University Physics IV · topic UPIV-07.08
The rectangular barrier: tunnelling and resonant transmission
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 rectangular barrier and tunnelling: predict a finite change
Barrier tunnelling is the transmission of a particle through a region where its energy lies below the potential, at a rate fixed by how far the wavefunction's real exponential decays across the width. Use the supplied topic model Y_out = 1/(1+((exp(3.6225*sqrt(v)*d)-exp(-3.6225*sqrt(v)*d))/2)^2), where Y_out is T AT E = V₀/2 (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; exp(x) means eˣ. v is Barrier height V₀, measured in eV. Keep d (Barrier width L) = 0.7 nm fixed. For this exercise use only 0.765625 eV ≤ v ≤ 2.73438 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 v = 1.19875 eV and 1.47437 eV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δv across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
T AT E = V₀/2 against v
Predictions from the supplied The rectangular barrier and tunnelling 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): T AT E = V₀/2
Horizontal axis (x): v: Barrier height V₀ / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| v: Barrier height V₀ / eV | T AT E = V₀/2 |
|---|---|
| 0.765625 | 0.046199843900237764 |
| 1.2578125 | 0.01345682152108596 |
| 1.75 | 0.0048678257776563275 |
| 2.2421875 | 0.0020117631255674655 |
| 2.734375 | 0.0009115478881454766 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of v separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At v = 1.19875 eV, substitute into Y_out = 1/(1+((exp(3.6225*sqrt(v)*d)-exp(-3.6225*sqrt(v)*d))/2)^2): Y₀ = 0.015389.
- At v = 1.47437 eV, the same fixed-parameter model gives Y₁ = 0.00843012.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.00695891; the input change is Δv = 0.275625 eV.
- Average sensitivity is ΔY_out/Δv = -0.0252477 1 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 = 0.015389, Y₁ = 0.00843012; ΔY_out = -0.00695891; average sensitivity = -0.0252477 1 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.
The rectangular barrier and tunnelling: infer an input with uncertainty
Barrier tunnelling is the transmission of a particle through a region where its energy lies below the potential, at a rate fixed by how far the wavefunction's real exponential decays across the width. Use the supplied topic model Y_out = 1/(1+((exp(3.6225*sqrt(v)*d)-exp(-3.6225*sqrt(v)*d))/2)^2), where Y_out is T AT E = V₀/2 (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; exp(x) means eˣ. v is Barrier height V₀, measured in eV. Keep d (Barrier width L) = 0.7 nm fixed. For this exercise use only 0.765625 eV ≤ v ≤ 2.73438 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 = 0.0032558173, with a bounded reading uncertainty ±0.00035164. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown v by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000196875 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.
T AT E = V₀/2 against v
Predictions from the supplied The rectangular barrier and tunnelling 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): T AT E = V₀/2
Horizontal axis (x): v: Barrier height V₀ / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| v: Barrier height V₀ / eV | T AT E = V₀/2 |
|---|---|
| 0.765625 | 0.046199843900237764 |
| 1.2578125 | 0.01345682152108596 |
| 1.75 | 0.0048678257776563275 |
| 2.2421875 | 0.0020117631255674655 |
| 2.734375 | 0.0009115478881454766 |
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(v) = 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(0.765625) = 0.0461998 and Y_out(2.73438) = 0.000911548. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives v_line = 0.765625 + (0.00325582 − 0.0461998)(2.73438 − 0.765625)/(0.000911548 − 0.0461998) = 2.63247 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 v = 1.96656 eV; substituting back gives Y_out = 0.0032558173.
- The interpolation error in the input is |v_line − v| = 0.665904 eV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.0029041773 and 0.0036074573. Sort the roots even if the response decreases: 1.91016 eV ≤ v ≤ 2.03038 eV.
- The lower and upper input deviations are 0.0564056 eV and 0.0638143 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 v ≈ 1.96656 eV; compatible interval [1.91016, 2.03038] eV. Endpoint interpolation gives 2.63247 eV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.