• Support PF! Buy your school textbooks, materials and every day products Here!

Exam preparation question

  • Thread starter Void123
  • Start date
  • #1
141
0

Homework Statement



I was reviewing some homework problems and looking at the solutions. There is one problem with a tiny step I just cannot rationalize and I am hoping someone can point me in the right direction.

I have a spherical finite well:

[tex] V = {- V_{0}: 0 < r < a}[/tex],

[tex] = {0: r \geq a} [/tex]

[tex]- k_{2} = k_{1} cot (k_{1} a)[/tex] (1)

Refining the notation,

[tex]\alpha = a \sqrt{(2m(E + V_{0})}/hbar = k_{1} a[/tex]

[tex]R = a \sqrt{(2m(V_{0})}/hbar[/tex] and [tex]k_{2} = \sqrt{(2m(V_{0})}/hbar[/tex]

So (1) may be rewritten as [tex] \sqrt{R^{2} - \alpha^{2}} = - \alpha cot (\alpha)[/tex]



Homework Equations



From part 1.



The Attempt at a Solution



I don't understand how at [tex] R = \pi/2 [/tex] there are no bound states.

Also, I am given this restriction: [tex] -V_{0} < E < 0 [/tex]

How is this justified and how is the precise range of bound states determined?
 

Answers and Replies

  • #2
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,617
1,252
When [itex]R=\pi/2[/itex], the only solutions are at [itex]\alpha=\pm\pi/2[/itex]. In these cases, you get k2=0.
 

Related Threads on Exam preparation question

Replies
0
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
2K
Replies
0
Views
939
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
13
Views
793
  • Last Post
Replies
1
Views
820
  • Last Post
Replies
1
Views
1K
Replies
1
Views
641
Replies
2
Views
942
Top