An Electron in a Finite Well — Question 5

Section 39-4 · Physics pages 6480–6481

Question 5 Conditional penetration depth follows the probability decay rather than amplitude decay A finite-well bound state has a right-hand tail ψ(x) = Be−κ(x−a) for x > a, with κ = 4.00 nm−1. Condition on the electron being detected to the right of the well, and set y = x −a ≥0. y f(y) d Schematic; labels, not drawing lengths, specify the data. (a) Normalize the conditional density of y and find its mean. (b) Find the conditional probability of penetrating more than .300 nm.
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Question 5 - Solution Conditional penetration depth follows the probability decay rather than amplitude decay y f(y) d (a) Method and calculation f(y | x > a) = |B|2e−2κy |B|2/(2κ) = 2κe−2κy . ⟨y⟩= ∫∞ 0 y(2κ)e−2κydy = 1 2κ = .125 nm . (b) Interpretation and further calculation P(y > d | x > a) = ∫∞ d 2κe−2κydy = e−2κd. P(y > .300 nm | x > a) = e−2.4 = .0907180 . Physical check This is conditional on being outside on the right. The unconditional probability also includes the probability of reaching that side of the tail.

Question 5

Conditional penetration depth follows the probability decay rather than amplitude decay

A finite-well bound state has a right-hand tail ψ(x)=Be−κ(x−a)\psi(x)=Be^{-\kappa(x-a)} for x>ax>a, with κ=4.00nm−1\kappa=4.00\,\mathrm{nm}^{-1}. Condition on the electron being detected to the right of the well, and set y=x−a≥0y=x-a\ge0.

Diagram shown in the original SVG worksheet.

  1. Normalize the conditional density of y and find its mean.

  2. Find the conditional probability of penetrating more than .300nm.300\,\mathrm{nm}.

Question 5 - Solution

Conditional penetration depth follows the probability decay rather than amplitude decay

Diagram shown in the original SVG worksheet.

(a) Method and calculation

f(y∣x>a)=|B|2e−2κy|B|2/(2κ)=2κe−2κy.f(y\mid x>a)=\frac{|B|^2e^{-2\kappa y}}{|B|^2/(2\kappa)}=\boxed{2\kappa e^{-2\kappa y}}.⟨y⟩=∫0∞y(2κ)e−2κydy=12κ=.125nm.\boxed{\langle y\rangle=\int_0^\infty y(2\kappa)e^{-2\kappa y}dy=\frac1{2\kappa}=.125\,\mathrm{nm}}.

(b) Interpretation and further calculation

P(y>d∣x>a)=∫d∞2κe−2κydy=e−2κd.P(y>d\mid x>a)=\int_d^\infty2\kappa e^{-2\kappa y}dy=e^{-2\kappa d}.P(y>.300nm∣x>a)=e−2.4=.0907180.\boxed{P(y>.300\,\mathrm{nm}\mid x>a)=e^{-2.4}=.0907180}.

Physical check

This is conditional on being outside on the right. The unconditional probability also includes the probability of reaching that side of the tail.