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University Physics III · unit UPIII-09

Quantum Physics

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.

Extra challenging · 01

Blackbody radiation and Planck's quantum hypothesis: infer an input with uncertainty

Planck's quantum hypothesis is that a cavity mode of frequency ν can hold energy only in whole steps of hν, so at temperature T its mean energy is hν/(e^(hν/kT) − 1) rather than the classical kT. Use the supplied topic model Y_out = (nu/(0.0208366*tk))/(exp((nu/(0.0208366*tk)))-1), where Y_out is ⟨E⟩ PER MODE ÷ kT (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. nu is Marker frequency ν, measured in THz. Keep tk (Cavity temperature T) = 5800 K fixed. For this exercise use only 115.188 THz ≤ nu ≤ 894.812 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 = 0.037109965, with a bounded reading uncertainty ±0.00489. 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 0.0779625 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.

Explore the question data

⟨E⟩ PER MODE ÷ kT against nu

Predictions from the supplied Blackbody radiation and Planck's quantum 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): ⟨E⟩ PER MODE ÷ kT

⟨E⟩ PER MODE ÷ kT against nuPredictions from the supplied Blackbody radiation and Planck's quantum 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. Horizontal axis: nu: Marker frequency ν / THz. Vertical axis: ⟨E⟩ PER MODE ÷ kT. Exact values are available in the data table below.-0.019230.1410.30130.46150.621884294.5505715.5926xy

Horizontal axis (x): nu: Marker frequency ν / THz

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1nu: Marker frequency ν / THz: 115.2⟨E⟩ PER MODE ÷ kT: 0.598
Read the exact data as a table
⟨E⟩ PER MODE ÷ kT against nu — supplied values
nu: Marker frequency ν / THz⟨E⟩ PER MODE ÷ kT
115.18750.5980189752041845
310.093750.21360627873002808
504.999999999999940.06500909926097907
699.90624999999990.017739105057302855
894.81249999999990.004509677031257769

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
  1. Endpoint predictions are Y_out(115.188) = 0.598019 and Y_out(894.812) = 0.00450968. The target lies between them. On the selected branch the response is strictly decreasing.
  2. Endpoint interpolation gives nu_line = 115.188 + (0.03711 − 0.598019)(894.812 − 115.188)/(0.00450968 − 0.598019) = 851.989 THz. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
  3. Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives nu = 590.759 THz; substituting back gives Y_out = 0.037109965.
  4. The interpolation error in the input is |nu_line − nu| = 261.23 THz. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 0.032219965 and 0.041999965. Sort the roots even if the response decreases: 572.067 THz ≤ nu ≤ 611.946 THz.
  6. The lower and upper input deviations are 18.6915 THz and 21.1873 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 nu ≈ 590.759 THz; compatible interval [572.067, 611.946] THz. Endpoint interpolation gives 851.989 THz.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

Extra challenging · 02

Photons and the photoelectric effect: infer an input with uncertainty

The photoelectric effect is the emission of electrons from a surface by light, in which each electron absorbs a single photon and leaves with at most K_max = hf − φ, however intense the light. Use the supplied topic model Y_out = 1239.84/phi, where Y_out is THRESHOLD λ₀ = hc/φ (nm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. phi is Work function φ, measured in eV. Keep lam (Wavelength λ) = 400 nm; pw (Beam power P) = 1 mW fixed. For this exercise use only 1.925 eV ≤ phi ≤ 5.075 eV 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 = 322.32939 nm, with a bounded reading uncertainty ±11.704 nm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown phi 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 eV, 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.

Explore the question data

THRESHOLD λ₀ = hc/φ against phi

Predictions from the supplied Photons and the photoelectric effect 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): THRESHOLD λ₀ = hc/φ / nm

THRESHOLD λ₀ = hc/φ against phiPredictions from the supplied Photons and the photoelectric effect 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. Horizontal axis: phi: Work function φ / eV. Vertical axis: THRESHOLD λ₀ = hc/φ / nm. Exact values are available in the data table below.228.3336.3444.2552.1660.11.7992.6493.54.355.201xy

Horizontal axis (x): phi: Work function φ / eV

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1phi: Work function φ / eV: 1.925THRESHOLD λ₀ = hc/φ / nm: 644.1
Read the exact data as a table
THRESHOLD λ₀ = hc/φ against phi — supplied values
phi: Work function φ / eVTHRESHOLD λ₀ = hc/φ / nm
1.925644.0727272727272
2.7125457.0838709677419
3.5354.23999999999995
4.2875289.17551020408166
5.074999999999999244.30344827586208

