University Quantum Mechanics I · unit UQMI-01
Foundations of Quantum Mechanics
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.
Where the classical account fails: infer an input with uncertainty
The classical account fails wherever a motion's characteristic energy hν is not small beside k_BT, or its action is not large beside Planck's constant h: there its predictions miss the measurements by factors, not percent. Use the supplied topic model Y_out = 0.531977*r0^1.5*(1-sqrt(max(1-t/(1.05016e-4*r0^3),0))), where Y_out is ORBITS SO FAR (thousand). 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. r0 is Starting orbit radius r₀, measured in pm. Keep t (Time since start t) = 8 ps fixed. For this exercise use only 42.65 pm ≤ r0 ≤ 65.75 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 = 53.76463 thousand, with a bounded reading uncertainty ±1.92 thousand. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown r0 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00231 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.
ORBITS SO FAR against r0
Predictions from the supplied Where the classical account fails 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): ORBITS SO FAR / thousand
Horizontal axis (x): r0: Starting orbit radius r₀ / pm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r0: Starting orbit radius r₀ / pm | ORBITS SO FAR / thousand |
|---|---|
| 42.65 | 128.25169035237255 |
| 48.425 | 76.4181735058303 |
| 54.2 | 58.97240809055958 |
| 59.975 | 48.35776019659553 |
| 65.75 | 40.96448004401035 |
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(r0) = 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(42.65) = 128.252 thousand and Y_out(65.75) = 40.9645 thousand. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives r0_line = 42.65 + (53.7646 − 128.252)(65.75 − 42.65)/(40.9645 − 128.252) = 62.3625 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 r0 = 56.741 pm; substituting back gives Y_out = 53.76463 thousand.
- The interpolation error in the input is |r0_line − r0| = 5.62152 pm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 51.84463 thousand and 55.68463 thousand. Sort the roots even if the response decreases: 55.7463 pm ≤ r0 ≤ 57.8115 pm.
- The lower and upper input deviations are 0.994737 pm and 1.07051 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 r0 ≈ 56.741 pm; compatible interval [55.7463, 57.8115] pm. Endpoint interpolation gives 62.3625 pm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Blackbody radiation and the Planck hypothesis: infer an input with uncertainty
The Planck hypothesis says an oscillator of frequency ν can hold energy only in whole multiples of hν; applied to cavity radiation, it replaces the classical kT per mode with hν/(e^(hν/kT) − 1) and reproduces the measured blackbody spectrum. Use the supplied topic model Y_out = hs/6.62607*(exp(11038.26/temp)-1)/(exp(1665.883*hs/temp)-1), where Y_out is TRIAL ÷ DATA AT 230 THz (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. hs is Trial Planck constant h, measured in 10⁻³⁴ J s. Keep temp (Cavity temperature T) = 1500 K fixed. For this exercise use only 5.425 10⁻³⁴ J s ≤ hs ≤ 8.575 10⁻³⁴ J s 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 = 0.49795855, with a bounded reading uncertainty ±0.052419. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown hs by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000315 10⁻³⁴ J s, 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.
TRIAL ÷ DATA AT 230 THz against hs
Predictions from the supplied Blackbody radiation and the Planck hypothesis 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): TRIAL ÷ DATA AT 230 THz
Horizontal axis (x): hs: Trial Planck constant h / 10⁻³⁴ J s
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| hs: Trial Planck constant h / 10⁻³⁴ J s | TRIAL ÷ DATA AT 230 THz |
|---|---|
| 5.425 | 3.1133085094666892 |
| 6.2124999999999995 | 1.4847272085089358 |
| 7 | 0.6972585708319023 |
| 7.7875 | 0.32341369130892206 |
| 8.575 | 0.14849816155590756 |
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(hs) = 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(5.425) = 3.11331 and Y_out(8.575) = 0.148498. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives hs_line = 5.425 + (0.497959 − 3.11331)(8.575 − 5.425)/(0.148498 − 3.11331) = 8.20371 10⁻³⁴ J s. 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 hs = 7.3465 10⁻³⁴ J s; substituting back gives Y_out = 0.49795855.
- The interpolation error in the input is |hs_line − hs| = 0.857211 10⁻³⁴ J s. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.44553955 and 0.55037755. Sort the roots even if the response decreases: 7.24372 10⁻³⁴ J s ≤ hs ≤ 7.46049 10⁻³⁴ J s.
- The lower and upper input deviations are 0.102776 10⁻³⁴ J s and 0.113988 10⁻³⁴ J s. 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 hs ≈ 7.3465 10⁻³⁴ J s; compatible interval [7.24372, 7.46049] 10⁻³⁴ J s. Endpoint interpolation gives 8.20371 10⁻³⁴ J s.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The photoelectric effect and the photon hypothesis: infer an input with uncertainty
The photoelectric effect is the emission of electrons from a surface under light, in which each electron absorbs the whole energy hν of one quantum, so the fastest leave with hν − φ however bright the light. Use the supplied topic model Y_out = 2998/nu, where Y_out is WAVELENGTH λ (nm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. nu is Frequency ν, measured in 10¹⁴ Hz. Keep phi (Work function φ) = 2.3 eV; pw (Beam power P) = 1 μW fixed. For this exercise use only 4.85 10¹⁴ Hz ≤ nu ≤ 11.15 10¹⁴ Hz 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 = 344.87519 nm, with a bounded reading uncertainty ±11.399 nm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown nu by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00063 10¹⁴ Hz, 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.
WAVELENGTH λ against nu
Predictions from the supplied The photoelectric effect and the photon hypothesis 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): WAVELENGTH λ / nm
Horizontal axis (x): nu: Frequency ν / 10¹⁴ Hz
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| nu: Frequency ν / 10¹⁴ Hz | WAVELENGTH λ / nm |
|---|---|
| 4.85 | 618.1443298969073 |
| 6.424999999999999 | 466.614785992218 |
| 7.999999999999999 | 374.75000000000006 |
| 9.575 | 313.1070496083551 |
| 11.149999999999999 | 268.8789237668162 |
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(nu) = 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(4.85) = 618.144 nm and Y_out(11.15) = 268.879 nm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives nu_line = 4.85 + (344.875 − 618.144)(11.15 − 4.85)/(268.879 − 618.144) = 9.77919 10¹⁴ Hz. 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 nu = 8.693 10¹⁴ Hz; substituting back gives Y_out = 344.87519 nm.
- The interpolation error in the input is |nu_line − nu| = 1.08619 10¹⁴ Hz. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 333.47619 nm and 356.27419 nm. Sort the roots even if the response decreases: 8.41487 10¹⁴ Hz ≤ nu ≤ 8.99015 10¹⁴ Hz.
- The lower and upper input deviations are 0.278133 10¹⁴ Hz and 0.297147 10¹⁴ Hz. 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 nu ≈ 8.693 10¹⁴ Hz; compatible interval [8.41487, 8.99015] 10¹⁴ Hz. Endpoint interpolation gives 9.77919 10¹⁴ Hz.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
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.