University Physics IV · unit UPIV-04
Photons and Matter Waves
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.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
Photon energy, momentum and the massless limit: infer an input with uncertainty
A photon is the massless quantum of the electromagnetic field: because m = 0, its energy and momentum are locked together as E = pc, so fixing E = hf fixes p = h/λ, and neither can change without the other. Use the supplied topic model Y_out = p/max(sqrt(p^2+m^2),0.01), where Y_out is β = pc/E (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. p is Lab-frame momentum pc, measured in MeV. Keep m (Rest energy mc²) = 0.6 MeV; b (Observer boost β = v/c) = 0.3 fixed. For this exercise use only 0.301875 MeV ≤ p ≤ 1.79812 MeV 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 = 0.89657043, with a bounded reading uncertainty ±0.0078021. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown p by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000149625 MeV, 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.
β = pc/E against p
Predictions from the supplied Photon energy, momentum and the massless limit 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): β = pc/E
Horizontal axis (x): p: Lab-frame momentum pc / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| p: Lab-frame momentum pc / MeV | β = pc/E |
|---|---|
| 0.301875 | 0.4494454708717874 |
| 0.6759375 | 0.7478669678549078 |
| 1.0499999999999998 | 0.8682431421244593 |
| 1.4240624999999998 | 0.9215439910169114 |
| 1.7981249999999998 | 0.9485843377372829 |
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(p) = 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(0.301875) = 0.449445 and Y_out(1.79812) = 0.948584. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives p_line = 0.301875 + (0.89657 − 0.449445)(1.79812 − 0.301875)/(0.948584 − 0.449445) = 1.6422 MeV. 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 p = 1.21459 MeV; substituting back gives Y_out = 0.89657043.
- The interpolation error in the input is |p_line − p| = 0.427617 MeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.88876833 and 0.90437253. Sort the roots even if the response decreases: 1.16342 MeV ≤ p ≤ 1.27154 MeV.
- The lower and upper input deviations are 0.0511683 MeV and 0.0569563 MeV. 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 p ≈ 1.21459 MeV; compatible interval [1.16342, 1.27154] MeV. Endpoint interpolation gives 1.6422 MeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Radiation pressure and photon flux: infer an input with uncertainty
Radiation pressure is the momentum a beam delivers per second per unit area of surface — I/c when the surface absorbs the light head-on, 2I/c when it reflects it, and (1 + r)I cos²θ/c in general. Use the supplied topic model Y_out = 100/sqrt(max(2.6782*w*cosd(th),1)), where Y_out is SHOT NOISE 1/√N (%). 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; sind/cosd/tand use degrees. th is Incidence angle θ from normal, measured in °. Keep w (Beam intensity I) = 1360 W/m²; r (Power reflectance r) = 1 fixed. For this exercise use only 8.5 ° ≤ th ≤ 71.5 ° 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 = 2.005087 %, with a bounded reading uncertainty ±0.050844 %. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown th by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0063 °, 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.
SHOT NOISE 1/√N against th
Predictions from the supplied Radiation pressure and photon flux 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): SHOT NOISE 1/√N / %
Horizontal axis (x): th: Incidence angle θ from normal / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| th: Incidence angle θ from normal / ° | SHOT NOISE 1/√N / % |
|---|---|
| 8.5 | 1.6661243601839753 |
| 24.25 | 1.735274914702826 |
| 40 | 1.8931370703716506 |
| 55.75 | 2.2086662623158344 |
| 71.5 | 2.9415133264013904 |
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(th) = 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(8.5) = 1.66612 % and Y_out(71.5) = 2.94151 %. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives th_line = 8.5 + (2.00509 − 1.66612)(71.5 − 8.5)/(2.94151 − 1.66612) = 25.2436 °. 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 th = 46.93 °; substituting back gives Y_out = 2.005087 %.
- The interpolation error in the input is |th_line − th| = 21.6864 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.954243 % and 2.055931 %. Sort the roots even if the response decreases: 44.0373 ° ≤ th ≤ 49.4936 °.
