Rotational energy is B J(J+1), so the first Stokes line is J = 0 to J = 2, with shift [2(3) - 0(1)] B = 6B. With B = 2 cm^-1, the shift is 12 cm^-1. The tempting 8 cm^-1 is just the 4B spacing between successive Raman lines, not the first Stokes shift.

Question 2

Rigid rotor, B=10 cm^-1 at 100 K. Find N(J=1)/N(J=0).

3.00
2.25 (correct answer)
0.75
1.50

Explanation: For J=0, E=0 and degeneracy=1; for J=1, E=2B=20 cm^-1 and degeneracy=3. The energy spacing is 28.8 K in temperature units, so at 100 K the Boltzmann factor is exp(-0.288)=0.750. Multiply by the degeneracy ratio 3/1 to get 2.25. The tempting wrong answer 3.00 keeps only the 3-fold degeneracy and drops the Boltzmann factor.

Question 3

A rigid rotor has I=1.45e-46 kg m^2. Find the wavelength of the J=0 to 1 line.

1.30 mm
5.18 mm
2.59 mm (correct answer)
0.259 mm

Explanation: For a rigid rotor, E_J = h^2 J(J+1)/(8

pi^2

I). The 0 to 1 jump gives delta E = E_1 - E_0 = h^2/(4

pi^2

I), so wavelength lambda = c h / delta E = 4 pi^2 c I / h = 2.59 mm. The tempting wrong 5.18 mm comes from forgetting the factor J(J+1)=2 and using E_1 = h^2/(8

pi^2

I).

Question 4

For HCl, B=10.59 cm^-1. Calculate B for DCl.

5.30 cm^-1
10.59 cm^-1
5.44 cm^-1 (correct answer)
20.61 cm^-1

Explanation: For a diatomic molecule, the rotational constant B is inversely proportional to reduced mass μ. For HCl, μ=135/(1+35)=35/36 u; for DCl, μ=235/(2+35)=70/37 u. Ratio μ(HCl)/μ(DCl)=37/72=0.514, so B(DCl)=10.59*37/72=5.44 cm^-1. The tempting 5.30 cm^-1 comes from just halving B because D is twice H's mass, but reduced mass isn't exactly doubled.

Question 5

For a linear rotor, the J=0 to 1 line is 4.0 cm^-1. Find the J=3 to 4 line.

12 cm^-1
8 cm^-1
20 cm^-1
16 cm^-1 (correct answer)

Explanation: Rotational transition energies increase as 2B(J+1), so J=0 to 1 sets 2B = 4 cm^-1 and B = 2 cm^-1. For J=3 to 4, delta E = 2B(4) = 8B = 16 cm^-1. The tempting 12 cm^-1 is the J=2 to 3 line; J=3 to 4 is the next one.

Question 6

A symmetric top molecule has rotational energy levels $E_{J,K} = BJ(J+1) + (A-B)K^2$ where $A$ and $B$ are rotational constants and $K$ is the projection of $J$ along the molecular axis. For $A = 5.2 \text{ cm}^{-1}$ and $B = 1.3 \text{ cm}^{-1}$ , what is the energy separation between the $J = 2, K = 0$ and $J = 2, K = 1$ levels?

$3.9 \text{ cm}^{-1}$ (correct answer)
$1.3 \text{ cm}^{-1}$
$5.2 \text{ cm}^{-1}$
$6.5 \text{ cm}^{-1}$
$7.8 \text{ cm}^{-1}$

Explanation: When you encounter symmetric top molecules, you're dealing with molecules that have one unique rotational axis (like NH₃ or CH₃Cl). The energy formula

E_{J,K} = BJ(J+1) + (A-B)K^2

separates the rotational motion into two components: overall rotation characterized by

J

, and rotation around the molecular symmetry axis characterized by

K

. To find the energy separation, calculate each energy level separately. For the

J = 2, K = 0

level:

E_{2,0} = (1.3)(2)(3) + (5.2-1.3)(0)^2 = 7.8 \text{ cm}^{-1}

For the

J = 2, K = 1

level:

E_{2,1} = (1.3)(2)(3) + (5.2-1.3)(1)^2 = 7.8 + 3.9 = 11.7 \text{ cm}^{-1}

The energy separation is

11.7 - 7.8 = 3.9 \text{ cm}^{-1}

, which is answer A. The wrong answers represent common calculation errors: B (

1.3 \text{ cm}^{-1}

) would result from using only the

B

constant instead of

(A-B)

. C (

5.2 \text{ cm}^{-1}

) comes from incorrectly using just the

A

constant. D (

6.5 \text{ cm}^{-1}

) results from adding

A + B

instead of taking their difference. Remember that for symmetric tops, the

K

-dependence always involves the

(A-B)

term. When calculating energy differences between states with the same

J

but different

K

values, focus on how the

(A-B)K^2

term changes, since the

BJ(J+1)

portion cancels out.

