University Physics V · topic UPV-05.02
Boundary conditions as the infinite-wall limit
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 infinite-wall limit: predict a finite change
A Dirichlet boundary condition sets the wavefunction itself to zero at a wall, and it is what the finite well's matching condition ψ′(0) = κψ(0) becomes once the barrier is high enough that the decay length 1/κ is negligible against the width. Use the supplied topic model Y_out = 0.0381*(pi/(w+2*d))*(pi/(w+2*d))/max(sin(pi*d/(w+2*d))*sin(pi*d/(w+2*d)),0.0001), where Y_out is BARRIER DEPTH V₀ (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; unmarked trigonometric functions use radians. d is Extrapolation length δₑ, measured in nm. Keep w (Well width L) = 1 nm fixed. For this exercise use only 0.04975 nm ≤ d ≤ 0.27025 nm 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 = 0.09826 nm and 0.12913 nm. 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.
BARRIER DEPTH V₀ against d
Predictions from the supplied The infinite-wall limit 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): BARRIER DEPTH V₀ / eV
Horizontal axis (x): d: Extrapolation length δₑ / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Extrapolation length δₑ / nm | BARRIER DEPTH V₀ / eV |
|---|---|
| 0.049749999999999996 | 15.49765533902073 |
| 0.104875 | 3.5509577665029566 |
| 0.16 | 1.562354111168292 |
| 0.21512499999999998 | 0.887382634827974 |
| 0.27025 | 0.5778551797077442 |
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 = 0.09826 nm, substitute into Y_out = 0.0381*(pi/(w+2*d))*(pi/(w+2*d))/max(sin(pi*d/(w+2*d))*sin(pi*d/(w+2*d)),0.0001): Y₀ = 4.03486 eV.
- At d = 0.12913 nm, the same fixed-parameter model gives Y₁ = 2.36576 eV.
- Signed change: ΔY_out = Y₁ − Y₀ = -1.6691 eV; the input change is Δd = 0.03087 nm.
- Average sensitivity is ΔY_out/Δd = -54.0686 eV per nm. 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 = 4.03486 eV, Y₁ = 2.36576 eV; ΔY_out = -1.6691 eV; average sensitivity = -54.0686 eV per nm.
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 infinite-wall limit: infer an input with uncertainty
A Dirichlet boundary condition sets the wavefunction itself to zero at a wall, and it is what the finite well's matching condition ψ′(0) = κψ(0) becomes once the barrier is high enough that the decay length 1/κ is negligible against the width. Use the supplied topic model Y_out = 0.0381*(pi/(w+2*d))*(pi/(w+2*d))/max(sin(pi*d/(w+2*d))*sin(pi*d/(w+2*d)),0.0001), where Y_out is BARRIER DEPTH V₀ (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; unmarked trigonometric functions use radians. d is Extrapolation length δₑ, measured in nm. Keep w (Well width L) = 1 nm fixed. For this exercise use only 0.04975 nm ≤ d ≤ 0.27025 nm 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 = 1.1916343 eV, with a bounded reading uncertainty ±0.092067 eV. 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 0.00002205 nm, 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.
BARRIER DEPTH V₀ against d
Predictions from the supplied The infinite-wall limit 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): BARRIER DEPTH V₀ / eV
Horizontal axis (x): d: Extrapolation length δₑ / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Extrapolation length δₑ / nm | BARRIER DEPTH V₀ / eV |
|---|---|
| 0.049749999999999996 | 15.49765533902073 |
| 0.104875 | 3.5509577665029566 |
| 0.16 | 1.562354111168292 |
| 0.21512499999999998 | 0.887382634827974 |
| 0.27025 | 0.5778551797077442 |
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(0.04975) = 15.4977 eV and Y_out(0.27025) = 0.577855 eV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives d_line = 0.04975 + (1.19163 − 15.4977)(0.27025 − 0.04975)/(0.577855 − 15.4977) = 0.261179 nm. 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 = 0.184255 nm; substituting back gives Y_out = 1.1916343 eV.
- The interpolation error in the input is |d_line − d| = 0.0769239 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.0995673 eV and 1.2837013 eV. Sort the roots even if the response decreases: 0.177228 nm ≤ d ≤ 0.19218 nm.
- The lower and upper input deviations are 0.00702744 nm and 0.00792459 nm. 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 ≈ 0.184255 nm; compatible interval [0.177228, 0.19218] nm. Endpoint interpolation gives 0.261179 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.