Derivation of formula for orbital ranges in hydrogen atom

  • Thread starter jjr
  • Start date
  • #1
jjr
51
1
I almost have the answer, I'm sure there's just a minor flaw in my reasoning. Here it goes:
We're given that the angular momentum of the atom is integer multiples of h-bar (n[itex]\hbar[/itex]) (integer depending on the orbit). Now the centripetal force is given by F = [itex]\frac{mv^2}{r}[/itex] = [itex]\frac{p^2}{mr}[/itex] = n^2[itex]\hbar[/itex]^2/mr where m is the electron mass, v is the velocity, p is the angular momentum and r is the range. This force equals the attractive coloumb force between the proton and electron, so: [itex]\frac{p^2}{mr}[/itex] = [itex]\frac{k(q^2)}{r^2}[/itex] => r = k(q^2)m/(n^2[itex]\hbar[/itex]^2) where k is coulombs constant, q is the charge of electron/proton.

The problem is obvious, seeing as how the radius drops with higher n's. The answer is in fact the reciprocal of the right part of the last equation. Can anyone spot my error?

Thanks, J
 

Answers and Replies

  • #2
jjr
51
1
Bump. If something is confusing, I'll be happy to elaborate
 
  • #3
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,040
1,611
p=mv is linear momentum, not angular momentum.
 

Related Threads on Derivation of formula for orbital ranges in hydrogen atom

Replies
2
Views
9K
Replies
15
Views
346
Replies
2
Views
3K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
1
Views
1K
Replies
5
Views
3K
  • Last Post
Replies
1
Views
617
  • Last Post
Replies
9
Views
13K
Top