University Quantum Mechanics I · topic UQMI-15.01
The Coulomb potential and the one-electron model of hydrogen
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.
Further practice
Work through a fresh case before tackling the extension.
The Coulomb potential and the one-electron model of hydrogen: predict a finite change
The one-electron model of hydrogen is a single spinless, non-relativistic electron moving in the Coulomb potential energy V(r) = −e²/(4πε₀r) of a fixed point proton, so the whole atom is H = −(ħ²/2mₑ)∇² − e²/(4πε₀r). Use the supplied topic model Y_out = 13.606/(mu*b^2), where Y_out is KINETIC ⟨T⟩ (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. b is Cloud size b, measured in a₀. Keep mu (Reduced mass μ) = 1 mₑ fixed. For this exercise use only 1.05062 a₀ ≤ b ≤ 3.64937 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. Calculate Y_out at b = 1.62235 a₀ and 1.98617 a₀. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δb across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
KINETIC ⟨T⟩ against b
Predictions from the supplied The Coulomb potential and the one-electron model of hydrogen 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): KINETIC ⟨T⟩ / eV
Horizontal axis (x): b: Cloud size b / a₀
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Cloud size b / a₀ | KINETIC ⟨T⟩ / eV |
|---|---|
| 1.050625 | 12.326364473145468 |
| 1.7003124999999997 | 4.706228087565529 |
| 2.3499999999999996 | 2.463739248528747 |
| 2.9996874999999994 | 1.5120927806998343 |
| 3.649374999999999 | 1.0216296330363415 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of b separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At b = 1.62235 a₀, substitute into Y_out = 13.606/(mu*b^2): Y₀ = 5.16941 eV.
- At b = 1.98617 a₀, the same fixed-parameter model gives Y₁ = 3.44902 eV.
- Signed change: ΔY_out = Y₁ − Y₀ = -1.7204 eV; the input change is Δb = 0.363825 a₀.
- Average sensitivity is ΔY_out/Δb = -4.72864 eV per a₀. Its negative sign means that this response decreases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 5.16941 eV, Y₁ = 3.44902 eV; ΔY_out = -1.7204 eV; average sensitivity = -4.72864 eV per a₀.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
The Coulomb potential and the one-electron model of hydrogen: infer an input with uncertainty
The one-electron model of hydrogen is a single spinless, non-relativistic electron moving in the Coulomb potential energy V(r) = −e²/(4πε₀r) of a fixed point proton, so the whole atom is H = −(ħ²/2mₑ)∇² − e²/(4πε₀r). Use the supplied topic model Y_out = 13.606/(mu*b^2), where Y_out is KINETIC ⟨T⟩ (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. b is Cloud size b, measured in a₀. Keep mu (Reduced mass μ) = 1 mₑ fixed. For this exercise use only 1.05062 a₀ ≤ b ≤ 3.64937 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 = 1.9583259 eV, with a bounded reading uncertainty ±0.1405 eV. 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 0.000259875 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.
KINETIC ⟨T⟩ against b
Predictions from the supplied The Coulomb potential and the one-electron model of hydrogen 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): KINETIC ⟨T⟩ / eV
Horizontal axis (x): b: Cloud size b / a₀
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Cloud size b / a₀ | KINETIC ⟨T⟩ / eV |
|---|---|
| 1.050625 | 12.326364473145468 |
| 1.7003124999999997 | 4.706228087565529 |
| 2.3499999999999996 | 2.463739248528747 |
| 2.9996874999999994 | 1.5120927806998343 |
| 3.649374999999999 | 1.0216296330363415 |
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(1.05062) = 12.3264 eV and Y_out(3.64937) = 1.02163 eV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives b_line = 1.05062 + (1.95833 − 12.3264)(3.64937 − 1.05062)/(1.02163 − 12.3264) = 3.43405 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 b = 2.63586 a₀; substituting back gives Y_out = 1.9583259 eV.
- The interpolation error in the input is |b_line − b| = 0.798183 a₀. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.8178259 eV and 2.0988259 eV. Sort the roots even if the response decreases: 2.54611 a₀ ≤ b ≤ 2.73583 a₀.
- The lower and upper input deviations are 0.0897533 a₀ and 0.0999674 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 b ≈ 2.63586 a₀; compatible interval [2.54611, 2.73583] a₀. Endpoint interpolation gives 3.43405 a₀.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.