University Quantum Mechanics II · topic UQMII-02.02
The dimensionless radial equation for the Coulomb potential
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 dimensionless Coulomb radial equation: predict a finite change
The dimensionless Coulomb radial equation is the radial equation for u = rR with r measured in units of 1/κ = ħ/√(−2μE), which leaves l and one pure number, ρ₀ = 2Z/(κa), as its only parameters, the unknown energy being carried inside ρ₀. Use the supplied topic model Y_out = -54.393/rho0^2, where Y_out is HYDROGEN ENERGY −4ℛ/ρ₀² (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. rho0 is Coulomb parameter ρ₀, dimensionless. Keep l (Orbital quantum number l) = 1 fixed. For this exercise use only 1.95625 ≤ rho0 ≤ 9.04375 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 rho0 = 3.5155 and 4.50775. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δrho0 across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
HYDROGEN ENERGY −4ℛ/ρ₀² against rho0
Predictions from the supplied The dimensionless 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): HYDROGEN ENERGY −4ℛ/ρ₀² / eV
Horizontal axis (x): rho0: Coulomb parameter ρ₀
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rho0: Coulomb parameter ρ₀ | HYDROGEN ENERGY −4ℛ/ρ₀² / eV |
|---|---|
| 1.9562499999999998 | -14.21327971092897 |
| 3.7281249999999995 | -3.91347065762913 |
| 5.5 | -1.7981157024793388 |
| 7.271875 | -1.0286087222934965 |
| 9.04375 | -0.665037164325877 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of rho0 separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At rho0 = 3.5155, substitute into Y_out = -54.393/rho0^2: Y₀ = -4.40118 eV.
- At rho0 = 4.50775, the same fixed-parameter model gives Y₁ = -2.67685 eV.
- Signed change: ΔY_out = Y₁ − Y₀ = 1.72433 eV; the input change is Δrho0 = 0.99225.
- Average sensitivity is ΔY_out/Δrho0 = 1.7378 eV per unit input. Its positive sign means that this response increases 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 = -4.40118 eV, Y₁ = -2.67685 eV; ΔY_out = 1.72433 eV; average sensitivity = 1.7378 eV per unit input.
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 dimensionless Coulomb radial equation: infer an input with uncertainty
The dimensionless Coulomb radial equation is the radial equation for u = rR with r measured in units of 1/κ = ħ/√(−2μE), which leaves l and one pure number, ρ₀ = 2Z/(κa), as its only parameters, the unknown energy being carried inside ρ₀. Use the supplied topic model Y_out = -54.393/rho0^2, where Y_out is HYDROGEN ENERGY −4ℛ/ρ₀² (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. rho0 is Coulomb parameter ρ₀, dimensionless. Keep l (Orbital quantum number l) = 1 fixed. For this exercise use only 1.95625 ≤ rho0 ≤ 9.04375 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 = -1.3793535 eV, with a bounded reading uncertainty ±0.10715 eV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown rho0 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00070875, 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.
HYDROGEN ENERGY −4ℛ/ρ₀² against rho0
Predictions from the supplied The dimensionless 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): HYDROGEN ENERGY −4ℛ/ρ₀² / eV
Horizontal axis (x): rho0: Coulomb parameter ρ₀
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rho0: Coulomb parameter ρ₀ | HYDROGEN ENERGY −4ℛ/ρ₀² / eV |
|---|---|
| 1.9562499999999998 | -14.21327971092897 |
| 3.7281249999999995 | -3.91347065762913 |
| 5.5 | -1.7981157024793388 |
| 7.271875 | -1.0286087222934965 |
| 9.04375 | -0.665037164325877 |
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(rho0) = 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(1.95625) = -14.2133 eV and Y_out(9.04375) = -0.665037 eV. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives rho0_line = 1.95625 + (-1.37935 − -14.2133)(9.04375 − 1.95625)/(-0.665037 − -14.2133) = 8.67007. 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 rho0 = 6.27963; substituting back gives Y_out = -1.3793535 eV.
- The interpolation error in the input is |rho0_line − rho0| = 2.39044. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -1.4865035 eV and -1.2722035 eV. Sort the roots even if the response decreases: 6.04907 ≤ rho0 ≤ 6.53873.
- The lower and upper input deviations are 0.230556 and 0.259102. 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 rho0 ≈ 6.27963; compatible interval [6.04907, 6.53873]. Endpoint interpolation gives 8.67007.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.