University Physics V · topic UPV-12.06
Screening, shell structure, and the periodic table
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.
Screening and shell structure: predict a finite change
Screening is the reduction of the nuclear charge felt by an outer electron by the electrons lying between it and the nucleus, leaving an effective charge Z_eff(r) that falls from Z at the nucleus towards the net core charge far out. Use the supplied topic model Y_out = (z-1)*exp(-1)/max(d,0.05), where Y_out is CORE PULL AT r = d (Ha). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ; max(a,b) selects the larger value. d is Core screening radius d, measured in a₀. Keep z (Core charge at the nucleus Z) = 11 e; l (Orbital quantum number l) = 0 fixed. For this exercise use only 0.338125 a₀ ≤ d ≤ 1.36187 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 d = 0.56335 a₀ and 0.706675 a₀. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δd across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
CORE PULL AT r = d against d
Predictions from the supplied Screening and shell structure 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): CORE PULL AT r = d / Ha
Horizontal axis (x): d: Core screening radius d / a₀
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Core screening radius d / a₀ | CORE PULL AT r = d / Ha |
|---|---|
| 0.338125 | 10.879983472722879 |
| 0.5940624999999999 | 6.192605006568204 |
| 0.85 | 4.327993425546381 |
| 1.1059375 | 3.326403537012194 |
| 1.361875 | 2.7012717112175664 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of d separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At d = 0.56335 a₀, substitute into Y_out = (z-1)*exp(-1)/max(d,0.05): Y₀ = 6.53021 Ha.
- At d = 0.706675 a₀, the same fixed-parameter model gives Y₁ = 5.20578 Ha.
- Signed change: ΔY_out = Y₁ − Y₀ = -1.32443 Ha; the input change is Δd = 0.143325 a₀.
- Average sensitivity is ΔY_out/Δd = -9.24076 Ha 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 = 6.53021 Ha, Y₁ = 5.20578 Ha; ΔY_out = -1.32443 Ha; average sensitivity = -9.24076 Ha 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.
Screening and shell structure: infer an input with uncertainty
Screening is the reduction of the nuclear charge felt by an outer electron by the electrons lying between it and the nucleus, leaving an effective charge Z_eff(r) that falls from Z at the nucleus towards the net core charge far out. Use the supplied topic model Y_out = (z-1)*exp(-1)/max(d,0.05), where Y_out is CORE PULL AT r = d (Ha). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ; max(a,b) selects the larger value. d is Core screening radius d, measured in a₀. Keep z (Core charge at the nucleus Z) = 11 e; l (Orbital quantum number l) = 0 fixed. For this exercise use only 0.338125 a₀ ≤ d ≤ 1.36187 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 = 3.8216774 Ha, with a bounded reading uncertainty ±0.16806 Ha. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown d by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000102375 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.
CORE PULL AT r = d against d
Predictions from the supplied Screening and shell structure 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): CORE PULL AT r = d / Ha
Horizontal axis (x): d: Core screening radius d / a₀
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Core screening radius d / a₀ | CORE PULL AT r = d / Ha |
|---|---|
| 0.338125 | 10.879983472722879 |
| 0.5940624999999999 | 6.192605006568204 |
| 0.85 | 4.327993425546381 |
| 1.1059375 | 3.326403537012194 |
| 1.361875 | 2.7012717112175664 |
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(d) = 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(0.338125) = 10.88 Ha and Y_out(1.36187) = 2.70127 Ha. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives d_line = 0.338125 + (3.82168 − 10.88)(1.36187 − 0.338125)/(2.70127 − 10.88) = 1.22163 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 d = 0.962612 a₀; substituting back gives Y_out = 3.8216774 Ha.
- The interpolation error in the input is |d_line − d| = 0.259018 a₀. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 3.6536174 Ha and 3.9897374 Ha. Sort the roots even if the response decreases: 0.922064 a₀ ≤ d ≤ 1.00689 a₀.
- The lower and upper input deviations are 0.0405482 a₀ and 0.0442785 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 d ≈ 0.962612 a₀; compatible interval [0.922064, 1.00689] a₀. Endpoint interpolation gives 1.22163 a₀.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.