# Derivation of formula for orbital ranges in hydrogen atom

1. Feb 25, 2012

### jjr

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$\hbar$) (integer depending on the orbit). Now the centripetal force is given by F = $\frac{mv^2}{r}$ = $\frac{p^2}{mr}$ = n^2$\hbar$^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: $\frac{p^2}{mr}$ = $\frac{k(q^2)}{r^2}$ => r = k(q^2)m/(n^2$\hbar$^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

2. Feb 26, 2012

### jjr

Bump. If something is confusing, I'll be happy to elaborate

3. Feb 26, 2012

### vela

Staff Emeritus
p=mv is linear momentum, not angular momentum.