1 / 10

E 1

E 1. E 2. E 3. Graphically solve for E n ’s, now know α n ’s and k n ’s. E 1 =0.0675eV k 1 =1.366E9 α 1 =4.9656E9 E 2 =0.291eV k 2 =2.7789E9 α 2 =4.3298E9 E 3 =0.714eV k 3 =4.345E9 α 3 =2.75E9. Psi1, not normalized. Psi1, normalized. Psi2, Normalized.

stewart-english
Download Presentation

E 1

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. E1 E2 E3

  2. Graphically solve for En’s, now know αn’s and kn’s • E1=0.0675eV • k1=1.366E9 • α1=4.9656E9 • E2=0.291eV • k2=2.7789E9 • α2=4.3298E9 • E3=0.714eV • k3=4.345E9 • α3=2.75E9

  3. Psi1, not normalized

  4. Psi1, normalized

  5. Psi2, Normalized

  6. To find area outside of well, simply integrate Psi*Psidx outside of well • For Psi1: Probability outside of well is 1.13% • For Psi2: Probability outside of well is 3.79%

  7. To find <x> and <p> use the operator expression and integrate • For Psi1, <x1> = 0.979nm …should be 1nm • For Psi2, <x2> = .917nm, …should be 1nm • For Psi1, <p1> = i*3.13E-27 • For Psi2, <p2> = i*1.38E-26

  8. Problem 2 • Use notes from lecture to determine E1,E2,E3 and Psi1, Psi2, Psi3.

  9. Perform Integration to find Mmn • M21= 0.36nm • M31=1.4x10-8nm ~ 0 • M32=0.39nm

More Related