University Physics IV · topic UPIV-12.03
Rotational spectra and the rigid rotor
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Further practice
Work through a fresh case before tackling the extension.
Rotational spectra and the rigid rotor: predict a finite change
A rigid rotor is a two-atom molecule modelled as two point masses held a fixed distance apart and free to tumble; its allowed energies are E_J = J(J+1)ħ²/2I, where I = μR² is the moment of inertia about the centre of mass. Use the supplied topic model Y_out = 156.81/sqrt(b), where Y_out is BOND LENGTH WITH CO MASSES (pm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. b is Rotational constant B̃, measured in cm⁻¹. Keep t (Temperature T) = 300 K; q (Distortion D̃/B̃) = 0 ×10⁻⁴ fixed. For this exercise use only 1.58688 cm⁻¹ ≤ b ≤ 3.71312 cm⁻¹ on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. Calculate Y_out at b = 2.05465 cm⁻¹ and 2.35232 cm⁻¹. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δb across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
BOND LENGTH WITH CO MASSES against b
Predictions from the supplied Rotational spectra and the rigid rotor model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): BOND LENGTH WITH CO MASSES / pm
Horizontal axis (x): b: Rotational constant B̃ / cm⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Rotational constant B̃ / cm⁻¹ | BOND LENGTH WITH CO MASSES / pm |
|---|---|
| 1.586875 | 124.48080715887716 |
| 2.1184374999999998 | 107.73726018713388 |
| 2.65 | 96.32761727073189 |
| 3.1815625 | 87.91308599143515 |
| 3.713125 | 81.37742995890225 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of b separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At b = 2.05465 cm⁻¹, substitute into Y_out = 156.81/sqrt(b): Y₀ = 109.397 pm.
- At b = 2.35232 cm⁻¹, the same fixed-parameter model gives Y₁ = 102.241 pm.
- Signed change: ΔY_out = Y₁ − Y₀ = -7.15585 pm; the input change is Δb = 0.297675 cm⁻¹.
- Average sensitivity is ΔY_out/Δb = -24.0391 pm per cm⁻¹. Its negative sign means that this response decreases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 109.397 pm, Y₁ = 102.241 pm; ΔY_out = -7.15585 pm; average sensitivity = -24.0391 pm per cm⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
Rotational spectra and the rigid rotor: infer an input with uncertainty
A rigid rotor is a two-atom molecule modelled as two point masses held a fixed distance apart and free to tumble; its allowed energies are E_J = J(J+1)ħ²/2I, where I = μR² is the moment of inertia about the centre of mass. Use the supplied topic model Y_out = 156.81/sqrt(b), where Y_out is BOND LENGTH WITH CO MASSES (pm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. b is Rotational constant B̃, measured in cm⁻¹. Keep t (Temperature T) = 300 K; q (Distortion D̃/B̃) = 0 ×10⁻⁴ fixed. For this exercise use only 1.58688 cm⁻¹ ≤ b ≤ 3.71312 cm⁻¹ on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 92.338879 pm, with a bounded reading uncertainty ±1.6442 pm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown b by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000212625 cm⁻¹, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
BOND LENGTH WITH CO MASSES against b
Predictions from the supplied Rotational spectra and the rigid rotor model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): BOND LENGTH WITH CO MASSES / pm
Horizontal axis (x): b: Rotational constant B̃ / cm⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Rotational constant B̃ / cm⁻¹ | BOND LENGTH WITH CO MASSES / pm |
|---|---|
| 1.586875 | 124.48080715887716 |
| 2.1184374999999998 | 107.73726018713388 |
| 2.65 | 96.32761727073189 |
| 3.1815625 | 87.91308599143515 |
| 3.713125 | 81.37742995890225 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(b) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(1.58688) = 124.481 pm and Y_out(3.71312) = 81.3774 pm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives b_line = 1.58688 + (92.3389 − 124.481)(3.71312 − 1.58688)/(81.3774 − 124.481) = 3.17241 cm⁻¹. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives b = 2.88389 cm⁻¹; substituting back gives Y_out = 92.338879 pm.
- The interpolation error in the input is |b_line − b| = 0.288519 cm⁻¹. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 90.694679 pm and 93.983079 pm. Sort the roots even if the response decreases: 2.78387 cm⁻¹ ≤ b ≤ 2.9894 cm⁻¹.
- The lower and upper input deviations are 0.100022 cm⁻¹ and 0.105512 cm⁻¹. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer b ≈ 2.88389 cm⁻¹; compatible interval [2.78387, 2.9894] cm⁻¹. Endpoint interpolation gives 3.17241 cm⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.