University Physics V · unit UPV-02
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.
Four quantitative failures of the classical account: infer an input with uncertainty
The classical failure inventory is the list of sharp wrong predictions — atomic collapse in picoseconds, harmonic spectra, temperature-blind heat capacities, an undefined entropy constant — that classical mechanics, electrodynamics and statistical mechanics make about ordinary matter. Use the supplied topic model Y_out = 1.5+(tr/t)^2*exp(-tr/t)/max((1-exp(-tr/t))^2,0.000000001)+(tv/t)^2*exp(-tv/t)/max((1-exp(-tv/t))^2,0.000000001), where Y_out is C_V AT T (R). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ; max(a,b) selects the larger value. tv is Vibration quantum θ_vib, measured in K. Keep tr (Rotation quantum θ_rot) = 85 K; t (Gas temperature T) = 300 K fixed. For this exercise use only 1384.38 K ≤ tv ≤ 5715.63 K 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 = 2.4936041 R, with a bounded reading uncertainty ±0.000039783 R. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown tv by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.433125 K, 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.
C_V AT T against tv
Predictions from the supplied Four quantitative failures of the classical account 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): C_V AT T / R
Horizontal axis (x): tv: Vibration quantum θ_vib / K
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| tv: Vibration quantum θ_vib / K | C_V AT T / R |
|---|---|
| 1384.375 | 2.7085280607810933 |
| 2467.1875 | 2.511482706259243 |
| 3550 | 2.49435336253227 |
| 4632.8125 | 2.4933838079367443 |
| 5715.625 | 2.4933388825594234 |
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(tv) = 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(1384.38) = 2.70853 R and Y_out(5715.63) = 2.49334 R. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives tv_line = 1384.38 + (2.4936 − 2.70853)(5715.63 − 1384.38)/(2.49334 − 2.70853) = 5710.29 K. 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 tv = 4026.42 K; substituting back gives Y_out = 2.4936041 R.
- The interpolation error in the input is |tv_line − tv| = 1683.86 K. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 2.4935643 R and 2.4936439 R. Sort the roots even if the response decreases: 3977.43 K ≤ tv ≤ 4083.2 K.
- The lower and upper input deviations are 48.9911 K and 56.775 K. 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 tv ≈ 4026.42 K; compatible interval [3977.43, 4083.2] K. Endpoint interpolation gives 5710.29 K.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Cavity modes and the Planck law: infer an input with uncertainty
Planck's law u(ν,T) = (8πhν³/c³)/(e^(hν/kT) − 1) is the energy per volume per frequency of thermal radiation: the cavity's mode count multiplied by the mean energy the Boltzmann-weighted ladder Eₙ = nhν allows each mode. Use the supplied topic model Y_out = 1/max(exp(47.99*f/t)-1, 0.000001), where Y_out is PROBE n̄ = 1/(eˣ − 1) (photons). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ; max(a,b) selects the larger value. f is Probe frequency ν, measured in THz. Keep t (Temperature T) = 6000 K fixed. For this exercise use only 193.438 THz ≤ f ≤ 1256.56 THz 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 = 0.0011910066 photons, with a bounded reading uncertainty ±0.00017218 photons. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown f by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.106313 THz, 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.
PROBE n̄ = 1/(eˣ − 1) against f
Predictions from the supplied Cavity modes and the Planck law 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): PROBE n̄ = 1/(eˣ − 1) / photons
Horizontal axis (x): f: Probe frequency ν / THz
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| f: Probe frequency ν / THz | PROBE n̄ = 1/(eˣ − 1) / photons |
|---|---|
| 193.4375 | 0.27040245345998226 |
| 459.21875 | 0.026062548271025568 |
| 725 | 0.0030404314539991552 |
| 990.78125 | 0.0003618658852608267 |
| 1256.5625 | 0.00004317009447302449 |
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(f) = 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(193.438) = 0.270402 photons and Y_out(1256.56) = 0.0000431701 photons. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives f_line = 193.438 + (0.00119101 − 0.270402)(1256.56 − 193.438)/(0.0000431701 − 0.270402) = 1252.05 THz. 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 f = 841.944 THz; substituting back gives Y_out = 0.0011910066 photons.
- The interpolation error in the input is |f_line − f| = 410.105 THz. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.0010188266 photons and 0.0013631866 photons. Sort the roots even if the response decreases: 825.083 THz ≤ f ≤ 861.445 THz.