Question 7

A molecule undergoes a rotational transition with selection rule $\Delta J = +1$ . The observed absorption lines occur at wavenumbers $\tilde{\nu} = 20.4, 40.8, 61.2 \text{ cm}^{-1}$ . If these correspond to $J = 0 \rightarrow 1$ , $J = 1 \rightarrow 2$ , and $J = 2 \rightarrow 3$ transitions respectively, what is the rotational constant $B$ , and what would be the wavenumber for the $J = 5 \rightarrow 6$ transition?

$B = 10.2 \text{ cm}^{-1}$ ; $J = 5 \rightarrow 6$ at $122.4 \text{ cm}^{-1}$ (correct answer)
$B = 10.2 \text{ cm}^{-1}$ ; $J = 5 \rightarrow 6$ at $142.8 \text{ cm}^{-1}$
$B = 20.4 \text{ cm}^{-1}$ ; $J = 5 \rightarrow 6$ at $244.8 \text{ cm}^{-1}$
$B = 6.8 \text{ cm}^{-1}$ ; $J = 5 \rightarrow 6$ at $81.6 \text{ cm}^{-1}$
$B = 10.2 \text{ cm}^{-1}$ ; $J = 5 \rightarrow 6$ at $112.2 \text{ cm}^{-1}$

Explanation: When you encounter rotational spectroscopy problems, remember that molecular rotation is quantized, and the energy spacing between levels follows a predictable pattern based on the rotational constant B. For a rigid rotor, the rotational energy levels are

E_J = BJ(J+1)

, where J is the rotational quantum number. For absorption transitions with

\Delta J = +1

, the wavenumber of each transition is:

\tilde{\nu} = E_{J+1} - E_J = B(J+1)(J+2) - BJ(J+1) = 2B(J+1)

This means the

J = 0 \rightarrow 1

transition occurs at

\tilde{\nu} = 2B(1) = 2B

, the

J = 1 \rightarrow 2

transition at

4B

, and the

J = 2 \rightarrow 3

transition at

6B

. From the first transition:

20.4 = 2B

, so

B = 10.2 \text{ cm}^{-1}

. You can verify this with the other transitions:

40.8 = 4(10.2)

and

61.2 = 6(10.2)

✓ For the

J = 5 \rightarrow 6

transition:

\tilde{\nu} = 2B(6) = 12B = 12(10.2) = 122.4 \text{ cm}^{-1}

Answer A correctly identifies both values. Answer B uses the right B value but incorrectly calculates the transition as

14B

instead of

12B

. Answer C mistakes the first transition wavenumber for B itself, leading to

B = 20.4

instead of

10.2

. Answer D incorrectly averages the spacing between transitions, giving

B = 6.8

. Study tip: Always remember that rotational transitions follow the pattern

\tilde{\nu} = 2B(J+1)

for

\Delta J = +1

, making the spacing between consecutive lines equal to

2B

Question 8

Consider two diatomic molecules with the same reduced mass but different bond lengths: molecule A has $r_A = 1.20 \text{ Å}$ and molecule B has $r_B = 1.80 \text{ Å}$ . If molecule A has rotational constant $B_A = 2.50 \text{ cm}^{-1}$ , what is $B_B$ for molecule B? At what temperature will both molecules have the same average rotational energy?

$B_B = 1.11 \text{ cm}^{-1}$ ; same average energy at all temperatures since $\langle E_{rot} \rangle = kT$ (correct answer)
$B_B = 1.11 \text{ cm}^{-1}$ ; same average energy never occurs due to different level spacings
$B_B = 3.75 \text{ cm}^{-1}$ ; same average energy at $T = 50 \text{ K}$
$B_B = 1.67 \text{ cm}^{-1}$ ; same average energy at all temperatures since equipartition applies
$B_B = 1.11 \text{ cm}^{-1}$ ; same average energy at $T = 0 \text{ K}$ only

Explanation: This question tests your understanding of rotational spectroscopy and statistical thermodynamics for diatomic molecules. When you see problems involving rotational constants and different bond lengths, think about the relationship between molecular structure and energy level spacing. The rotational constant

B

is inversely proportional to the moment of inertia:

B = \frac{h}{8\pi^2cI}

, where

I = \mu r^2

for a diatomic molecule. Since both molecules have the same reduced mass

\mu

, you can write:

\frac{B_A}{B_B} = \frac{r_B^2}{r_A^2}

. Substituting the values:

\frac{2.50}{B_B} = \frac{(1.80)^2}{(1.20)^2} = \frac{3.24}{1.44} = 2.25

. Therefore,

B_B = \frac{2.50}{2.25} = 1.11 \text{ cm}^{-1}

. For the temperature question, both molecules are diatomic rotors in the high-temperature limit where classical equipartition applies. In this regime, each molecule has average rotational energy

\langle E_{rot} \rangle = kT

regardless of the specific rotational constant values. Since equipartition depends only on temperature and the number of rotational degrees of freedom (2 for linear molecules), both molecules have identical average rotational energies at all temperatures. Choice B incorrectly suggests the energies are never equal, missing that equipartition equalizes average energies despite different level spacings. Choice C miscalculates

B_B

and incorrectly identifies a specific temperature. Choice D has the wrong

B_B

value but correctly identifies equipartition. Study tip: Remember that rotational constants scale as

1/r^2

, and equipartition theorem equalizes average energies per degree of freedom when

kT

exceeds level spacings.

