Rotational Energy Levels - Physical Chemistry 2

1. 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?
   Answer: 7 levels with equal spacing proportional to $m_J$ 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.
2. 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?
   Answer: $\Delta E = 0$ (no first-order shift for $m_J = 0$) 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.
3. 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 $\tilde{\nu}_0 = 2150 \text{ cm}^{-1}$?
   Answer: $\Delta E_{rot} = 11.6 \text{ cm}^{-1}$; absorption at $2161.6 \text{ cm}^{-1}$ 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.
4. A linear triatomic molecule ABC has moments of inertia $I_A = 15.2 \times 10^{-47} \text{ kg}\cdot\text{m}^2$ 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?
   Answer: $\lambda = 1.18 \text{ mm}$ 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.
5. A linear triatomic molecule XYX has two equivalent bonds. When analyzing its rotational spectrum, both the fundamental vibrational frequency $\omega_e = 1200 \text{ cm}^{-1}$ and the rotational constant $B_v = 0.95 \text{ cm}^{-1}$ are observed to depend on the vibrational state $v$. If $\alpha_e = 0.012 \text{ cm}^{-1}$ describes the vibration-rotation coupling, what is the effective rotational constant for the $v = 2$ vibrational level?
   Answer: $B_2 = 0.926 \text{ cm}^{-1}$ When you encounter vibration-rotation coupling in molecular spectroscopy, you're dealing with how molecular vibrations affect the rotational constant. As molecules vibrate, bond lengths change, which alters the moment of inertia and thus the rotational constant. The effective rotational constant for vibrational level $v$ follows the relationship: $B_v = B_e - \alpha_e(v + \frac{1}{2})$ where $B_e$ is the equilibrium rotational constant and $\alpha_e$ is the vibration-rotation coupling constant. Since we're given $B_v = 0.95 \text{ cm}^{-1}$ (which appears to be $B_0$ for $v = 0$), we can find $B_e$: $0.95 = B_e - 0.012(0 + 0.5)$ $B_e = 0.95 + 0.006 = 0.956 \text{ cm}^{-1}$ Now for $v = 2$: $B_2 = 0.956 - 0.012(2 + 0.5) = 0.956 - 0.012(2.5) = 0.956 - 0.030 = 0.926 \text{ cm}^{-1}$ Answer A ($0.926 \text{ cm}^{-1}$) is correct. Answer B ($0.974 \text{ cm}^{-1}$) incorrectly adds the coupling correction instead of subtracting it. Answer C ($0.950 \text{ cm}^{-1}$) likely assumes no vibration-rotation coupling. Answer D ($0.902 \text{ cm}^{-1}$) appears to use an incorrect value of $v$ or misapply the coupling constant. Remember: vibration-rotation coupling almost always decreases the rotational constant with increasing vibrational quantum number because molecular vibration increases average bond lengths, increasing the moment of inertia.
6. 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?
   Answer: $T = 145 \text{ K}$; $\frac{N_3}{N_1} = 1.04$ 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.
7. 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?
   Answer: The nuclei have spin $I = \frac{1}{2}$ and the molecule has identical nuclei with antisymmetric nuclear wavefunctions 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.
8. A rigid rotor molecule undergoes a transition from $J = 5$ to $J = 6$. If this transition has a wavenumber of 36.0 $\text{cm}^{-1}$, what would be the wavenumber of the $J = 0 \rightarrow J = 1$ transition?
