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Solutions of the Schrodinger equation for hydrogen

  • Thread starter atarr3
  • Start date
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1. Homework Statement
Consider an electron in the hydrogen atom with radial wave function [tex]R_{31}[/tex] (n=3, l=1). Please verify that this radial function verifies the radial equation.



2. Homework Equations
The radial equation

[tex]\frac{1}{r^{2}}[/tex][tex]\frac{d}{dr}[/tex][tex]\left(r^{2}\frac{dR}{dr}\right)[/tex] + [tex]\frac{2\mu}{h^{2}}[/tex][tex]\left[E-V-\frac{h^{2}}{2\mu}\frac{l\left(l+1\right)}{r^{2}}\right][/tex]R = 0


3. The Attempt at a Solution

Ok so I found the corresponding solution for the given radial wave funtion, and I think I'm supposed to set that equal to A, some constant, times [tex]e^{\frac{-r}{3a_{0}}}[/tex]
and then plug that into the original radial wave function? I'm not really sure of what I'm supposed to do here.
 
Last edited:
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Oh and those h's are supposed to be h bars. I don't know how to do that in latex.
 

gabbagabbahey

Homework Helper
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Just use the equation for [itex]R_{31}[/itex] that is in your text/notes, and substitute it into the Differential equation...

P.S. To write [itex]\hbar[/itex] in [itex]\LaTeX[/itex], just use \hbar
 
76
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You mean like an equation like this?

[tex]\frac{1}{a_{0}^{3/2}}\frac{4}{81\sqrt{6}}\left(6-\frac{r}{a_{0}}\right)\frac{r}{a_{0}}e^{-r/3a_{0}}[/tex]

I tried using that and plugging it into the radial equation, but it gets really messy and I'm not sure if I know how to simplify it. I also don't know what to do with the V and E quantities.
 
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And I assumed that the stuff not depending on R was equal to some constant A to help make it easier... would that screw my answer up?
 

gabbagabbahey

Homework Helper
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You mean like an equation like this?

[tex]\frac{1}{a_{0}^{3/2}}\frac{4}{81\sqrt{6}}\left(6-\frac{r}{a_{0}}\right)\frac{r}{a_{0}}e^{-r/3a_{0}}[/tex]
Yup.

I tried using that and plugging it into the radial equation, but it gets really messy and I'm not sure if I know how to simplify it. I also don't know what to do with the V and E quantities.
[itex]V[/itex] is just the Coulomb potential, and if the electron is in the [itex]n=3[/itex] state, shouldn't [itex]E[/itex] be [itex]E_3[/itex] (which you should have an equation for)?
 
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Ok so [tex]V =\frac{1}{4\pi\epsilon_{0}}\frac{-e^{2}}{r}[/tex] and E is just [tex]\frac{-E_{0}}{n^{2}}[/tex]? And that will all cancel out if I plug everything in?
 

gabbagabbahey

Homework Helper
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Yup.
 
76
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Wow. Ok. Thank you so much! You've saved me a great deal of work.
 
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Also, try finding [tex] \frac {2\mu V}{\hbar ^2} [/tex] and [tex] \frac {2\mu E_n}{\hbar^2}[/tex] in terms of [itex]a_0[/itex] and [itex]r[/itex]. It might make it easier.
 
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Just to verify that this is correct, I'm getting [tex]\frac{2\mu V}{\hbar^{2}}=\frac{-2}{a_{0}r}[/tex] and [tex]\frac{2\mu E}{\hbar^{2}}=\frac{-1}{9a_{0}^{2}}[/tex] I'm getting almost everything to cancel out, but not quite everything. There might be an error in my derivatives.
 
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Ok I just got the answer. Thank you all so much for your help!
 

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