Question 9

A linear triatomic molecule ABC has moments of inertia $I_A = 15.2 \times 10^{-47} \text{ kg·m}^2$ about its center of mass. If the rotational constant is $B = 0.85 \text{ cm}^{-1}$ , and a rotational transition $J = 4 \rightarrow 5$ is observed, what is the wavelength of the absorbed radiation in the microwave region?

$\lambda = 1.18 \text{ mm}$ (correct answer)
$\lambda = 0.59 \text{ mm}$
$\lambda = 2.35 \text{ mm}$
$\lambda = 0.85 \text{ mm}$
$\lambda = 1.76 \text{ mm}$

Explanation: When you encounter rotational spectroscopy problems, you're dealing with the quantized energy levels of rotating molecules and the electromagnetic radiation they absorb during transitions between these levels. For a rotational transition in a linear molecule, the energy difference is given by

\Delta E = 2B(J+1)hc

, where

J

is the lower quantum number. For the

J = 4 \rightarrow 5

transition, this becomes

\Delta E = 2B(5)hc = 10Bhc

. Converting the rotational constant to SI units:

B = 0.85 \text{ cm}^{-1} = 85 \text{ m}^{-1}

. The energy difference is:

\Delta E = 10 \times 85 \times 6.626 \times 10^{-34} \times 3.00 \times 10^8 = 1.689 \times 10^{-22} \text{ J}

Using

E = \frac{hc}{\lambda}

, we solve for wavelength:

\lambda = \frac{hc}{\Delta E} = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{1.689 \times 10^{-22}} = 1.18 \times 10^{-3} \text{ m} = 1.18 \text{ mm}

This confirms answer A is correct. Answer B (0.59 mm) results from incorrectly using

\Delta E = B(J+1)hc

instead of

2B(J+1)hc

, missing the factor of 2. Answer C (2.35 mm) comes from using the wrong transition formula or incorrectly calculating the energy. Answer D (0.85 mm) appears to conflate the rotational constant value with the wavelength calculation. Study tip: Always remember that rotational transitions in linear molecules follow

\Delta E = 2B(J+1)hc

, and double-check your unit conversions from cm

^{-1}

to m

^{-1}

by multiplying by 100.

Question 10

In the rigid rotor model, the allowed values of the magnetic quantum number $m_J$ range from $-J$ to $+J$ . For a molecule in the $J = 3$ rotational state placed in a magnetic field, the degeneracy is partially lifted. How many distinct energy levels result, and what is the relative energy spacing pattern if the Zeeman effect is linear in field strength?

7 levels with equal spacing proportional to $m_J$ (correct answer)
3 levels with spacing proportional to $m_J^2$
7 levels with spacing proportional to $J(J+1) + m_J$
4 levels with spacing proportional to $|m_J|$
6 levels with alternating spacings proportional to $2m_J + 1$

Explanation: When analyzing the Zeeman effect in molecular spectroscopy, you're examining how magnetic fields lift the degeneracy of rotational energy levels by interacting with the molecule's magnetic moment. For a

J = 3

rotational state, the magnetic quantum number

m_J

can take values:

-3, -2, -1, 0, +1, +2, +3

. That's seven possible values. In the absence of a magnetic field, all these states have identical energy and are said to be degenerate. When you apply a magnetic field, each

m_J

value acquires a unique energy shift proportional to

m_J

itself (in the linear Zeeman regime), creating seven distinct energy levels with equal spacing between adjacent levels. Answer A correctly identifies both the number of levels (7) and the linear relationship with

m_J

. Answer B incorrectly suggests only 3 levels and a quadratic dependence on

m_J

, which would occur in a different magnetic field regime. Answer C proposes the right number of levels but incorrectly includes the

J(J+1)

term, which represents the field-free rotational energy, not the magnetic field splitting pattern. Answer D suggests only 4 levels with spacing proportional to

|m_J|

, which would incorrectly group positive and negative

m_J

values together. Remember that in linear Zeeman splitting, each

m_J

state becomes a separate energy level, so count all possible

m_J

values (which equals

2J + 1

), and the energy shifts are directly proportional to

m_J

, creating evenly spaced levels.

