University Quantum Mechanics II · topic UQMII-10.05
Helium with an effective nuclear charge
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.
Helium with an effective nuclear charge: predict a finite change
An effective nuclear charge is the variational parameter ζ of a trial that puts both electrons in the hydrogenic 1s orbital of charge ζ instead of Z; minimising ⟨H⟩ over ζ picks the orbital size that best balances kinetic energy against attraction and repulsion. Use the supplied topic model Y_out = 27.2114*(zeta^2-3.375*zeta)+79.0144, where Y_out is ABOVE E₀ = −79.01 eV (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. zeta is Floating charge ζ (Z_eff), dimensionless. Keep no other adjustable input fixed. For this exercise use only 1.7055 ≤ zeta ≤ 2.2725 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 zeta = 1.83024 and 1.90962. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δzeta across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
ABOVE E₀ = −79.01 eV against zeta
Predictions from the supplied Helium with an effective nuclear charge 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): ABOVE E₀ = −79.01 eV / eV
Horizontal axis (x): zeta: Floating charge ζ (Z_eff)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| zeta: Floating charge ζ (Z_eff) | ABOVE E₀ = −79.01 eV / eV |
|---|---|
| 1.7054999999999998 | 1.5345032123499749 |
| 1.8472499999999998 | 2.22012334746249 |
| 1.9889999999999999 | 3.999264204399992 |
| 2.13075 | 6.871925783162496 |
| 2.2725 | 10.838108083749987 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of zeta separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At zeta = 1.83024, substitute into Y_out = 27.2114*(zeta^2-3.375*zeta)+79.0144: Y₀ = 2.08011 eV.
- At zeta = 1.90962, the same fixed-parameter model gives Y₁ = 2.86822 eV.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.788113 eV; the input change is Δzeta = 0.07938.
- Average sensitivity is ΔY_out/Δzeta = 9.92835 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 = 2.08011 eV, Y₁ = 2.86822 eV; ΔY_out = 0.788113 eV; average sensitivity = 9.92835 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.
Helium with an effective nuclear charge: infer an input with uncertainty
An effective nuclear charge is the variational parameter ζ of a trial that puts both electrons in the hydrogenic 1s orbital of charge ζ instead of Z; minimising ⟨H⟩ over ζ picks the orbital size that best balances kinetic energy against attraction and repulsion. Use the supplied topic model Y_out = 27.2114*(zeta^2-3.375*zeta)+79.0144, where Y_out is ABOVE E₀ = −79.01 eV (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. zeta is Floating charge ζ (Z_eff), dimensionless. Keep no other adjustable input fixed. For this exercise use only 1.7055 ≤ zeta ≤ 2.2725 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 = 5.1285135 eV, with a bounded reading uncertainty ±0.5391 eV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown zeta by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000567, 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.
ABOVE E₀ = −79.01 eV against zeta
Predictions from the supplied Helium with an effective nuclear charge 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): ABOVE E₀ = −79.01 eV / eV
Horizontal axis (x): zeta: Floating charge ζ (Z_eff)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| zeta: Floating charge ζ (Z_eff) | ABOVE E₀ = −79.01 eV / eV |
|---|---|
| 1.7054999999999998 | 1.5345032123499749 |
| 1.8472499999999998 | 2.22012334746249 |
| 1.9889999999999999 | 3.999264204399992 |
| 2.13075 | 6.871925783162496 |
| 2.2725 | 10.838108083749987 |
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(zeta) = 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(1.7055) = 1.5345 eV and Y_out(2.2725) = 10.8381 eV. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives zeta_line = 1.7055 + (5.12851 − 1.5345)(2.2725 − 1.7055)/(10.8381 − 1.5345) = 1.92453. 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 zeta = 2.05137; substituting back gives Y_out = 5.1285135 eV.
- The interpolation error in the input is |zeta_line − zeta| = 0.126836. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 4.5894135 eV and 5.6676135 eV. Sort the roots even if the response decreases: 2.02304 ≤ zeta ≤ 2.07764.
- The lower and upper input deviations are 0.0283259 and 0.0262748. 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 zeta ≈ 2.05137; compatible interval [2.02304, 2.07764]. Endpoint interpolation gives 1.92453.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.