University Quantum Mechanics I · topic UQMI-08.06
Barriers that are not rectangular: stacking slabs into an integral
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.
Smooth barriers: stacking slabs into an integral: predict a finite change
For a smooth barrier the transmission probability is T ≈ e^(−G), where the barrier exponent G = 2∫κ(x) dx adds up the local decay constant κ(x) = √(2m(V(x) − E))/ħ across the forbidden region between the two turning points. Use the supplied topic model Y_out = 2*b*sqrt(1-en/3), where Y_out is WIDTH x₂ − x₁ (nm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. en is Electron energy E, measured in eV. Keep b (Half-width at the base b) = 0.5 nm; n (Number of slabs N) = 4 fixed. For this exercise use only 0.486875 eV ≤ en ≤ 2.61312 eV 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 en = 0.95465 eV and 1.25232 eV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δen across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
WIDTH x₂ − x₁ against en
Predictions from the supplied Smooth barriers: stacking slabs into an integral 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): WIDTH x₂ − x₁ / nm
Horizontal axis (x): en: Electron energy E / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| en: Electron energy E / eV | WIDTH x₂ − x₁ / nm |
|---|---|
| 0.486875 | 0.9152640784677029 |
| 1.0184374999999999 | 0.8127243280063255 |
| 1.5499999999999998 | 0.695221787153807 |
| 2.0815625 | 0.5533044671185416 |
| 2.6131249999999997 | 0.35910769044025415 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of en separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At en = 0.95465 eV, substitute into Y_out = 2*b*sqrt(1-en/3): Y₀ = 0.825702 nm.
- At en = 1.25232 eV, the same fixed-parameter model gives Y₁ = 0.763255 nm.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.0624466 nm; the input change is Δen = 0.297675 eV.
- Average sensitivity is ΔY_out/Δen = -0.209781 nm per eV. 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 = 0.825702 nm, Y₁ = 0.763255 nm; ΔY_out = -0.0624466 nm; average sensitivity = -0.209781 nm per eV.
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.
Smooth barriers: stacking slabs into an integral: infer an input with uncertainty
For a smooth barrier the transmission probability is T ≈ e^(−G), where the barrier exponent G = 2∫κ(x) dx adds up the local decay constant κ(x) = √(2m(V(x) − E))/ħ across the forbidden region between the two turning points. Use the supplied topic model Y_out = 2*b*sqrt(1-en/3), where Y_out is WIDTH x₂ − x₁ (nm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. en is Electron energy E, measured in eV. Keep b (Half-width at the base b) = 0.5 nm; n (Number of slabs N) = 4 fixed. For this exercise use only 0.486875 eV ≤ en ≤ 2.61312 eV 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 = 0.63668739 nm, with a bounded reading uncertainty ±0.041637 nm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown en by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000212625 eV, 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.
WIDTH x₂ − x₁ against en
Predictions from the supplied Smooth barriers: stacking slabs into an integral 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): WIDTH x₂ − x₁ / nm
Horizontal axis (x): en: Electron energy E / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| en: Electron energy E / eV | WIDTH x₂ − x₁ / nm |
|---|---|
| 0.486875 | 0.9152640784677029 |
| 1.0184374999999999 | 0.8127243280063255 |
| 1.5499999999999998 | 0.695221787153807 |
| 2.0815625 | 0.5533044671185416 |
| 2.6131249999999997 | 0.35910769044025415 |
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(en) = 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(0.486875) = 0.915264 nm and Y_out(2.61312) = 0.359108 nm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives en_line = 0.486875 + (0.636687 − 0.915264)(2.61312 − 0.486875)/(0.359108 − 0.915264) = 1.55191 eV. 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 en = 1.78389 eV; substituting back gives Y_out = 0.63668739 nm.
- The interpolation error in the input is |en_line − en| = 0.231982 eV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.59505039 nm and 0.67832439 nm. Sort the roots even if the response decreases: 1.61963 eV ≤ en ≤ 1.93775 eV.
- The lower and upper input deviations are 0.164259 eV and 0.153858 eV. 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 en ≈ 1.78389 eV; compatible interval [1.61963, 1.93775] eV. Endpoint interpolation gives 1.55191 eV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.