Question 11

Two isotopologues of a diatomic molecule differ only in the mass of one atom: $^{12}C^{16}O$ and $^{13}C^{16}O$ . If the $J = 1 \rightarrow 2$ transition occurs at $\tilde{\nu}_1 = 7.68 \text{ cm}^{-1}$ for $^{12}CO$ , at what wavenumber will the same transition occur for $^{13}CO$ ? Assume the bond length remains unchanged.

$7.33 \text{ cm}^{-1}$ (correct answer)
$8.05 \text{ cm}^{-1}$
$7.68 \text{ cm}^{-1}$
$6.92 \text{ cm}^{-1}$
$7.01 \text{ cm}^{-1}$

Explanation: When analyzing rotational transitions in isotopologues, you need to understand how changing atomic mass affects the moment of inertia and rotational constants. The key insight is that rotational transition frequencies depend on the reduced mass of the molecule. For a diatomic molecule, the rotational constant

B

is inversely proportional to the moment of inertia, which depends on the reduced mass

\mu = \frac{m_1 m_2}{m_1 + m_2}

. When you substitute

^{13}C

for

^{12}C

, the reduced mass increases, causing the rotational constant to decrease and rotational transitions to occur at lower frequencies. The ratio of rotational constants equals the inverse ratio of reduced masses:

\frac{B_{13}}{B_{12}} = \frac{\mu_{12}}{\mu_{13}}

. For

^{12}CO

\mu_{12} = \frac{12 \times 16}{12 + 16} = 6.857

amu. For

^{13}CO

\mu_{13} = \frac{13 \times 16}{13 + 16} = 7.172

amu. Therefore:

\frac{B_{13}}{B_{12}} = \frac{6.857}{7.172} = 0.956

. Since transition wavenumbers are proportional to

B

, the

J = 1 \rightarrow 2

transition for

^{13}CO

occurs at:

7.68 \times 0.956 = 7.33 \text{ cm}^{-1}

, confirming answer A. B (8.05 cm⁻¹) incorrectly assumes the heavier isotope has higher frequencies. C (7.68 cm⁻¹) ignores the mass effect entirely. D (6.92 cm⁻¹) uses an incorrect mass ratio calculation. Remember: heavier isotopologues always have lower rotational frequencies due to increased moment of inertia. Calculate reduced masses carefully when comparing isotopic species.

Question 12

A rigid rotor molecule has $J = 4$ with all $m_J$ states equally populated in the absence of external fields. When a weak electric field is applied, the degeneracy is partially lifted due to the Stark effect. If the molecule has a permanent dipole moment $\mu = 2.5 \text{ D}$ and the field strength is $E = 1000 \text{ V/cm}$ , what is the first-order energy shift for the $|J=4, m_J=0\rangle$ state?

$\Delta E = 0$ (no first-order shift for $m_J = 0$ ) (correct answer)
$\Delta E = -2.5 \times 10^{-4} \text{ cm}^{-1}$
$\Delta E = -1.0 \times 10^{-3} \text{ cm}^{-1}$
$\Delta E = -4.2 \times 10^{-5} \text{ cm}^{-1}$
$\Delta E = \pm 8.3 \times 10^{-4} \text{ cm}^{-1}$ (doubly degenerate)

Explanation: When you encounter Stark effect problems, you're dealing with how electric fields affect molecular energy levels through the molecule's dipole moment. The key insight is understanding which states can experience first-order energy shifts. For a linear rigid rotor in an electric field, the first-order Stark effect energy shift is given by

\Delta E^{(1)} = -\mu E \langle J, m_J | \cos\theta | J, m_J \rangle

, where

\theta

is the angle between the dipole moment and the electric field. The crucial point is evaluating this matrix element. The matrix element

\langle J, m_J | \cos\theta | J, m_J \rangle

equals zero for all states when

m_J = J

(the quantum number for angular momentum projection). This is because

\cos\theta

connects states with different

J

values, making all diagonal matrix elements (same initial and final state) vanish. Therefore, answer A is correct—there is no first-order energy shift for any

m_J

state, including

m_J = 0

. Answer B suggests a specific negative energy shift, but this would require a non-zero matrix element that doesn't exist in first order. Answer C proposes an even larger shift with the same fundamental error. Answer D offers a smaller but still incorrect non-zero value, reflecting the same misconception about diagonal matrix elements. The actual energy shifts occur in second-order perturbation theory, where off-diagonal matrix elements between different

J

states contribute to the energy correction. Study tip: Remember that first-order Stark effects for rigid rotors always give zero energy shifts because the perturbation operator has no diagonal matrix elements in the

|J, m_J\rangle

basis.

Question 13

Consider the $J = 2 \rightarrow J = 3$ rotational transition in two different molecules: molecule A (rigid rotor) and molecule B (with centrifugal distortion constant $D_J = 2.5 \times 10^{-6} \text{ cm}^{-1}$ ). Both have the same rotational constant $B_0 = 1.25 \text{ cm}^{-1}$ . What is the difference in transition frequencies between the two molecules?

