University Physics IV · unit UPIV-10
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.
Central potentials and the 3D Schrödinger equation: infer an input with uncertainty
A central potential depends only on the distance r between two particles, which lets the two-body problem reduce exactly to one particle of reduced mass μ moving in V(r), with a wavefunction that factors as ψ = R(r)Y_lm(θ, φ). Use the supplied topic model Y_out = 10^k, where Y_out is MASS RATIO M/m (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. k is Mass ratio, log₁₀(M/m), dimensionless. Keep t (Orbital phase) = 45 ° fixed. For this exercise use only 0.3825 ≤ k ≤ 3.2175 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 = 129.37489, with a bounded reading uncertainty ±19.044. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown k by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0002835, 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.
MASS RATIO M/m against k
Predictions from the supplied Central potentials and the 3D Schrödinger equation 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): MASS RATIO M/m
Horizontal axis (x): k: Mass ratio, log₁₀(M/m)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| k: Mass ratio, log₁₀(M/m) | MASS RATIO M/m |
|---|---|
| 0.3825 | 2.4126815325916335 |
| 1.09125 | 12.338148697760213 |
| 1.8 | 63.09573444801933 |
| 2.5087499999999996 | 322.6636185911479 |
| 3.2175 | 1650.061001320227 |
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(k) = 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(0.3825) = 2.41268 and Y_out(3.2175) = 1650.06. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives k_line = 0.3825 + (129.375 − 2.41268)(3.2175 − 0.3825)/(1650.06 − 2.41268) = 0.600956. 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 k = 2.11185; substituting back gives Y_out = 129.37489.
- The interpolation error in the input is |k_line − k| = 1.51089. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 110.33089 and 148.41889. Sort the roots even if the response decreases: 2.0427 ≤ k ≤ 2.17149.
- The lower and upper input deviations are 0.0691529 and 0.0596392. 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 k ≈ 2.11185; compatible interval [2.0427, 2.17149]. Endpoint interpolation gives 0.600956.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The angular equation and spherical harmonics: infer an input with uncertainty
The angular equation is what is left of a central-potential Schrödinger equation once the radial part is separated off; its normalisable solutions are the spherical harmonics Y_l^m(θ, φ), eigenfunctions of L² with eigenvalue l(l+1)ħ² and of L_z with eigenvalue mħ. Use the supplied topic model Y_out = l*(l+1)-m*m, where Y_out is L⊥² = l(l+1) − m² (ħ²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. m is Magnetic quantum number m, dimensionless. Keep l (Orbital quantum number l) = 2 fixed. For this exercise use only -2 ≤ m ≤ 0, 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 = 5 ħ², with a bounded reading uncertainty ±0.32 ħ². Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. Determine every allowed value of m 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.
L⊥² = l(l+1) − m² against m
Predictions from the supplied The angular equation and spherical harmonics 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): L⊥² = l(l+1) − m² / ħ²
Horizontal axis (x): m: Magnetic quantum number m
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| m: Magnetic quantum number m | L⊥² = l(l+1) − m² / ħ² |
|---|---|
| -2 | 2 |
| -1 | 5 |
| 0 | 6 |
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) = 2 ħ² and Y_out(0) = 6 ħ². The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives m_line = -2 + (5 − 2)(0 − -2)/(6 − 2) = -0.5. 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 m = -1; only the allowed settings -1 lie inside the stated reading bound.
- The interpolation error in the input is |m_line − m| = 0.5.
- The compatible allowed values span -1 to -1. 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 m: -1. Endpoint interpolation predicts -0.5; check admissibility against the discrete settings.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The radial equation and effective potential: infer an input with uncertainty
The radial equation is the one-dimensional Schrödinger equation obeyed by u(r) = rR(r) on the half-line r ≥ 0, in an effective potential that adds the centrifugal barrier ħ²l(l+1)/2μr² to the actual central potential. Use the supplied topic model Y_out = sign(max(n^2 - l*(l+1), 0))*(n^2 - n*sqrt(max(n^2 - l*(l+1), 0))), where Y_out is INNER TURNING r− (a₀). 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. n is Principal quantum number n, dimensionless. Keep l (Orbital quantum number l) = 1 fixed. For this exercise use only 2 ≤ n ≤ 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 = 1.0627461 a₀, with a bounded reading uncertainty ±0.011056 a₀. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. Determine every allowed value of n 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.
INNER TURNING r− against n
Predictions from the supplied The radial equation and effective potential 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): INNER TURNING r− / a₀
Horizontal axis (x): n: Principal quantum number n
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| n: Principal quantum number n | INNER TURNING r− / a₀ |
|---|---|
| 2 | 1.1715728752538097 |
| 3 | 1.0627460668062279 |
| 4 | 1.0333704529042347 |
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) = 1.17157 a₀ and Y_out(4) = 1.03337 a₀. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives n_line = 2 + (1.06275 − 1.17157)(4 − 2)/(1.03337 − 1.17157) = 3.57489. 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 n = 3; only the allowed settings 3 lie inside the stated reading bound.
- The interpolation error in the input is |n_line − n| = 0.57489.
- 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 n: 3. Endpoint interpolation predicts 3.57489; check admissibility against the discrete settings.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Solving the Coulomb radial equation: infer an input with uncertainty
Quantisation from normalisability is the demand that the power series in the hydrogen radial solution terminate, because an unterminated series rebuilds the exponentially growing solution and destroys the normalisation integral. Use the supplied topic model Y_out = -13.606/max(n*n,0.25), where Y_out is TRIAL ENERGY (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. n is Trial n (rho0 = 2n), dimensionless. Keep l (Orbital number l) = 0 fixed. For this exercise use only 1.31875 ≤ n ≤ 3.68125 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 = -1.7862887 eV, with a bounded reading uncertainty ±0.11734 eV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown n by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00023625, 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 ENERGY against n
Predictions from the supplied Solving the Coulomb radial equation 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 ENERGY / eV
Horizontal axis (x): n: Trial n (rho0 = 2n)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| n: Trial n (rho0 = 2n) | TRIAL ENERGY / eV |
|---|---|
| 1.31875 | -7.823579883650411 |
| 1.909375 | -3.732054719664846 |
| 2.5 | -2.17696 |
| 3.090625 | -1.4244192691906215 |
| 3.68125 | -1.004014170373082 |
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(n) = 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(1.31875) = -7.82358 eV and Y_out(3.68125) = -1.00401 eV. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives n_line = 1.31875 + (-1.78629 − -7.82358)(3.68125 − 1.31875)/(-1.00401 − -7.82358) = 3.41025. 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 n = 2.75987; substituting back gives Y_out = -1.7862887 eV.
- The interpolation error in the input is |n_line − n| = 0.650372. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -1.9036287 eV and -1.6689487 eV. Sort the roots even if the response decreases: 2.67346 ≤ n ≤ 2.85525.
- The lower and upper input deviations are 0.0864124 and 0.0953724. 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 n ≈ 2.75987; compatible interval [2.67346, 2.85525]. Endpoint interpolation gives 3.41025.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.