University Quantum Mechanics I · topic UQMI-08.08
Scanning tunnelling microscopy, tunnel diodes, and gate leakage
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.
Scanning tunnelling microscopy, tunnel diodes, gate leakage: predict a finite change
In device tunnelling each electron is taken to cross alone and at fixed energy, so the current counts filled states that face empty states at the same energy and weights each by a transmission falling as e^(−2κz) with the gap z. Use the supplied topic model Y_out = 10/(2*5.1232*sqrt(phi)), where Y_out is HEIGHT FOR 1% OF CURRENT (pm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. phi is Barrier height φ, measured in eV. Keep rtip (Apex radius R) = 0.5 nm; z (Gap z) = 0.6 nm fixed. For this exercise use only 1.53125 eV ≤ phi ≤ 5.46875 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 phi = 2.3975 eV and 2.94875 eV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δphi across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
HEIGHT FOR 1% OF CURRENT against phi
Predictions from the supplied Scanning tunnelling microscopy, tunnel diodes, gate leakage 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): HEIGHT FOR 1% OF CURRENT / pm
Horizontal axis (x): phi: Barrier height φ / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| phi: Barrier height φ / eV | HEIGHT FOR 1% OF CURRENT / pm |
|---|---|
| 1.53125 | 0.7886887449658111 |
| 2.515625 | 0.6153266757019836 |
| 3.5 | 0.5216685702537953 |
| 4.484375 | 0.4608692505472793 |
| 5.46875 | 0.41733485620303623 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of phi separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At phi = 2.3975 eV, substitute into Y_out = 10/(2*5.1232*sqrt(phi)): Y₀ = 0.630303 pm.
- At phi = 2.94875 eV, the same fixed-parameter model gives Y₁ = 0.568342 pm.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.0619611 pm; the input change is Δphi = 0.55125 eV.
- Average sensitivity is ΔY_out/Δphi = -0.112401 pm 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 = 0.630303 pm, Y₁ = 0.568342 pm; ΔY_out = -0.0619611 pm; average sensitivity = -0.112401 pm 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.
Scanning tunnelling microscopy, tunnel diodes, gate leakage: infer an input with uncertainty
In device tunnelling each electron is taken to cross alone and at fixed energy, so the current counts filled states that face empty states at the same energy and weights each by a transmission falling as e^(−2κz) with the gap z. Use the supplied topic model Y_out = 10/(2*5.1232*sqrt(phi)), where Y_out is HEIGHT FOR 1% OF CURRENT (pm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. phi is Barrier height φ, measured in eV. Keep rtip (Apex radius R) = 0.5 nm; z (Gap z) = 0.6 nm fixed. For this exercise use only 1.53125 eV ≤ phi ≤ 5.46875 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 = 0.49210731 pm, with a bounded reading uncertainty ±0.011216 pm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown phi by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00039375 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.
HEIGHT FOR 1% OF CURRENT against phi
Predictions from the supplied Scanning tunnelling microscopy, tunnel diodes, gate leakage 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): HEIGHT FOR 1% OF CURRENT / pm
Horizontal axis (x): phi: Barrier height φ / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| phi: Barrier height φ / eV | HEIGHT FOR 1% OF CURRENT / pm |
|---|---|
| 1.53125 | 0.7886887449658111 |
| 2.515625 | 0.6153266757019836 |
| 3.5 | 0.5216685702537953 |
| 4.484375 | 0.4608692505472793 |
| 5.46875 | 0.41733485620303623 |
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(phi) = 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(1.53125) = 0.788689 pm and Y_out(5.46875) = 0.417335 pm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives phi_line = 1.53125 + (0.492107 − 0.788689)(5.46875 − 1.53125)/(0.417335 − 0.788689) = 4.67593 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 phi = 3.93313 eV; substituting back gives Y_out = 0.49210731 pm.
- The interpolation error in the input is |phi_line − phi| = 0.742806 eV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.48089131 pm and 0.50332331 pm. Sort the roots even if the response decreases: 3.75979 eV ≤ phi ≤ 4.11873 eV.
- The lower and upper input deviations are 0.173338 eV and 0.185607 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 phi ≈ 3.93313 eV; compatible interval [3.75979, 4.11873] eV. Endpoint interpolation gives 4.67593 eV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.