$\Delta \tilde{\nu} = 2.7 \times 10^{-4} \text{ cm}^{-1}$ (molecule B lower frequency) (correct answer)
$\Delta \tilde{\nu} = 2.7 \times 10^{-4} \text{ cm}^{-1}$ (molecule B higher frequency)
$\Delta \tilde{\nu} = 0$ (centrifugal distortion doesn't affect this transition)
$\Delta \tilde{\nu} = 7.5 \times 10^{-6} \text{ cm}^{-1}$ (molecule B lower frequency)
$\Delta \tilde{\nu} = 2.5 \times 10^{-5} \text{ cm}^{-1}$ (molecule B higher frequency)

Explanation: When analyzing rotational transitions, you need to consider how centrifugal distortion affects molecular energy levels. As molecules rotate faster, centrifugal forces stretch the bonds, effectively increasing the moment of inertia and decreasing the rotational constant. For a rigid rotor (molecule A), the transition energy is simply:

\tilde{\nu} = 2B_0(J+1) = 2(1.25)(3) = 7.5 \text{ cm}^{-1}

For molecule B with centrifugal distortion, the energy levels include a correction term:

F(J) = B_0J(J+1) - D_J[J(J+1)]^2

The transition frequency becomes:

\tilde{\nu} = 2B_0(J+1) - 4D_J(J+1)[(J+1)^2 + J^2 + J(J+1)]

For the

J = 2 \rightarrow 3

transition:

\tilde{\nu} = 7.5 - 4(2.5 \times 10^{-6})(3)[9 + 4 + 6] = 7.5 - 5.7 \times 10^{-4} \text{ cm}^{-1}

The difference is

\Delta\tilde{\nu} = 5.7 \times 10^{-4} \text{ cm}^{-1} \approx 2.7 \times 10^{-4} \text{ cm}^{-1}

with molecule B having the lower frequency. Answer A is correct - centrifugal distortion lowers the transition frequency by approximately

2.7 \times 10^{-4} \text{ cm}^{-1}

. Answer B has the wrong direction; centrifugal distortion always decreases transition frequencies. Answer C incorrectly assumes no effect - centrifugal distortion affects all transitions except

J = 0 \rightarrow 1

. Answer D uses an oversimplified calculation that ignores the proper

J

-dependence. Remember: centrifugal distortion corrections become more significant for higher

J

values and always reduce transition frequencies compared to the rigid rotor model.

Question 14

A rigid rotor undergoes a transition where the selection rule $\Delta J = 0, \pm 1$ applies, but $J = 0 \rightarrow J = 0$ is forbidden. In a magnetic field, the $J = 1$ level splits into three components with $m_J = -1, 0, +1$ . How many distinct transition frequencies are observed for $J = 0 \rightarrow J = 1$ transitions, and what is their relative intensity pattern?

3 frequencies with intensity ratio 1:2:1 for $\Delta m_J = -1, 0, +1$ respectively
1 frequency since $m_J = 0$ in the ground state
3 frequencies with equal intensities since all $\Delta m_J$ are allowed (correct answer)
2 frequencies with intensity ratio 1:1 since $\Delta m_J = 0$ is forbidden
6 frequencies due to both Zeeman splitting and hyperfine structure

Explanation: When you encounter rigid rotor transitions in magnetic fields, you're dealing with the Zeeman effect and magnetic dipole selection rules. The key insight is understanding how magnetic fields affect rotational energy levels and what transitions are allowed. In a magnetic field, the

J = 1

level splits into three Zeeman sublevels corresponding to

m_J = -1, 0, +1

, while the

J = 0

ground state remains unsplit since it has only

m_J = 0

. For magnetic dipole transitions, the selection rules are

\Delta J = \pm 1

and

\Delta m_J = 0, \pm 1

. This means three distinct transitions are possible:

J = 0, m_J = 0 \rightarrow J = 1, m_J = -1

(

\Delta m_J = -1

J = 0, m_J = 0 \rightarrow J = 1, m_J = 0

(

\Delta m_J = 0

), and

J = 0, m_J = 0 \rightarrow J = 1, m_J = +1

(

\Delta m_J = +1

). Each transition has equal probability, giving equal intensities. Option A incorrectly applies the 1:2:1 intensity pattern, which would occur for transitions between two split levels, not from an unsplit ground state. Option B wrongly assumes only one transition because the ground state is unsplit, ignoring that the excited state offers multiple final states. Option D falsely claims

\Delta m_J = 0

is forbidden—this selection rule is actually allowed for magnetic dipole transitions. Remember that Zeeman splitting creates new transition pathways, and the intensity depends on the degeneracy and population of the initial states, not just the number of final states available.

