University Physics IV · topic UPIV-12.04
Vibrational and rotation-vibration spectra
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.
Vibrational and rotation-vibration spectra: predict a finite change
A vibrational spectrum is the set of transitions between the quantised stretching levels of a bond, E_v = (v+½)ħω; in a gas of Σ-state diatomics such as HCl each one must be accompanied by a rotational change, so it appears as a band of lines rather than a single line. Use the supplied topic model Y_out = (sqrt(max(t,1)/(2*1.4388*max(b0,0.1)))-0.5), where Y_out is BRIGHTEST J″ (before display rounding) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value. b0 is Rotational constant B₀ (v = 0), measured in cm⁻¹. Keep d (Bond stretch B₀ − B₁) = 0.3 cm⁻¹; t (Temperature T) = 300 K fixed. For this exercise use only 5.74375 cm⁻¹ ≤ b0 ≤ 11.2562 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 b0 = 6.9565 cm⁻¹ and 7.72825 cm⁻¹. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δb0 across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
BRIGHTEST J″ (before display rounding) against b0
Predictions from the supplied Vibrational and rotation-vibration spectra 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): BRIGHTEST J″ (before display rounding)
Horizontal axis (x): b0: Rotational constant B₀ (v = 0) / cm⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b0: Rotational constant B₀ (v = 0) / cm⁻¹ | BRIGHTEST J″ (before display rounding) |
|---|---|
| 5.74375 | 3.760373263861288 |
| 7.121875 | 3.326028859184726 |
| 8.5 | 3.002159747431779 |
| 9.878125 | 2.748689136681243 |
| 11.25625 | 2.5433265154180007 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of b0 separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At b0 = 6.9565 cm⁻¹, substitute into Y_out = (sqrt(max(t,1)/(2*1.4388*max(b0,0.1)))-0.5): Y₀ = 3.37124.
- At b0 = 7.72825 cm⁻¹, the same fixed-parameter model gives Y₁ = 3.17286.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.198375; the input change is Δb0 = 0.77175 cm⁻¹.
- Average sensitivity is ΔY_out/Δb0 = -0.257046 1 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 = 3.37124, Y₁ = 3.17286; ΔY_out = -0.198375; average sensitivity = -0.257046 1 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.
Vibrational and rotation-vibration spectra: infer an input with uncertainty
A vibrational spectrum is the set of transitions between the quantised stretching levels of a bond, E_v = (v+½)ħω; in a gas of Σ-state diatomics such as HCl each one must be accompanied by a rotational change, so it appears as a band of lines rather than a single line. Use the supplied topic model Y_out = (sqrt(max(t,1)/(2*1.4388*max(b0,0.1)))-0.5), where Y_out is BRIGHTEST J″ (before display rounding) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value. b0 is Rotational constant B₀ (v = 0), measured in cm⁻¹. Keep d (Bond stretch B₀ − B₁) = 0.3 cm⁻¹; t (Temperature T) = 300 K fixed. For this exercise use only 5.74375 cm⁻¹ ≤ b0 ≤ 11.2562 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 = 2.8835504, with a bounded reading uncertainty ±0.051034. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown b0 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00055125 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.
BRIGHTEST J″ (before display rounding) against b0
Predictions from the supplied Vibrational and rotation-vibration spectra 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): BRIGHTEST J″ (before display rounding)
Horizontal axis (x): b0: Rotational constant B₀ (v = 0) / cm⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b0: Rotational constant B₀ (v = 0) / cm⁻¹ | BRIGHTEST J″ (before display rounding) |
|---|---|
| 5.74375 | 3.760373263861288 |
| 7.121875 | 3.326028859184726 |
| 8.5 | 3.002159747431779 |
| 9.878125 | 2.748689136681243 |
| 11.25625 | 2.5433265154180007 |
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(b0) = 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(5.74375) = 3.76037 and Y_out(11.2562) = 2.54333. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives b0_line = 5.74375 + (2.88355 − 3.76037)(11.2562 − 5.74375)/(2.54333 − 3.76037) = 9.71524 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 b0 = 9.10637 cm⁻¹; substituting back gives Y_out = 2.8835504.
- The interpolation error in the input is |b0_line − b0| = 0.608863 cm⁻¹. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 2.8325164 and 2.9345844. Sort the roots even if the response decreases: 8.83776 cm⁻¹ ≤ b0 ≤ 9.38742 cm⁻¹.
- The lower and upper input deviations are 0.26861 cm⁻¹ and 0.281045 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 b0 ≈ 9.10637 cm⁻¹; compatible interval [8.83776, 9.38742] cm⁻¹. Endpoint interpolation gives 9.71524 cm⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.