Aula 11 - Potenciais Unidimenssionais.pdf
This document was uploaded by user and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this DMCA report form. Report DMCA
Overview
Download & View Aula 11 - Potenciais Unidimenssionais.pdf as PDF for free.
More details
- Words: 301
- Pages: 18
Cap. 1
08-10-2018
Prof. Dr. Elias Oliveira Serqueira 63 [email protected]
64
65
β2 π 2 Ο π₯ β + π π₯ Ο π₯ = πΈΟ π₯ 2π ππ₯ 2
β2 π 2 Ο π₯ β + π0 Ο π₯ = πΈΟ π₯ 2π ππ₯ 2 β2 π 2 Ο π₯ β = πΈΟ π₯ 2 2π ππ₯
66
β2 π 2 Ο π₯ β = πΈΟ π₯ 2π ππ₯ 2
β2 π 2 Ο π₯ β + π0 Ο π₯ = πΈΟ π₯ 2π ππ₯ 2
πΈ
π2 Ο π₯ 2ππΈ =β 2 Ο π₯ 2 ππ₯ β π2 Ο π₯ 2Ο π₯ = βπ ππ₯ 2
π2 Ο π₯ 2π = β 2 πΈ β π0 Ο π₯ ππ₯ 2 β π2 Ο π₯ 2π = 2 π0 β πΈ Ο π₯ 2 ππ₯ β 2π 2 π = 2 π0 β πΈ β π2 Ο π₯ 2Ο π₯ = π ππ₯ 2
A solução:
Ο π₯ = π΄ cos ππ₯ + π΅ sin ππ₯ Ο π₯ = π΄ π πππ₯ + π΅π βπππ₯
Ο π₯ = πΆπ ππ₯ + π·π βππ₯
67
68
69
70
71
72
73
β2 π 2 Ο π₯ β = πΈΟ π₯ 2π ππ₯ 2
β2 π 2 Ο π₯ β + π0 Ο π₯ = πΈΟ π₯ 2π ππ₯ 2
πΈ
π2 Ο π₯ 2π = β 2 πΈ β π0 Ο π₯ ππ₯ 2 β π2 Ο π₯ 2ππΈ =β 2 Ο π₯ 2 ππ₯ β π2 Ο π₯ 2Ο π₯ = βπ ππ₯ 2
π2
2π = 2 πΈ β π0 β
π2 Ο π₯ 2Ο π₯ = βπ ππ₯ 2
A solução:
Ο π₯ =π΄
π πππ₯
+
π΅π βπππ₯
Ο π₯ = πΆπ πππ₯ + π·π βπππ₯
74
75
Resulta
76
Ou ainda
77
Note que 78
79
80
08-10-2018
Prof. Dr. Elias Oliveira Serqueira 63 [email protected]
64
65
β2 π 2 Ο π₯ β + π π₯ Ο π₯ = πΈΟ π₯ 2π ππ₯ 2
β2 π 2 Ο π₯ β + π0 Ο π₯ = πΈΟ π₯ 2π ππ₯ 2 β2 π 2 Ο π₯ β = πΈΟ π₯ 2 2π ππ₯
66
β2 π 2 Ο π₯ β = πΈΟ π₯ 2π ππ₯ 2
β2 π 2 Ο π₯ β + π0 Ο π₯ = πΈΟ π₯ 2π ππ₯ 2
πΈ
π2 Ο π₯ 2ππΈ =β 2 Ο π₯ 2 ππ₯ β π2 Ο π₯ 2Ο π₯ = βπ ππ₯ 2
π2 Ο π₯ 2π = β 2 πΈ β π0 Ο π₯ ππ₯ 2 β π2 Ο π₯ 2π = 2 π0 β πΈ Ο π₯ 2 ππ₯ β 2π 2 π = 2 π0 β πΈ β π2 Ο π₯ 2Ο π₯ = π ππ₯ 2
A solução:
Ο π₯ = π΄ cos ππ₯ + π΅ sin ππ₯ Ο π₯ = π΄ π πππ₯ + π΅π βπππ₯
Ο π₯ = πΆπ ππ₯ + π·π βππ₯
67
68
69
70
71
72
73
β2 π 2 Ο π₯ β = πΈΟ π₯ 2π ππ₯ 2
β2 π 2 Ο π₯ β + π0 Ο π₯ = πΈΟ π₯ 2π ππ₯ 2
πΈ
π2 Ο π₯ 2π = β 2 πΈ β π0 Ο π₯ ππ₯ 2 β π2 Ο π₯ 2ππΈ =β 2 Ο π₯ 2 ππ₯ β π2 Ο π₯ 2Ο π₯ = βπ ππ₯ 2
π2
2π = 2 πΈ β π0 β
π2 Ο π₯ 2Ο π₯ = βπ ππ₯ 2
A solução:
Ο π₯ =π΄
π πππ₯
+
π΅π βπππ₯
Ο π₯ = πΆπ πππ₯ + π·π βπππ₯
74
75
Resulta
76
Ou ainda
77
Note que 78
79
80
Related Documents
Aula 11 - Potenciais Unidimenssionais.pdf
October 2019 38
Aula 12 - Potenciais Unidimensionais.pdf
October 2019 45
Aula 15 - Potenciais Unidimensionais..pdf
October 2019 30
Aula 13 - Potenciais Unidimensionais..pdf
October 2019 64
Aula 11: Modelizacao
May 2020 12
Aula 11 - Pdf.pdf
December 2019 15More Documents from "Washington Dias Silva"
Aula 13 - Potenciais Unidimensionais..pdf
October 2019 64
Aula 09 - Teoria De Shrodinger.pdf
October 2019 42
Songbook - Edu Lobo.pdf
April 2020 43
Oficio
May 2020 37