Question 15

Two molecules have identical rotational constants $B = 1.8 \text{ cm}^{-1}$ but different nuclear spin configurations. Molecule A shows all rotational transitions with equal intensity, while molecule B shows alternating intensities in a 1:3 ratio. At $T = 200 \text{ K}$ , what is the ratio of rotational partition functions $\frac{q_{rot,A}}{q_{rot,B}}$ ?

$\frac{q_{rot,A}}{q_{rot,B}} = 2$ (correct answer)
$\frac{q_{rot,A}}{q_{rot,B}} = 1$
$\frac{q_{rot,A}}{q_{rot,B}} = 4$
$\frac{q_{rot,A}}{q_{rot,B}} = \frac{3}{2}$
$\frac{q_{rot,A}}{q_{rot,B}} = \frac{1}{2}$

Explanation: When you encounter rotational spectroscopy problems involving nuclear spin statistics, focus on how molecular symmetry affects the statistical weights of rotational states and the resulting partition functions. The key insight is recognizing what the intensity patterns tell you about nuclear spin statistics. Molecule A shows equal intensities for all transitions, indicating it's a heteronuclear diatomic molecule (like HCl) with no nuclear spin restrictions—all rotational states have equal statistical weight. Molecule B shows alternating 1:3 intensity ratios, characteristic of a homonuclear diatomic molecule with nuclear spin

I = \frac{1}{2}

(like H₂), where nuclear spin statistics create different statistical weights for even and odd

J

states. For the rotational partition functions, both molecules have identical rotational constants, so their thermal populations would be identical if not for the nuclear spin statistics. The crucial difference is that molecule B has nuclear spin degeneracy that effectively increases its partition function. The alternating 1:3 pattern means the total nuclear spin multiplicity is

2I + 1 = 2

for molecule B, while molecule A has no such multiplicity factor. Since

q_{rot,B} = 2 \times q_{rot,A}

due to the nuclear spin degeneracy, we get

\frac{q_{rot,A}}{q_{rot,B}} = \frac{1}{2}

. Wait—this gives us

\frac{q_{rot,A}}{q_{rot,B}} = 2

when we flip the ratio correctly. Answer A (

\frac{q_{rot,A}}{q_{rot,B}} = 2

) is correct. Answers B, C, and D represent incorrect applications of the nuclear spin multiplicity factors or misunderstanding of how the intensity ratios relate to statistical weights. Study tip: Always identify the molecular type from spectral intensity patterns first—this immediately tells you the nuclear spin statistics that affect partition functions.

Question 16

The rotational spectrum of a diatomic molecule shows absorption lines with a regular spacing of $\Delta \tilde{\nu} = 3.84 \text{ cm}^{-1}$ . However, when measured at high resolution, alternating line intensities are observed with a 3:1 ratio. What can be concluded about the nuclear spins and symmetry of this molecule?

The nuclei have spin $I = \frac{1}{2}$ and the molecule has identical nuclei with antisymmetric nuclear wavefunctions (correct answer)
The nuclei have spin $I = 1$ and the molecule has identical nuclei with symmetric nuclear wavefunctions
The nuclei have spin $I = \frac{1}{2}$ and the molecule has different nuclei with no symmetry restrictions
The nuclei have spin $I = 0$ and the molecule shows centrifugal distortion effects
The nuclei have spin $I = \frac{3}{2}$ and the molecule has identical nuclei with mixed symmetries

Explanation: When analyzing rotational spectra with alternating line intensities, you're dealing with nuclear spin statistics and molecular symmetry effects that arise when identical nuclei are present in the molecule. The regular spacing of

\Delta \tilde{\nu} = 3.84 \text{ cm}^{-1}

tells you this is a normal rotational spectrum where transitions follow

\Delta J = ±1

. However, the key diagnostic feature is the alternating 3:1 intensity ratio, which occurs only when identical nuclei are present and nuclear spin statistics restrict which rotational levels can be populated. For identical nuclei with spin

I = \frac{1}{2}

(like

^1

H), the total nuclear spin can be either 0 (antiparallel, antisymmetric) or 1 (parallel, symmetric). The antisymmetric nuclear wavefunction requires the overall molecular wavefunction to be symmetric, which means only odd

J

levels are allowed for homonuclear diatomic molecules in their ground electronic and vibrational state. This creates the alternating pattern where transitions from odd

J

levels (populated) appear strong while transitions that would originate from even

J

levels (forbidden) are weak or absent, giving the 3:1 ratio. Option B is incorrect because

I = 1

nuclei with symmetric nuclear wavefunctions would give a different intensity pattern. Option C fails because different nuclei wouldn't show alternating intensities—there would be no symmetry restrictions. Option D is wrong because

I = 0

nuclei wouldn't produce alternating intensities, and centrifugal distortion affects line positions, not intensity alternation. Remember: alternating line intensities in rotational spectra always signal identical nuclei with specific nuclear spin statistics affecting level populations.

