University Physics IV · topic UPIV-11.08
Selection rules, line intensities, and lifetimes
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.
Selection rules, intensities and lifetimes: predict a finite change
An electric-dipole selection rule is the statement that the matrix element ⟨f|(−er)|i⟩ vanishes identically between two states, so the transition rate between them is zero at leading order and the line is missing from the spectrum. Use the supplied topic model Y_out = 1000000000/(1063200*w*w*w*d*d), where Y_out is LIFETIME τ = 1/A (ns). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. w is Photon energy ħω, measured in eV. Keep d (Dipole element |⟨f|r|i⟩|) = 0.75 a₀; q (Radiator size a) = 1 a₀ fixed. For this exercise use only 2.16875 eV ≤ w ≤ 10.8312 eV 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 w = 4.0745 eV and 5.28725 eV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δw across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
LIFETIME τ = 1/A against w
Predictions from the supplied Selection rules, intensities and lifetimes 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): LIFETIME τ = 1/A / ns
Horizontal axis (x): w: Photon energy ħω / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| w: Photon energy ħω / eV | LIFETIME τ = 1/A / ns |
|---|---|
| 2.16875 | 163.9207863915401 |
| 4.334375 | 20.534449145686462 |
| 6.499999999999999 | 6.08866998598122 |
| 8.665624999999999 | 2.569584071696344 |
| 10.831249999999997 | 1.3159117508414313 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of w separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At w = 4.0745 eV, substitute into Y_out = 1000000000/(1063200*w*w*w*d*d): Y₀ = 24.7195 ns.
- At w = 5.28725 eV, the same fixed-parameter model gives Y₁ = 11.3129 ns.
- Signed change: ΔY_out = Y₁ − Y₀ = -13.4066 ns; the input change is Δw = 1.21275 eV.
- Average sensitivity is ΔY_out/Δw = -11.0547 ns per eV. 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 = 24.7195 ns, Y₁ = 11.3129 ns; ΔY_out = -13.4066 ns; average sensitivity = -11.0547 ns per eV.
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.
Selection rules, intensities and lifetimes: infer an input with uncertainty
An electric-dipole selection rule is the statement that the matrix element ⟨f|(−er)|i⟩ vanishes identically between two states, so the transition rate between them is zero at leading order and the line is missing from the spectrum. Use the supplied topic model Y_out = 1000000000/(1063200*w*w*w*d*d), where Y_out is LIFETIME τ = 1/A (ns). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. w is Photon energy ħω, measured in eV. Keep d (Dipole element |⟨f|r|i⟩|) = 0.75 a₀; q (Radiator size a) = 1 a₀ fixed. For this exercise use only 2.16875 eV ≤ w ≤ 10.8312 eV 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 = 4.0391594 ns, with a bounded reading uncertainty ±0.40849 ns. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown w 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 eV, 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.
LIFETIME τ = 1/A against w
Predictions from the supplied Selection rules, intensities and lifetimes 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): LIFETIME τ = 1/A / ns
Horizontal axis (x): w: Photon energy ħω / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| w: Photon energy ħω / eV | LIFETIME τ = 1/A / ns |
|---|---|
| 2.16875 | 163.9207863915401 |
| 4.334375 | 20.534449145686462 |
| 6.499999999999999 | 6.08866998598122 |
| 8.665624999999999 | 2.569584071696344 |
| 10.831249999999997 | 1.3159117508414313 |
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(w) = 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(2.16875) = 163.921 ns and Y_out(10.8312) = 1.31591 ns. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives w_line = 2.16875 + (4.03916 − 163.921)(10.8312 − 2.16875)/(1.31591 − 163.921) = 10.6862 eV. 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 w = 7.45288 eV; substituting back gives Y_out = 4.0391594 ns.
- The interpolation error in the input is |w_line − w| = 3.2333 eV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 3.6306694 ns and 4.4476494 ns. Sort the roots even if the response decreases: 7.21734 eV ≤ w ≤ 7.72251 eV.
- The lower and upper input deviations are 0.235532 eV and 0.269637 eV. 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 w ≈ 7.45288 eV; compatible interval [7.21734, 7.72251] eV. Endpoint interpolation gives 10.6862 eV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.