   Answer: $6.0 \text{ cm}^{-1}$ When you encounter rigid rotor spectroscopy problems, you're dealing with the quantum mechanical description of molecular rotation. The key insight is that rotational energy levels follow a specific pattern that makes transitions predictable. For a rigid rotor, the rotational energy levels are given by $E_J = BJ(J+1)$, where $B$ is the rotational constant and $J$ is the rotational quantum number. The wavenumber for any transition from $J$ to $J+1$ is $\tilde{\nu} = 2B(J+1)$. Given that the $J = 5 \rightarrow J = 6$ transition has a wavenumber of 36.0 cm⁻¹, you can write: $36.0 = 2B(6) = 12B$. Solving for $B$: $B = 3.0 \text{ cm}^{-1}$. For the $J = 0 \rightarrow J = 1$ transition: $\tilde{\nu} = 2B(1) = 2(3.0) = 6.0 \text{ cm}^{-1}$. This confirms answer D is correct. Looking at the wrong answers: A (18.0 cm⁻¹) represents exactly half of the given transition wavenumber, which might tempt you if you incorrectly assume a simple inverse relationship with $J$ values. B (3.0 cm⁻¹) is the rotational constant $B$ itself, not the transition wavenumber. C (12.0 cm⁻¹) equals $12B$, which you might get if you confuse the energy level formula with the transition formula. Remember that in rigid rotor spectroscopy, once you determine $B$ from any transition, you can calculate any other transition using $\tilde{\nu} = 2B(J+1)$. The rotational constant is the fundamental parameter that governs all rotational transitions.
9. Rigid rotor, B=10 $cm^{-1}$ at 100 K. Find N(J=1)/N(J=0).
   Answer: 2.25 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.
10. 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?
    Answer: $B = 10.2 \text{ cm}^{-1}$; $J = 5 \rightarrow 6$ at $122.4 \text{ cm}^{-1}$ 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 $2B$.
11. A diatomic molecule shows rotational absorption lines at wavenumbers following the pattern $\tilde{\nu} = 2B(J+1)$ where $J$ is the lower rotational quantum number. If the $J = 4 \rightarrow J = 5$ transition occurs at 30.0 $\text{cm}^{-1}$, at what wavenumber will the $J = 1 \rightarrow J = 2$ transition appear?
    Answer: $12.0 \text{ cm}^{-1}$ For $J = 4 \rightarrow J = 5$: $\tilde{\nu} = 2B(4+1) = 10B = 30.0 \text{ cm}^{-1}$, so $B = 3.0 \text{ cm}^{-1}$. For $J = 1 \rightarrow J = 2$: $\tilde{\nu} = 2B(1+1) = 4B = 4(3.0) = 12.0 \text{ cm}^{-1}$. Choice A uses $2B$. Choice B uses $3B$. Choice D uses $5B$.
12. Consider a rigid rotor where the energy levels are given by $E_J = BJ(J+1)$. If the spacing between consecutive absorption lines in the microwave spectrum is $\Delta\tilde{\nu} = 4.2 \text{ cm}^{-1}$, what is the bond length of the diatomic molecule? (Assume equal masses $m = 16 \text{ amu}$, $1 \text{ amu} = 1.66 \times 10^{-27} \text{ kg}$)
    Answer: $2.05 \times 10^{-10} \text{ m}$ Line spacing gives $2B = 4.2 \text{ cm}^{-1}$, so $B = 2.1 \text{ cm}^{-1}$. For equal masses, the reduced mass is $\mu = \frac{m}{2} = 8 \times 1.66 \times 10^{-27} = 1.33 \times 10^{-26} \text{ kg}$. The moment of inertia is $I = \mu r^2$. Using $B = \frac{h}{8\pi^2Ic}$: $I = \frac{6.626 \times 10^{-34}}{8\pi^2 \times 2.1 \times 3.00 \times 10^{10}} = 1.33 \times 10^{-46} \text{ kg}\cdot\text{m}^2$. Then $r = \sqrt{\frac{I}{\mu}} = 2.05 \times 10^{-10} \text{ m}$. Other choices use incorrect moment of inertia expressions or reduced mass calculations.
13. A rigid rotor has I=1.45e-46 kg $m^2$. Find the wavelength of the J=0 to 1 line.
    Answer: 2.59 mm 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).
14. For a rigid rotor with rotational constant $B = 1.8 \text{ cm}^{-1}$, what is the frequency of the $J = 2 \rightarrow J = 3$ transition in Hz?