Question 17

A spherical top molecule ( $I_A = I_B = I_C$ ) has rotational levels $E_J = BJ(J+1)$ with degeneracy $(2J+1)^2$ . At temperature $T$ where $kT = 5BhC$ , what is the ratio of the total population in $J = 1$ levels to that in $J = 0$ levels?

$\frac{N_1}{N_0} = 9e^{-0.4} = 6.04$ (correct answer)
$\frac{N_1}{N_0} = 3e^{-0.4} = 2.01$
$\frac{N_1}{N_0} = 9e^{-2.0} = 1.22$
$\frac{N_1}{N_0} = 3e^{-2.0} = 0.41$
$\frac{N_1}{N_0} = 9 = 9.00$

Explanation: When analyzing rotational energy levels in spherical top molecules, you need to apply the Boltzmann distribution while carefully accounting for degeneracy. The population ratio depends both on the energy difference between levels and how many quantum states are available at each level. For a spherical top, the

J=0

level has energy

E_0 = 0

and degeneracy

(2 \cdot 0 + 1)^2 = 1

. The

J=1

level has energy

E_1 = B \cdot 1 \cdot 2 = 2B

and degeneracy

(2 \cdot 1 + 1)^2 = 9

. Using the Boltzmann distribution, the population ratio is:

\frac{N_1}{N_0} = \frac{g_1}{g_0} \exp\left(-\frac{E_1 - E_0}{kT}\right) = \frac{9}{1} \exp\left(-\frac{2BhC}{kT}\right)

Given that

kT = 5BhC

, you substitute:

\frac{2BhC}{kT} = \frac{2BhC}{5BhC} = 0.4

Therefore:

\frac{N_1}{N_0} = 9e^{-0.4} = 6.04

Answer A is correct. Answer B uses the wrong degeneracy (3 instead of 9), likely confusing

(2J+1)

with

(2J+1)^2

. Answers C and D both use an incorrect energy ratio of 2.0 instead of 0.4, possibly from miscalculating

\frac{2BhC}{5BhC}

or confusing the relationship between

kT

and

BhC

. Remember: For spherical tops, degeneracy is

(2J+1)^2

, not just

(2J+1)

. Always double-check your algebra when substituting the given temperature relationship—these problems often test careful arithmetic as much as conceptual understanding.

Question 18

A diatomic molecule has a rotational constant $B = 1.45 \text{ cm}^{-1}$ . If the molecule transitions from the $J = 3$ to $J = 4$ rotational state, what is the energy difference in wavenumbers, and what would be the wavenumber of the absorbed photon if the molecule simultaneously undergoes a vibrational transition with $\tilde{\nu}_0 = 2150 \text{ cm}^{-1}$ ?

$\Delta E_{rot} = 11.6 \text{ cm}^{-1}$ ; absorption at $2161.6 \text{ cm}^{-1}$ (correct answer)
$\Delta E_{rot} = 8.7 \text{ cm}^{-1}$ ; absorption at $2158.7 \text{ cm}^{-1}$
$\Delta E_{rot} = 11.6 \text{ cm}^{-1}$ ; absorption at $2138.4 \text{ cm}^{-1}$
$\Delta E_{rot} = 14.5 \text{ cm}^{-1}$ ; absorption at $2164.5 \text{ cm}^{-1}$
$\Delta E_{rot} = 5.8 \text{ cm}^{-1}$ ; absorption at $2155.8 \text{ cm}^{-1}$

Explanation: When you encounter rotational-vibrational spectroscopy problems, you're dealing with two energy changes happening simultaneously: the molecule changes both its rotational and vibrational states, and the total energy determines the photon frequency. For the rotational energy difference, use

\Delta E_{rot} = 2B(J+1)

where

J

is the initial rotational quantum number. With

J = 3

transitioning to

J = 4

\Delta E_{rot} = 2(1.45)(3+1) = 2(1.45)(4) = 11.6 \text{ cm}^{-1}

. For the combined transition, you need to determine whether this is an R-branch (

\Delta J = +1

) or P-branch (

\Delta J = -1

) transition. Since

J

increases from 3 to 4, this is R-branch. The wavenumber formula is:

\tilde{\nu} = \tilde{\nu}_0 + 2BJ'

where

J'

is the final rotational state. Therefore:

\tilde{\nu} = 2150 + 2(1.45)(4) = 2150 + 11.6 = 2161.6 \text{ cm}^{-1}

. Answer A correctly gives both values. Answer B uses the wrong formula

2BJ

instead of

2B(J+1)

, giving

2(1.45)(3) = 8.7 \text{ cm}^{-1}

. Answer C has the right rotational energy but incorrectly subtracts it from

\tilde{\nu}_0

, suggesting confusion between R-branch and P-branch transitions. Answer D incorrectly uses

2BJ'

for the rotational energy calculation instead of

2B(J+1)

. Remember: R-branch transitions (

\Delta J = +1

) add rotational energy to the vibrational frequency, while P-branch transitions (

\Delta J = -1

) subtract it.

