University Quantum Mechanics I · topic UQMI-01.04
Compton scattering and photon momentum
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.
Compton scattering and photon momentum: predict a finite change
Compton scattering is the collision of a single photon with a free electron, in which the photon hands over part of its momentum and energy and leaves with a longer wavelength, λ′ = λ + (h/mₑc)(1 − cos θ). Use the supplied topic model Y_out = 1239.84*(1/lam-1/(lam+2.4263*(1-cosd(th)))), where Y_out is ELECTRON KINETIC ENERGY K (keV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. lam is Incident wavelength λ, measured in pm. Keep th (Photon scattering angle θ) = 90 ° fixed. For this exercise use only 7.26875 pm ≤ lam ≤ 53.7312 pm 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 lam = 17.4905 pm and 23.9952 pm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δlam across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
ELECTRON KINETIC ENERGY K against lam
Predictions from the supplied Compton scattering and photon momentum 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): ELECTRON KINETIC ENERGY K / keV
Horizontal axis (x): lam: Incident wavelength λ / pm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lam: Incident wavelength λ / pm | ELECTRON KINETIC ENERGY K / keV |
|---|---|
| 7.26875 | 42.687464169992886 |
| 18.884375 | 7.474984904366347 |
| 30.5 | 2.9954865339347267 |
| 42.115624999999994 | 1.6036068746537817 |
| 53.731249999999996 | 0.9969538279137481 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of lam separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At lam = 17.4905 pm, substitute into Y_out = 1239.84*(1/lam-1/(lam+2.4263*(1-cosd(th)))): Y₀ = 8.63552 keV.
- At lam = 23.9952 pm, the same fixed-parameter model gives Y₁ = 4.74489 keV.
- Signed change: ΔY_out = Y₁ − Y₀ = -3.89062 keV; the input change is Δlam = 6.50475 pm.
- Average sensitivity is ΔY_out/Δlam = -0.59812 keV per pm. 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 = 8.63552 keV, Y₁ = 4.74489 keV; ΔY_out = -3.89062 keV; average sensitivity = -0.59812 keV per pm.
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.
Compton scattering and photon momentum: infer an input with uncertainty
Compton scattering is the collision of a single photon with a free electron, in which the photon hands over part of its momentum and energy and leaves with a longer wavelength, λ′ = λ + (h/mₑc)(1 − cos θ). Use the supplied topic model Y_out = 1239.84*(1/lam-1/(lam+2.4263*(1-cosd(th)))), where Y_out is ELECTRON KINETIC ENERGY K (keV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. lam is Incident wavelength λ, measured in pm. Keep th (Photon scattering angle θ) = 90 ° fixed. For this exercise use only 7.26875 pm ≤ lam ≤ 53.7312 pm 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 = 2.2208501 keV, with a bounded reading uncertainty ±0.18358 keV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown lam by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00464625 pm, 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.
ELECTRON KINETIC ENERGY K against lam
Predictions from the supplied Compton scattering and photon momentum 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): ELECTRON KINETIC ENERGY K / keV
Horizontal axis (x): lam: Incident wavelength λ / pm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lam: Incident wavelength λ / pm | ELECTRON KINETIC ENERGY K / keV |
|---|---|
| 7.26875 | 42.687464169992886 |
| 18.884375 | 7.474984904366347 |
| 30.5 | 2.9954865339347267 |
| 42.115624999999994 | 1.6036068746537817 |
| 53.731249999999996 | 0.9969538279137481 |
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(lam) = 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(7.26875) = 42.6875 keV and Y_out(53.7312) = 0.996954 keV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives lam_line = 7.26875 + (2.22085 − 42.6875)(53.7312 − 7.26875)/(0.996954 − 42.6875) = 52.3673 pm. 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 lam = 35.6109 pm; substituting back gives Y_out = 2.2208501 keV.
- The interpolation error in the input is |lam_line − lam| = 16.7564 pm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 2.0372701 keV and 2.4044301 keV. Sort the roots even if the response decreases: 34.1788 pm ≤ lam ≤ 37.2325 pm.
- The lower and upper input deviations are 1.43209 pm and 1.62161 pm. 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 lam ≈ 35.6109 pm; compatible interval [34.1788, 37.2325] pm. Endpoint interpolation gives 52.3673 pm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.