An Electron in a Finite Well — Question 10

Section 39-4 · Physics pages 6490–6491

Question 10 Measured oscillation and decay scales reconstruct a finite-well ground state A symmetric finite electron well has an even nodeless bound state with measured interior wave number k = 4.00 nm−1 and exterior decay constant κ = 3.00 nm−1. Use k tan(ka) = κ, E = ℏ2k2/(2me) and V0 −E = ℏ2κ2/(2me), with ℏ2/(2me) = .0381 eV nm2. x V0 −a a k κ Schematic; labels, not drawing lengths, specify the data. (a) Find E and V0. (b) Infer the full width, choosing the correct branch for the ground state.
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Question 10 - Solution Measured oscillation and decay scales reconstruct a finite-well ground state x V0 −a a k κ (a) Method and calculation E = .0381(4.00)2 = .6096 eV . V0 = .0381(42 + 32) = .9525 eV . (b) Interpretation and further calculation 0 < ka < π/2 ⇒ka = arctan(κ/k) = arctan(.750). L = 2a = 2 arctan(.750) 4.00 nm = .321751 nm . Physical check Adding integer multiples of pi to the inverse tangent would describe other branches. Nodeless ground-state information removes that ambiguity.

Question 10

Measured oscillation and decay scales reconstruct a finite-well ground state

A symmetric finite electron well has an even nodeless bound state with measured interior wave number k=4.00nm−1k=4.00\,\mathrm{nm}^{-1} and exterior decay constant κ=3.00nm−1\kappa=3.00\,\mathrm{nm}^{-1}. Use ktan⁡(ka)=κk\tan(ka)=\kappa, E=ℏ2k2/(2me)E=\hbar^2k^2/(2m_e) and V0−E=ℏ2κ2/(2me)V_0-E=\hbar^2\kappa^2/(2m_e), with ℏ2/(2me)=.0381eVnm2\hbar^2/(2m_e)=.0381\,\mathrm{eV\,nm^2}.

Diagram shown in the original SVG worksheet.

  1. Find E and V0V_0.

  2. Infer the full width, choosing the correct branch for the ground state.

Question 10 - Solution

Measured oscillation and decay scales reconstruct a finite-well ground state

Diagram shown in the original SVG worksheet.

(a) Method and calculation

E=.0381(4.00)2=.6096eV.\boxed{E=.0381(4.00)^2=.6096\,\mathrm{eV}}.V0=.0381(42+32)=.9525eV.\boxed{V_0=.0381(4^2+3^2)=.9525\,\mathrm{eV}}.

(b) Interpretation and further calculation

0<ka<π/2⇒ka=arctan⁡(κ/k)=arctan⁡(.750).0<ka<\pi/2\Rightarrow ka=\arctan(\kappa/k)=\arctan(.750).L=2a=2arctan⁡(.750)4.00nm=.321751nm.\boxed{L=2a=\frac{2\arctan(.750)}{4.00}\,\mathrm{nm}=.321751\,\mathrm{nm}}.

Physical check

Adding integer multiples of pi to the inverse tangent would describe other branches. Nodeless ground-state information removes that ambiguity.