Hydrogen atom (HELP ME FAST PLEASE)

[aaaa202]

Hi, I'm trying to figure out the solution for the ground state of the hydrogen-atom, however it is not going well.
As far as i know, you can supress the angular dependence, because the states of hydrogen (or at least some of them) are spherically harmonic.
This way the schr. equation just reduces to a radial dependent equation, which has the solutions:
Ψ(r) = F(r)e^(-br), where b is a constant and F(r) a power series (the laguerre polynomials), which determines the number of the s-orbital such that if i just look at the ground state F(r) reduces to a constant. Now, am I right so far?
Well, if so, I'm finding it a little hard to determine these constant. I read somewhere, that you can do it using the normalization condition for the wave function, but isn't Ψ(r) also dependent on the other s-orbitals or how am I to solve the problem?

 

[phyzguy]

You're right so far. The ground state wave function looks like:
\psi(r) = A e^{-b r}
Now plug this into the time-independent Schrodinger equation:
\frac{-\hbar^2}{2m}\nabla^2 \psi(r)+\frac{e^2}{r}\psi(r) = E \psi(r)
This should give you two equations that allow you to solve for E and b.
Then you determine A by requiring that the integral of |psi^2| over all space is equal to 1.

 

[aaaa202]

Hmm, I seem to be getting multiple results when try to solve for b. What do I do with those?
Just to get it straigt, what I'm supposed to do is plug the ground-state wave function into the schrödinger-equation and solve for b right?
But again, I'm likely to have made a typo on my calculator..

 

[phyzguy]

If you do it right, you will get one term proportional to e^-r, and a second term proportional to (e^-r)/r. These two terms will allow you to solve for b and E.

 

[aaaa202]

Now this i don't understand. Don't u just plug the wave function in the schrödinger-equation. How does that give you two equation?
EDIT: I think i get it now