- The lower and upper input deviations are 2.89271 ° and 2.56362 °. 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 th ≈ 46.93 °; compatible interval [44.0373, 49.4936] °. Endpoint interpolation gives 25.2436 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The Compton shift, derived: infer an input with uncertainty
The Compton shift is the increase in a photon's wavelength when it scatters from a free electron at rest, equal to the Compton wavelength h/m_e c multiplied by (1 − cos θ), with θ the photon's deflection and no dependence on the incident wavelength. Use the supplied topic model Y_out = 100*2.4263*(1-cosd(th))/max(lam,0.01), where Y_out is Δλ / λ (%). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value; sind/cosd/tand use degrees. lam is Incident wavelength λ, measured in pm. Keep th (Scattering angle θ) = 90 ° fixed. For this exercise use only 41.25 pm ≤ lam ≤ 198.75 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 = 1.7668305 %, with a bounded reading uncertainty ±0.081908 %. 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 0.01575 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.
Δλ / λ against lam
Predictions from the supplied The Compton shift, derived 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): Δλ / λ / %
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 | Δλ / λ / % |
|---|---|
| 41.25 | 5.881939393939393 |
| 80.625 | 3.009364341085271 |
| 120 | 2.0219166666666664 |
| 159.375 | 1.52238431372549 |
| 198.75 | 1.2207798742138363 |
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(41.25) = 5.88194 % and Y_out(198.75) = 1.22078 %. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives lam_line = 41.25 + (1.76683 − 5.88194)(198.75 − 41.25)/(1.22078 − 5.88194) = 180.299 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 = 137.325 pm; substituting back gives Y_out = 1.7668305 %.
- The interpolation error in the input is |lam_line − lam| = 42.974 pm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.6849225 % and 1.8487385 %. Sort the roots even if the response decreases: 131.241 pm ≤ lam ≤ 144.001 pm.
- The lower and upper input deviations are 6.08416 pm and 6.67569 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 ≈ 137.325 pm; compatible interval [131.241, 144.001] pm. Endpoint interpolation gives 180.299 pm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Recoil kinematics and the Compton edge: infer an input with uncertainty
The recoil kinetic energy T(θ) = Eα(1 − cosθ)/[1 + α(1 − cosθ)] is what the Compton ledger hands the electron at each photon scattering angle, and its ceiling at θ = 180° is the Compton edge. Use the supplied topic model Y_out = 511*en/(511+en*(1-cosd(th))), where Y_out is SCATTERED E′(θ) (keV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. en is Photon energy E, measured in keV. Keep th (Scattering angle θ) = 120 ° fixed. For this exercise use only 230.375 keV ≤ en ≤ 1789.62 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 = 264.42505 keV, with a bounded reading uncertainty ±3.2646 keV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown en by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.155925 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 E′(θ) against en
Predictions from the supplied Recoil kinematics and the Compton edge 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 E′(θ) / keV
Horizontal axis (x): en: Photon energy E / keV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| en: Photon energy E / keV | SCATTERED E′(θ) / keV |
|---|---|
| 230.375 | 137.43495074790223 |
| 620.1875 | 219.88478133605082 |
| 1009.9999999999999 | 254.74333662388946 |
| 1399.8124999999998 | 273.9874555618065 |
| 1789.6249999999998 | 286.1887847908151 |
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(en) = 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(230.375) = 137.435 keV and Y_out(1789.62) = 286.189 keV. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives en_line = 230.375 + (264.425 − 137.435)(1789.62 − 230.375)/(286.189 − 137.435) = 1561.5 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 en = 1181.52 keV; substituting back gives Y_out = 264.42505 keV.
- The interpolation error in the input is |en_line − en| = 379.978 keV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 261.16045 keV and 267.68965 keV. Sort the roots even if the response decreases: 1119.02 keV ≤ en ≤ 1249.61 keV.
- The lower and upper input deviations are 62.5023 keV and 68.0943 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 en ≈ 1181.52 keV; compatible interval [1119.02, 1249.61] keV. Endpoint interpolation gives 1561.5 keV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.