- The lower and upper input deviations are 16.8603 THz and 19.501 THz. 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 f ≈ 841.944 THz; compatible interval [825.083, 861.445] THz. Endpoint interpolation gives 1252.05 THz.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The photoelectric effect and what it does not prove: infer an input with uncertainty
The photoelectric effect is the ejection of electrons from a surface by light above a threshold frequency; its threshold, promptness and h/e stopping-potential slope all follow from a quantised atom in a classical field, so they evidence atomic energy levels, not photons. Use the supplied topic model Y_out = 0.4136*nu, where Y_out is PHOTON ENERGY hν (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. nu is Drive frequency ν, measured in ×10¹⁴ Hz. Keep phi (Work function φ) = 2.28 eV; i (Relative intensity I) = 40 % fixed. For this exercise use only 4.0625 ×10¹⁴ Hz ≤ nu ≤ 11.9375 ×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 = 3.667081 eV, with a bounded reading uncertainty ±0.19054 eV. 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 0.0007875 ×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.
PHOTON ENERGY hν against nu
Predictions from the supplied The photoelectric effect and what it does not prove 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): PHOTON ENERGY hν / eV
Horizontal axis (x): nu: Drive 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: Drive frequency ν / ×10¹⁴ Hz | PHOTON ENERGY hν / eV |
|---|---|
| 4.0625 | 1.68025 |
| 6.03125 | 2.4945250000000003 |
| 8 | 3.3088 |
| 9.96875 | 4.123075 |
| 11.9375 | 4.93735 |
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(4.0625) = 1.68025 eV and Y_out(11.9375) = 4.93735 eV. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives nu_line = 4.0625 + (3.66708 − 1.68025)(11.9375 − 4.0625)/(4.93735 − 1.68025) = 8.86625 ×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 = 8.86625 ×10¹⁴ Hz; substituting back gives Y_out = 3.667081 eV.
- The interpolation error in the input is |nu_line − nu| = 3.55271e-15 ×10¹⁴ Hz. This is within the requested numerical tolerance.
- Invert both limits Y_out = 3.476541 eV and 3.857621 eV. Sort the roots even if the response decreases: 8.40556 ×10¹⁴ Hz ≤ nu ≤ 9.32694 ×10¹⁴ Hz.
- The lower and upper input deviations are 0.460687 ×10¹⁴ Hz and 0.460687 ×10¹⁴ Hz. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer nu ≈ 8.86625 ×10¹⁴ Hz; compatible interval [8.40556, 9.32694] ×10¹⁴ Hz. Endpoint interpolation gives 8.86625 ×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 elastic two-body collision of a single photon of four-momentum ħk with a quasi-free electron, whose four-momentum balance lengthens the photon's wavelength by (h/mₑc)(1 − cos θ). Use the supplied topic model Y_out = en/(1+(en/511)*(1-cosd(theta))), where Y_out is SCATTERED E′ (keV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. theta is Scattering angle θ, measured in deg. Keep en (Incident photon energy E) = 662 keV fixed. For this exercise use only 19.125 deg ≤ theta ≤ 160.875 deg 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 = 250.40463 keV, with a bounded reading uncertainty ±9.3464 keV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown theta by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.014175 deg, 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′ against theta
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): SCATTERED E′ / keV
Horizontal axis (x): theta: Scattering angle θ / deg
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| theta: Scattering angle θ / deg | SCATTERED E′ / keV |
|---|---|
| 19.125 | 617.8233375154246 |
| 54.5625 | 428.6593504259467 |
| 90 | 288.3904518329071 |
| 125.4375 | 217.28796902755607 |
| 160.875 | 188.09520285596054 |
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(theta) = 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(19.125) = 617.823 keV and Y_out(160.875) = 188.095 keV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives theta_line = 19.125 + (250.405 − 617.823)(160.875 − 19.125)/(188.095 − 617.823) = 140.322 deg. 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 theta = 105.593 deg; substituting back gives Y_out = 250.40463 keV.
- The interpolation error in the input is |theta_line − theta| = 34.7291 deg. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 241.05823 keV and 259.75103 keV. Sort the roots even if the response decreases: 101.266 deg ≤ theta ≤ 110.36 deg.
- The lower and upper input deviations are 4.32645 deg and 4.76743 deg. 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 theta ≈ 105.593 deg; compatible interval [101.266, 110.36] deg. Endpoint interpolation gives 140.322 deg.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.