University Physics V · topic UPV-13.01
Emission, absorption, and term analysis of a measured spectrum
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.
Emission, absorption and term analysis: predict a finite change
Term analysis is the least-squares inversion that turns a list of measured line wavenumbers into a set of term values T_n = −E_n/hc, each line modelled as a difference σ = T_l − T_u of two eigenvalues of one Hermitian atomic Hamiltonian. Use the supplied topic model Y_out = t2+g, where Y_out is σ 1→3 = σ 1→2 + σ 2→3 (10³ cm⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. g is Gap from level 2 up to level 3, measured in 10³ cm⁻¹. Keep t2 (Term of level 2 above ground) = 15.4 10³ cm⁻¹; c (Zero offset c added to every level) = 0 10³ cm⁻¹ fixed. For this exercise use only 6.16875 10³ cm⁻¹ ≤ g ≤ 14.8312 10³ 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 g = 8.0745 10³ cm⁻¹ and 9.28725 10³ cm⁻¹. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δg across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
σ 1→3 = σ 1→2 + σ 2→3 against g
Predictions from the supplied Emission, absorption and term analysis 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): σ 1→3 = σ 1→2 + σ 2→3 / 10³ cm⁻¹
Horizontal axis (x): g: Gap from level 2 up to level 3 / 10³ cm⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| g: Gap from level 2 up to level 3 / 10³ cm⁻¹ | σ 1→3 = σ 1→2 + σ 2→3 / 10³ cm⁻¹ |
|---|---|
| 6.16875 | 21.56875 |
| 8.334375 | 23.734375 |
| 10.5 | 25.9 |
| 12.665624999999999 | 28.065624999999997 |
| 14.831249999999997 | 30.231249999999996 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of g separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At g = 8.0745 10³ cm⁻¹, substitute into Y_out = t2+g: Y₀ = 23.4745 10³ cm⁻¹.
- At g = 9.28725 10³ cm⁻¹, the same fixed-parameter model gives Y₁ = 24.6872 10³ cm⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = 1.21275 10³ cm⁻¹; the input change is Δg = 1.21275 10³ cm⁻¹.
- Average sensitivity is ΔY_out/Δg = 1 10³ cm⁻¹ per 10³ cm⁻¹. Its positive sign means that this response increases 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 = 23.4745 10³ cm⁻¹, Y₁ = 24.6872 10³ cm⁻¹; ΔY_out = 1.21275 10³ cm⁻¹; average sensitivity = 1 10³ cm⁻¹ per 10³ 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.
Emission, absorption and term analysis: infer an input with uncertainty
Term analysis is the least-squares inversion that turns a list of measured line wavenumbers into a set of term values T_n = −E_n/hc, each line modelled as a difference σ = T_l − T_u of two eigenvalues of one Hermitian atomic Hamiltonian. Use the supplied topic model Y_out = t2+g, where Y_out is σ 1→3 = σ 1→2 + σ 2→3 (10³ cm⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. g is Gap from level 2 up to level 3, measured in 10³ cm⁻¹. Keep t2 (Term of level 2 above ground) = 15.4 10³ cm⁻¹; c (Zero offset c added to every level) = 0 10³ cm⁻¹ fixed. For this exercise use only 6.16875 10³ cm⁻¹ ≤ g ≤ 14.8312 10³ 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 = 26.852875 10³ cm⁻¹, with a bounded reading uncertainty ±0.50676 10³ cm⁻¹. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown g by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00086625 10³ 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.
σ 1→3 = σ 1→2 + σ 2→3 against g
Predictions from the supplied Emission, absorption and term analysis 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): σ 1→3 = σ 1→2 + σ 2→3 / 10³ cm⁻¹
Horizontal axis (x): g: Gap from level 2 up to level 3 / 10³ cm⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| g: Gap from level 2 up to level 3 / 10³ cm⁻¹ | σ 1→3 = σ 1→2 + σ 2→3 / 10³ cm⁻¹ |
|---|---|
| 6.16875 | 21.56875 |
| 8.334375 | 23.734375 |
| 10.5 | 25.9 |
| 12.665624999999999 | 28.065624999999997 |
| 14.831249999999997 | 30.231249999999996 |
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(g) = 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(6.16875) = 21.5688 10³ cm⁻¹ and Y_out(14.8312) = 30.2312 10³ cm⁻¹. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives g_line = 6.16875 + (26.8529 − 21.5688)(14.8312 − 6.16875)/(30.2312 − 21.5688) = 11.4529 10³ 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 g = 11.4529 10³ cm⁻¹; substituting back gives Y_out = 26.852875 10³ cm⁻¹.
- The interpolation error in the input is |g_line − g| = 0 10³ cm⁻¹. This is within the requested numerical tolerance.
- Invert both limits Y_out = 26.346115 10³ cm⁻¹ and 27.359635 10³ cm⁻¹. Sort the roots even if the response decreases: 10.9461 10³ cm⁻¹ ≤ g ≤ 11.9596 10³ cm⁻¹.
- The lower and upper input deviations are 0.50676 10³ cm⁻¹ and 0.50676 10³ cm⁻¹. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer g ≈ 11.4529 10³ cm⁻¹; compatible interval [10.9461, 11.9596] 10³ cm⁻¹. Endpoint interpolation gives 11.4529 10³ cm⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.