University Physics II · unit UPII-15
Introduction to Modern 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.
Blackbody Radiation & Planck's Hypothesis: infer an input with uncertainty
A blackbody is an ideal absorber whose thermal radiation spectrum depends only on temperature, and Planck's hypothesis is that a radiation mode of frequency ν exchanges energy only in whole multiples of hν. Use the supplied topic model Y_out = (14387.8/(10^q*T))/(exp(14387.8/(10^q*T))-1), where Y_out is PLANCK ÷ RAYLEIGH–JEANS (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. q is Probe q, where λ = 10^q μm, dimensionless. Keep T (Cavity temperature T) = 1500 K fixed. For this exercise use only -0.7875 ≤ q ≤ 0.7875 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 = 0.010327408, with a bounded reading uncertainty ±0.0015491. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown q by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0001575, 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.
PLANCK ÷ RAYLEIGH–JEANS against q
Predictions from the supplied Blackbody Radiation & Planck's 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): PLANCK ÷ RAYLEIGH–JEANS
Horizontal axis (x): q: Probe q, where λ = 10^q μm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| q: Probe q, where λ = 10^q μm | PLANCK ÷ RAYLEIGH–JEANS |
|---|---|
| -0.7875 | 1.7036096686236243e-24 |
| -0.39375000000000004 | 1.151882298208765e-9 |
| -1.1102230246251565e-16 | 0.0006549950593489061 |
| 0.39374999999999993 | 0.08219388564548756 |
| 0.7874999999999998 | 0.4138320731726888 |
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(q) = 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(-0.7875) = 1.70361e-24 and Y_out(0.7875) = 0.413832. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives q_line = -0.7875 + (0.0103274 − 1.70361e-24)(0.7875 − -0.7875)/(0.413832 − 1.70361e-24) = -0.748195. 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 q = 0.17325; substituting back gives Y_out = 0.010327408.
- The interpolation error in the input is |q_line − q| = 0.921445. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.008778308 and 0.011876508. Sort the roots even if the response decreases: 0.160513 ≤ q ≤ 0.184565.
- The lower and upper input deviations are 0.0127369 and 0.0113146. 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 q ≈ 0.17325; compatible interval [0.160513, 0.184565]. Endpoint interpolation gives -0.748195.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Special-relativity postulates: infer an input with uncertainty
The special-relativity postulates are two claims about inertial frames: every law of physics takes the same form in all of them, and light in vacuum travels at the same speed c in all of them, whatever the source or the receiver is doing. Use the supplied topic model Y_out = (1+b)/max(1-b,0.05), where Y_out is RATIO (1+β)/(1−β) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. b is Rocket speed v/c, measured in c. Keep len (Rocket length in station frame) = 240 m fixed. For this exercise use only 0.0584375 c ≤ b ≤ 0.491563 c 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 = 1.952656, with a bounded reading uncertainty ±0.12428. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown b by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000433125 c, 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.
RATIO (1+β)/(1−β) against b
Predictions from the supplied Special-relativity postulates 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): RATIO (1+β)/(1−β)
Horizontal axis (x): b: Rocket speed v/c / c
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Rocket speed v/c / c | RATIO (1+β)/(1−β) |
|---|---|
| 0.0584375 | 1.1241287753070028 |
| 0.16671875 | 1.400150009375586 |
| 0.275 | 1.7586206896551724 |
| 0.38328125 | 2.242969343805422 |
| 0.4915625 | 2.9336201598033194 |
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(b) = 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(0.0584375) = 1.12413 and Y_out(0.491563) = 2.93362. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives b_line = 0.0584375 + (1.95266 − 1.12413)(0.491563 − 0.0584375)/(2.93362 − 1.12413) = 0.256756 c. 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 b = 0.322644 c; substituting back gives Y_out = 1.952656.
- The interpolation error in the input is |b_line − b| = 0.0658876 c. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.828376 and 2.076936. Sort the roots even if the response decreases: 0.29288 c ≤ b ≤ 0.350003 c.
