University Physics V · unit UPV-11
The Hydrogen Atom
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.
The central-field Hamiltonian and its commuting set: infer an input with uncertainty
A central-field Hamiltonian is H = p²/2μ + V(r) with V depending only on the distance |r|; because it commutes with L² and L_z, its eigenvectors carry the labels |n l m⟩ and the three-dimensional eigenproblem splits into one half-line radial problem for each l. Use the supplied topic model Y_out = -13.606/max(a,0.05)^2, where Y_out is SPURIOUS E −ħ²/2μa² (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. a is Decay length a of e^(−r/a), measured in a₀. Keep k (Wavenumber k of sin(kr)) = 1.5 1/a₀ fixed. For this exercise use only 0.586875 a₀ ≤ a ≤ 2.71312 a₀ 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 = -3.8337213 eV, with a bounded reading uncertainty ±0.2978 eV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown a by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000212625 a₀, 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.
SPURIOUS E −ħ²/2μa² against a
Predictions from the supplied The central-field Hamiltonian and its commuting set 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): SPURIOUS E −ħ²/2μa² / eV
Horizontal axis (x): a: Decay length a of e^(−r/a) / a₀
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: Decay length a of e^(−r/a) / a₀ | SPURIOUS E −ħ²/2μa² / eV |
|---|---|
| 0.586875 | -39.50383397922925 |
| 1.1184375 | -10.876947353867415 |
| 1.65 | -4.99761248852158 |
| 2.1815625 | -2.8588748706482474 |
| 2.713125 | -1.8483782957810917 |
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(a) = 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(0.586875) = -39.5038 eV and Y_out(2.71312) = -1.84838 eV. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives a_line = 0.586875 + (-3.83372 − -39.5038)(2.71312 − 0.586875)/(-1.84838 − -39.5038) = 2.60102 a₀. 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 a = 1.88389 a₀; substituting back gives Y_out = -3.8337213 eV.
- The interpolation error in the input is |a_line − a| = 0.717133 a₀. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -4.1315213 eV and -3.5359213 eV. Sort the roots even if the response decreases: 1.81472 a₀ ≤ a ≤ 1.96162 a₀.
- The lower and upper input deviations are 0.069165 a₀ and 0.0777282 a₀. 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 a ≈ 1.88389 a₀; compatible interval [1.81472, 1.96162] a₀. Endpoint interpolation gives 2.60102 a₀.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Angular wavefunctions and the L², L_z eigenbasis: infer an input with uncertainty
The spherical harmonics Y_l^m(θ, φ) = ⟨n̂|l m⟩ are the simultaneous eigenfunctions of L² and L_z on the unit sphere, with eigenvalues ħ²l(l+1) and mħ, and they form a complete orthonormal basis for every square-integrable function of direction. Use the supplied topic model Y_out = deg(acos(clamp(m,-l,l)/max(sqrt(l*(l+1)),0.01))), where Y_out is TILT θ FROM z (°). 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; clamp(x,a,b) confines x to [a,b]; deg converts radians to degrees; unmarked trigonometric functions use radians. l is Angular-momentum label l, dimensionless. Keep m (Requested rung m (the ladder sticks at |m| = l)) = 2 fixed. For this exercise use only 2 ≤ l ≤ 4, at the specified test settings separated by 1. 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 = 54.73561 °, with a bounded reading uncertainty ±2.2536 °. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. Determine every allowed value of l consistent with that bound. Compare with a straight-line interpolation between the two endpoint predictions, and explain why a fractional setting may be inadmissible. Do not assume that the supplied calibration points prove monotonicity outside this branch.
TILT θ FROM z against l
Predictions from the supplied Angular wavefunctions and the L², L_z eigenbasis model, with all other inputs fixed as stated in the question. 3 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): TILT θ FROM z / °
Horizontal axis (x): l: Angular-momentum label l
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| l: Angular-momentum label l | TILT θ FROM z / ° |
|---|---|
| 2 | 35.26438968275464 |
| 3 | 54.735610317245346 |
| 4 | 63.43494882292201 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Enumerate the permitted settings in the stated range, evaluate Y_out at each, and retain every setting satisfying |Y_out − Y_measured| ≤ u. The plotted predictions are supplied calibration evidence.
Reveal the worked solution
- Endpoint predictions are Y_out(2) = 35.2644 ° and Y_out(4) = 63.4349 °. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives l_line = 2 + (54.7356 − 35.2644)(4 − 2)/(63.4349 − 35.2644) = 3.38238. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Evaluate the model at each permitted setting. The central reading corresponds to l = 3; only the allowed settings 3 lie inside the stated reading bound.
- The interpolation error in the input is |l_line − l| = 0.382381.
- The compatible allowed values span 3 to 3. Values between specified test settings are outside this exercise's input domain even if an interpolation passes through them.
- The quoted output bound restricts which specified settings remain compatible. It does not change physical restrictions on quantum numbers or counts, and the chosen grid alone does not establish physical quantisation.