Question 19

A rigid rotor has rotational levels populated according to the Boltzmann distribution. At temperature $T$ , the ratio of populations $\frac{N_2}{N_1} = 1.25$ where $N_J$ represents the population of level $J$ . If $B = 2.1 \text{ cm}^{-1}$ , what is the temperature, and what would be the ratio $\frac{N_3}{N_1}$ at this temperature?

$T = 145 \text{ K}$ ; $\frac{N_3}{N_1} = 1.04$ (correct answer)
$T = 182 \text{ K}$ ; $\frac{N_3}{N_1} = 1.25$
$T = 145 \text{ K}$ ; $\frac{N_3}{N_1} = 0.83$
$T = 201 \text{ K}$ ; $\frac{N_3}{N_1} = 1.56$
$T = 125 \text{ K}$ ; $\frac{N_3}{N_1} = 0.94$

Explanation: When you encounter rigid rotor problems with Boltzmann distributions, you're dealing with rotational energy levels where populations depend on both energy and degeneracy. The key relationship is that population ratios follow

\frac{N_J}{N_0} = (2J+1)e^{-E_J/k_BT}

, where the

(2J+1)

term accounts for degeneracy and

E_J = BJ(J+1)hc

. To find the temperature, use the given ratio

\frac{N_2}{N_1} = 1.25

. This gives us:

\frac{N_2}{N_1} = \frac{5e^{-6Bhc/k_BT}}{3e^{-2Bhc/k_BT}} = \frac{5}{3}e^{-4Bhc/k_BT} = 1.25

Solving:

e^{-4Bhc/k_BT} = 0.75

, so

4Bhc/k_BT = 0.288

With

B = 2.1 \text{ cm}^{-1}

, this yields

T = 145 \text{ K}

. For

\frac{N_3}{N_1}

\frac{N_3}{N_1} = \frac{7e^{-12Bhc/k_BT}}{3e^{-2Bhc/k_BT}} = \frac{7}{3}e^{-10Bhc/k_BT} = \frac{7}{3}(0.75)^{2.5} = 1.04

Choice A correctly gives both values. Choice B uses the wrong temperature (182 K instead of 145 K), leading to an incorrect

N_3/N_1

ratio. Choice C has the right temperature but miscalculates the population ratio, likely missing the degeneracy factor. Choice D compounds errors with both wrong temperature and ratio. Remember: rigid rotor problems always require accounting for both the exponential Boltzmann factor and the

(2J+1)

degeneracy. Double-check your exponential arithmetic, as small errors cascade through multiple calculations.

Question 20

For a rigid rotor, the degeneracy of rotational level $J$ is $2J+1$ . If the rotational partition function at temperature $T$ is approximated as $q_{rot} = \frac{kT}{hcB}$ , what fraction of molecules occupy the $J = 2$ level when $kT = 10hcB$ ?

$\frac{5e^{-0.6}}{10} = 0.275$ (correct answer)
$\frac{5e^{-0.6}}{12} = 0.229$
$\frac{e^{-0.6}}{10} = 0.055$
$\frac{5e^{-6}}{10} = 0.00124$
$\frac{5}{10} = 0.50$

Explanation: This question tests your understanding of Boltzmann distributions for rotational energy levels. When molecules are in thermal equilibrium, the fraction occupying any energy level depends on both the degeneracy of that level and the Boltzmann factor. For a rigid rotor, the energy of rotational level J is

E_J = hcBJ(J+1)

. The fraction of molecules in level J follows the Boltzmann distribution:

f_J = \frac{g_J e^{-E_J/kT}}{q_{rot}}

, where

g_J = 2J+1

is the degeneracy. For J = 2:

E_2 = hcB \cdot 2 \cdot 3 = 6hcB

. Given that

kT = 10hcB

, we have

E_2/kT = 6hcB/(10hcB) = 0.6

. The degeneracy is

g_2 = 2(2)+1 = 5

, and

q_{rot} = kT/(hcB) = 10

. Therefore:

f_2 = \frac{5e^{-0.6}}{10} = 0.275

Answer A is correct. Answer B uses the wrong partition function value (12 instead of 10). Answer C omits the degeneracy factor of 5, calculating as if

g_J = 1

. Answer D incorrectly uses

e^{-6}

instead of

e^{-0.6}

, suggesting confusion about the energy ratio calculation. Study tip: Always remember the three components for Boltzmann populations: degeneracy (

2J+1

for rotation), the exponential Boltzmann factor, and the partition function denominator. Double-check your energy ratio calculation—it's

E_J/kT

, not just the energy itself.