University Physics III · topic UPIII-09.03
Photon momentum and Compton scattering
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.
Photon momentum and Compton scattering: predict a finite change
Compton scattering is the collision of a photon with a free electron in which part of the photon's momentum h/λ and energy hc/λ passes to the electron, so the photon leaves at angle θ with a wavelength longer by (h/mₑc)(1 − cos θ). Use the supplied topic model Y_out = 1239.84/ek+2.4263*(1-cosd(theta)), where Y_out is SCATTERED λ′ (pm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. ek is Photon energy E, measured in keV. Keep theta (Scattering angle θ) = 90 ° fixed. For this exercise use only 115.188 keV ≤ ek ≤ 894.812 keV 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 ek = 286.705 keV and 395.852 keV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δek across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
SCATTERED λ′ against ek
Predictions from the supplied Photon momentum and Compton scattering 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): SCATTERED λ′ / pm
Horizontal axis (x): ek: Photon energy E / keV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| ek: Photon energy E / keV | SCATTERED λ′ / pm |
|---|---|
| 115.1875 | 13.189967932718393 |
| 310.09375 | 6.42457471530787 |
| 504.99999999999994 | 4.881428712871287 |
| 699.9062499999999 | 4.19773724605974 |
| 894.8124999999999 | 3.811886365858769 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of ek separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At ek = 286.705 keV, substitute into Y_out = 1239.84/ek+2.4263*(1-cosd(theta)): Y₀ = 6.75074 pm.
- At ek = 395.852 keV, the same fixed-parameter model gives Y₁ = 5.55838 pm.
- Signed change: ΔY_out = Y₁ − Y₀ = -1.19237 pm; the input change is Δek = 109.147 keV.
- Average sensitivity is ΔY_out/Δek = -0.0109244 pm per keV. 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 = 6.75074 pm, Y₁ = 5.55838 pm; ΔY_out = -1.19237 pm; average sensitivity = -0.0109244 pm per keV.
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.
Photon momentum and Compton scattering: infer an input with uncertainty
Compton scattering is the collision of a photon with a free electron in which part of the photon's momentum h/λ and energy hc/λ passes to the electron, so the photon leaves at angle θ with a wavelength longer by (h/mₑc)(1 − cos θ). Use the supplied topic model Y_out = 1239.84/ek+2.4263*(1-cosd(theta)), where Y_out is SCATTERED λ′ (pm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. ek is Photon energy E, measured in keV. Keep theta (Scattering angle θ) = 90 ° fixed. For this exercise use only 115.188 keV ≤ ek ≤ 894.812 keV 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 = 4.5250247 pm, with a bounded reading uncertainty ±0.10697 pm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown ek by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0779625 keV, 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.
SCATTERED λ′ against ek
Predictions from the supplied Photon momentum and Compton scattering 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): SCATTERED λ′ / pm
Horizontal axis (x): ek: Photon energy E / keV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| ek: Photon energy E / keV | SCATTERED λ′ / pm |
|---|---|
| 115.1875 | 13.189967932718393 |
| 310.09375 | 6.42457471530787 |
| 504.99999999999994 | 4.881428712871287 |
| 699.9062499999999 | 4.19773724605974 |
| 894.8124999999999 | 3.811886365858769 |
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(ek) = 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(115.188) = 13.19 pm and Y_out(894.812) = 3.81189 pm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives ek_line = 115.188 + (4.52502 − 13.19)(894.812 − 115.188)/(3.81189 − 13.19) = 835.527 keV. 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 ek = 590.759 keV; substituting back gives Y_out = 4.5250247 pm.
- The interpolation error in the input is |ek_line − ek| = 244.769 keV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 4.4180547 pm and 4.6319947 pm. Sort the roots even if the response decreases: 562.109 keV ≤ ek ≤ 622.486 keV.
- The lower and upper input deviations are 28.6501 keV and 31.7275 keV. 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 ek ≈ 590.759 keV; compatible interval [562.109, 622.486] keV. Endpoint interpolation gives 835.527 keV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.