Wave Functions of a Trapped Electron — Question 2

Section 39-3 · Physics pages 6454–6455

Question 2 The ground state favors the central half more than a uniform distribution does An electron occupies the ground mode. The infinite well occupies 0 < x < L; wave functions vanish outside. Its normalized modes are un(x) = √ 2/L sin(nπx/L) and energies En = n2E1. x 0 L Schematic; labels, not drawing lengths, specify the data. (a) Integrate the density over L/4 < x < 3L/4. (b) Find the probability in the two outer quarters and compare with a uniform spatial model.
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Question 2 - Solution The ground state favors the central half more than a uniform distribution does x 0 L (a) Method and calculation Pc = 2 L ∫3L/4 L/4 sin2(πx/L)dx = [ x L −sin(2πx/L) 2π ]3L/4 L/4 . Pc = 1 2 + 1 π = .818310 . (b) Interpretation and further calculation Pouter = 1 −Pc = 1 2 −1 π = .181690 . Pc,uniform = 1/2 ⇒Pc −Pc,uniform = 1/π . Physical check The ground-state density vanishes at the walls and peaks at the center. Equal interval lengths need not have equal probability.

Question 2

The ground state favors the central half more than a uniform distribution does

An electron occupies the ground mode. The infinite well occupies 0<x<L0<x<L; wave functions vanish outside. Its normalized modes are un(x)=2/Lsin⁡(nπx/L)u_n(x)=\sqrt{2/L}\sin(n\pi x/L) and energies En=n2E1E_n=n^2E_1.

Diagram shown in the original SVG worksheet.

  1. Integrate the density over L/4<x<3L/4L/4<x<3L/4.

  2. Find the probability in the two outer quarters and compare with a uniform spatial model.

Question 2 - Solution

The ground state favors the central half more than a uniform distribution does

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Pc=2L∫L/43L/4sin⁡2(πx/L)dx=[xL−sin⁡(2πx/L)2π]L/43L/4.P_c=\frac2L\int_{L/4}^{3L/4}\sin^2(\pi x/L)dx=\left[\frac xL-\frac{\sin(2\pi x/L)}{2\pi}\right]_{L/4}^{3L/4}.Pc=12+1π=.818310.\boxed{P_c=\frac12+\frac1\pi=.818310}.

(b) Interpretation and further calculation

Pouter=1−Pc=12−1π=.181690.\boxed{P_{\mathrm{outer}}=1-P_c=\frac12-\frac1\pi=.181690}.Pc,uniform=1/2⇒Pc−Pc,uniform=1/π.P_{c,\mathrm{uniform}}=1/2\Rightarrow\boxed{P_c-P_{c,\mathrm{uniform}}=1/\pi}.

Physical check

The ground-state density vanishes at the walls and peaks at the center. Equal interval lengths need not have equal probability.