- The lower and upper input deviations are 0.0297633 c and 0.027359 c. 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 b ≈ 0.322644 c; compatible interval [0.29288, 0.350003] c. Endpoint interpolation gives 0.256756 c.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Relativistic Time, Length, Momentum, Energy: infer an input with uncertainty
Relativistic mechanics rescales time, length, momentum and energy by the Lorentz factor γ = 1/√(1 − v²/c²): intervals between separated events stretch, lengths along the motion shrink, and p = γmv with E = γmc². Use the supplied topic model Y_out = 2.20/sqrt(1-b^2), where Y_out is 2.20 μs MUON IN LAB (μs). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. b is Speed v = βc, measured in c. Keep no other adjustable input fixed. For this exercise use only 0.100937 c ≤ b ≤ 0.849062 c 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 = 2.6495955 μs, with a bounded reading uncertainty ±0.065745 μs. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown b by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000748125 c, 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.
2.20 μs MUON IN LAB against b
Predictions from the supplied Relativistic Time, Length, Momentum, Energy 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): 2.20 μs MUON IN LAB / μs
Horizontal axis (x): b: Speed v = βc / c
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Speed v = βc / c | 2.20 μs MUON IN LAB / μs |
|---|---|
| 0.1009375 | 2.2112935879568183 |
| 0.28796875 | 2.297314694973386 |
| 0.475 | 2.500040354799399 |
| 0.66203125 | 2.9353832753557856 |
| 0.8490624999999999 | 4.164360410774019 |
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(b) = 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(0.100937) = 2.21129 μs and Y_out(0.849062) = 4.16436 μs. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives b_line = 0.100937 + (2.6496 − 2.21129)(0.849062 − 0.100937)/(4.16436 − 2.21129) = 0.26883 c. 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 b = 0.557294 c; substituting back gives Y_out = 2.6495955 μs.
- The interpolation error in the input is |b_line − b| = 0.288464 c. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 2.5838505 μs and 2.7153405 μs. Sort the roots even if the response decreases: 0.524448 c ≤ b ≤ 0.586138 c.
- The lower and upper input deviations are 0.0328457 c and 0.028844 c. 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 b ≈ 0.557294 c; compatible interval [0.524448, 0.586138] c. Endpoint interpolation gives 0.26883 c.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Photons and the photoelectric effect: infer an input with uncertainty
A photon is the quantum of an electromagnetic wave of frequency f: a massless packet carrying energy E = hf and momentum p = h/λ, absorbed or scattered whole rather than drained gradually. Use the supplied topic model Y_out = 1239.8/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 f (Photon frequency f) = 7.4 ×10¹⁴ Hz fixed. For this exercise use only 1.76563 eV ≤ phi ≤ 3.73438 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 = 417.92479 nm, with a bounded reading uncertainty ±12.889 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 0.000196875 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.
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
Horizontal axis (x): phi: Work function φ / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| phi: Work function φ / eV | THRESHOLD λ₀ = hc/φ / nm |
|---|---|
| 1.765625 | 702.187610619469 |
| 2.2578125 | 549.1155709342561 |
| 2.75 | 450.8363636363636 |
| 3.2421875 | 382.3961445783132 |
| 3.734375 | 331.99665271966524 |
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
- Endpoint predictions are Y_out(1.76563) = 702.188 nm and Y_out(3.73438) = 331.997 nm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives phi_line = 1.76563 + (417.925 − 702.188)(3.73438 − 1.76563)/(331.997 − 702.188) = 3.27739 eV. 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 phi = 2.96656 eV; substituting back gives Y_out = 417.92479 nm.
- The interpolation error in the input is |phi_line − phi| = 0.310829 eV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 405.03579 nm and 430.81379 nm. Sort the roots even if the response decreases: 2.87781 eV ≤ phi ≤ 3.06096 eV.
- The lower and upper input deviations are 0.088753 eV and 0.0944016 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 ≈ 2.96656 eV; compatible interval [2.87781, 3.06096] eV. Endpoint interpolation gives 3.27739 eV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.