    Answer: $3.24 \times 10^{11} \text{ Hz}$ For $J = 2 \rightarrow J = 3$: $\Delta E = B[3(4) - 2(3)] = 6B = 6(1.8) = 10.8 \text{ cm}^{-1}$. Converting to frequency: $\nu = \Delta E \times c = 10.8 \times 3.00 \times 10^{10} = 3.24 \times 10^{11} \text{ Hz}$. Choice B uses $\Delta E = 4B$. Choice C uses $\Delta E = 2B$. Choice D uses $\Delta E = B$.
15. 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?
    Answer: 3 frequencies with equal intensities since all $\Delta m_J$ are allowed 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.
16. 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.
    Answer: $7.33 \text{ cm}^{-1}$ 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.
17. A rigid rotor has rotational levels with degeneracies $g_J = 2J + 1$. At what temperature will the populations of the $J = 1$ and $J = 2$ levels be equal, given that $B = 5.0 \text{ cm}^{-1}$? ($k_B = 0.695 \text{ cm}^{-1}\text{K}^{-1}$)
    Answer: $43.2 \text{ K}$ When you encounter rigid rotor population problems, you're dealing with the Boltzmann distribution where both energy levels and degeneracies matter. The population of each rotational level depends on both the statistical weight (degeneracy) and the exponential energy factor. For a rigid rotor, the energy levels are $E_J = BJ(J+1)hc$ and degeneracies are $g_J = 2J + 1$. The population of level $J$ follows: $N_J \propto g_J e^{-E_J/k_BT}$. Setting populations equal: $N_1 = N_2$ $g_1 e^{-E_1/k_BT} = g_2 e^{-E_2/k_BT}$ Substituting values: $g_1 = 3$, $g_2 = 5$, $E_1 = 2Bhc$, $E_2 = 6Bhc$ $3e^{-2Bhc/k_BT} = 5e^{-6Bhc/k_BT}$ Rearranging: $\frac{3}{5} = e^{-(6-2)Bhc/k_BT} = e^{-4Bhc/k_BT}$ Taking natural logarithm: $\ln(3/5) = -4Bhc/k_BT$ Since $Bhc = 5.0 \text{ cm}^{-1}$ in wavenumber units: $T = \frac{4 \times 5.0}{k_B \ln(5/3)} = \frac{20.0}{0.695 \times 0.511} = 43.2 \text{ K}$ This confirms answer C. A (64.8 K) likely comes from using $\ln(5/3)$ incorrectly or wrong energy differences. B (21.6 K) suggests ignoring degeneracies entirely and only considering energy differences. D (30.0 K) might result from computational errors in the logarithm or constants. Strategy tip: Always remember that population depends on both degeneracy and energy. Higher $J$ levels have both higher degeneracy (favoring population) and higher energy (disfavoring population) – the temperature determines which effect dominates.
18. 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?
    Answer: $B_B = 1.11 \text{ cm}^{-1}$; same average energy at all temperatures since $\langle E_{rot} \rangle = kT$ 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.
19. In the rotational spectrum of a diatomic molecule, transitions are observed at 8.0, 16.0, 24.0, and 32.0 $\text{cm}^{-1}$. What is the energy of the $J = 3$ rotational level relative to the ground state?
    Answer: $48.0 \text{ cm}^{-1}$ The equal spacing of 8.0 cm⁻¹ indicates $2B = 8.0$, so $B = 4.0 \text{ cm}^{-1}$. The energy of $J = 3$ is $E_3 = BJ(J+1) = 4.0 \times 3 \times 4 = 48.0 \text{ cm}^{-1}$. Choice A incorrectly uses the wavenumber of the $J = 2 \rightarrow J = 3$ transition. Choice C uses $E = BJ^2$. Choice D uses $E = BJ$.
20. For a linear rotor with B=2 $cm^{-1}$, what is the first rotational Raman Stokes shift?
    Answer: 12 $cm^{-1}$ 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.

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