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(phi) = 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
  1. Endpoint predictions are Y_out(1.925) = 644.073 nm and Y_out(5.075) = 244.303 nm. The target lies between them. On the selected branch the response is strictly decreasing.
  2. Endpoint interpolation gives phi_line = 1.925 + (322.329 − 644.073)(5.075 − 1.925)/(244.303 − 644.073) = 4.46019 eV. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
  3. Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives phi = 3.8465 eV; substituting back gives Y_out = 322.32939 nm.
  4. The interpolation error in the input is |phi_line − phi| = 0.613691 eV. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 310.62539 nm and 334.03339 nm. Sort the roots even if the response decreases: 3.71172 eV ≤ phi ≤ 3.99143 eV.
  6. The lower and upper input deviations are 0.134775 eV and 0.144932 eV. 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 phi ≈ 3.8465 eV; compatible interval [3.71172, 3.99143] eV. Endpoint interpolation gives 4.46019 eV.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

Extra challenging · 03

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.

Explore the question data

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

SCATTERED λ′ against ekPredictions 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. Horizontal axis: ek: Photon energy E / keV. Vertical axis: SCATTERED λ′ / pm. Exact values are available in the data table below.3.4375.9698.50111.0313.5784294.5505715.5926xy

Horizontal axis (x): ek: Photon energy E / keV

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1ek: Photon energy E / keV: 115.2SCATTERED λ′ / pm: 13.19
Read the exact data as a table
SCATTERED λ′ against ek — supplied values
ek: Photon energy E / keVSCATTERED λ′ / pm
115.187513.189967932718393
310.093756.42457471530787
504.999999999999944.881428712871287
699.90624999999994.19773724605974
894.81249999999993.811886365858769

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
  1. 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.
  2. 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.
  3. 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.
  4. The interpolation error in the input is |ek_line − ek| = 244.769 keV. This is larger than the requested numerical tolerance.
  5. 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.
  6. 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.

Extra challenging · 04

Matter waves and electron diffraction: infer an input with uncertainty

A matter wave is the wave associated with any particle of momentum p, with de Broglie wavelength λ = h/p; it diffracts from a periodic structure exactly as light does, and visibly so only when λ is comparable to the structure's spacing. Use the supplied topic model Y_out = (1.2264/d)^2, where Y_out is CUTOFF VOLTAGE, λ = d (V). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Surface row spacing d, measured in nm. Keep v (Accelerating voltage V) = 54 V fixed. For this exercise use only 0.131875 nm ≤ d ≤ 0.368125 nm 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 = 19.746288 V, with a bounded reading uncertainty ±1.2971 V. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown d by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000023625 nm, 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.

Explore the question data

CUTOFF VOLTAGE, λ = d against d

Predictions from the supplied Matter waves and electron diffraction 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): CUTOFF VOLTAGE, λ = d / V

CUTOFF VOLTAGE, λ = d against dPredictions from the supplied Matter waves and electron diffraction 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. Horizontal axis: d: Surface row spacing d / nm. Vertical axis: CUTOFF VOLTAGE, λ = d / V. Exact values are available in the data table below.8.08328.4448.7969.1589.50.12240.18620.250.31380.3776xy

Horizontal axis (x): d: Surface row spacing d / nm

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1d: Surface row spacing d / nm: 0.1319CUTOFF VOLTAGE, λ = d / V: 86.48
Read the exact data as a table
CUTOFF VOLTAGE, λ = d against d — supplied values
d: Surface row spacing d / nmCUTOFF VOLTAGE, λ = d / V
0.1318750000000000286.4847109813346
0.1909375000000000441.25549666480051
0.2524.064911359999996
0.309062515.746051122918328
0.3681250000000000411.098739533207844

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(d) = 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
  1. Endpoint predictions are Y_out(0.131875) = 86.4847 V and Y_out(0.368125) = 11.0987 V. The target lies between them. On the selected branch the response is strictly decreasing.
  2. Endpoint interpolation gives d_line = 0.131875 + (19.7463 − 86.4847)(0.368125 − 0.131875)/(11.0987 − 86.4847) = 0.341025 nm. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
  3. Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives d = 0.275987 nm; substituting back gives Y_out = 19.746288 V.
  4. The interpolation error in the input is |d_line − d| = 0.0650372 nm. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 18.449188 V and 21.043388 V. Sort the roots even if the response decreases: 0.267346 nm ≤ d ≤ 0.285525 nm.
  6. The lower and upper input deviations are 0.00864112 nm and 0.00953709 nm. 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 d ≈ 0.275987 nm; compatible interval [0.267346, 0.285525] nm. Endpoint interpolation gives 0.341025 nm.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

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