Answer Compatible l: 3. Endpoint interpolation predicts 3.38238; check admissibility against the discrete settings.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Radial Hamiltonian and effective potential: infer an input with uncertainty
The radial Hamiltonian is the operator −(ℏ²/2μ)d²/dr² + V(r) + ℏ²l(l+1)/(2μr²) acting on u = rR on the half-line r > 0, made self-adjoint by the domain condition u(0) = 0. Use the supplied topic model Y_out = s*(s-1)-l*(l+1), where Y_out is INDICIAL RESIDUAL (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. s is Trial exponent s in u ~ r^s, dimensionless. Keep l (Orbital quantum number l) = 1; n (Principal quantum number n) = 2 fixed. For this exercise use only -2.25625 ≤ s ≤ 0.31625 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 = -0.84097165, with a bounded reading uncertainty ±0.20629. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown s by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00025725, 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.
INDICIAL RESIDUAL against s
Predictions from the supplied Radial Hamiltonian and effective potential 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): INDICIAL RESIDUAL
Horizontal axis (x): s: Trial exponent s in u ~ r^s
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Trial exponent s in u ~ r^s | INDICIAL RESIDUAL |
|---|---|
| -2.25625 | 5.346914062500001 |
| -1.6131250000000001 | 2.2152972656250007 |
| -0.97 | -0.08910000000000018 |
| -0.3268749999999998 | -1.5662777343750003 |
| 0.31625000000000014 | -2.2162359375 |
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(s) = 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(-2.25625) = 5.34691 and Y_out(0.31625) = -2.21624. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives s_line = -2.25625 + (-0.840972 − 5.34691)(0.31625 − -2.25625)/(-2.21624 − 5.34691) = -0.151527. 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 s = -0.687025; substituting back gives Y_out = -0.84097165.
- The interpolation error in the input is |s_line − s| = 0.535498. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -1.0472617 and -0.63468165. Sort the roots even if the response decreases: -0.770952 ≤ s ≤ -0.596694.
- The lower and upper input deviations are 0.0839267 and 0.0903307. 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 s ≈ -0.687025; compatible interval [-0.770952, -0.596694]. Endpoint interpolation gives -0.151527.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Radial wavefunctions and the Laguerre solution: infer an input with uncertainty
The radial wavefunction u_nl = r R_nl is the half-line eigenfunction of the radial Hamiltonian at fixed l: an r^(l+1) rise at the origin, an e^(−Zr/na₀) tail, and between them an associated Laguerre polynomial of degree n − l − 1. Use the supplied topic model Y_out = abs((2/(n*sqrt(n)))*rb*(1+(1-n)*rb/n+(1-n)*(2-n)*rb^2/(3*n^2)+(1-n)*(2-n)*(3-n)*rb^3/(18*n^3))*exp(-rb/n)), where Y_out is WALL AMPLITUDE |u(R)| (a_0^(-1/2)). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; exp(x) means eˣ; abs(x) means |x|. rb is Grid box radius R, measured in a₀. Keep n (Principal quantum number n) = 3 fixed. For this exercise use only 14.5725 a₀ ≤ rb ≤ 54.2625 a₀ 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 = 0.0031394565 a_0^(-1/2), with a bounded reading uncertainty ±0.00046292 a_0^(-1/2). Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown rb by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.003969 a₀, 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.
WALL AMPLITUDE |u(R)| against rb
Predictions from the supplied Radial wavefunctions and the Laguerre solution 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): WALL AMPLITUDE |u(R)| / a_0^(-1/2)
Horizontal axis (x): rb: Grid box radius R / a₀
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rb: Grid box radius R / a₀ | WALL AMPLITUDE |u(R)| / a_0^(-1/2) |
|---|---|
| 14.5725 | 0.30572928930086346 |
| 24.494999999999997 | 0.07807736936718887 |
| 34.4175 | 0.009076391329130584 |
| 44.339999999999996 | 0.0007615972842707709 |
| 54.262499999999996 | 0.000053312819680487595 |
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(rb) = 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(14.5725) = 0.305729 a_0^(-1/2) and Y_out(54.2625) = 0.0000533128 a_0^(-1/2). The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives rb_line = 14.5725 + (0.00313946 − 0.305729)(54.2625 − 14.5725)/(0.0000533128 − 0.305729) = 53.8618 a₀. 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 rb = 38.7834 a₀; substituting back gives Y_out = 0.0031394565 a_0^(-1/2).
- The interpolation error in the input is |rb_line − rb| = 15.0784 a₀. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.0026765365 a_0^(-1/2) and 0.0036023765 a_0^(-1/2). Sort the roots even if the response decreases: 38.2292 a₀ ≤ rb ≤ 39.4224 a₀.
- The lower and upper input deviations are 0.554158 a₀ and 0.639047 a₀. 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 rb ≈ 38.7834 a₀; compatible interval [38.2292, 39.4224] a₀. Endpoint interpolation gives 53.8618 a₀.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.