Solutions of the Schrodinger equation for hydrogen

[atarr3]

Homework Statement

Consider an electron in the hydrogen atom with radial wave function $$R_{31}$$ (n=3, l=1). Please verify that this radial function verifies the radial equation.

Homework Equations

The radial equation

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

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 $$e^{\frac{-r}{3a_{0}}}$$
and then plug that into the original radial wave function? I'm not really sure of what I'm supposed to do here.

 

[atarr3]

Oh and those h's are supposed to be h bars. I don't know how to do that in latex.

 

[gabbagabbahey]

Just use the equation for $R_{31}$ that is in your text/notes, and substitute it into the Differential equation...

P.S. To write $\hbar$ in $\LaTeX$, just use \hbar

 

[atarr3]

You mean like an equation like this?

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

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.

 

[atarr3]

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]

You mean like an equation like this?

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

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.

$V$ is just the Coulomb potential, and if the electron is in the $n=3$ state, shouldn't $E$ be $E_3$ (which you should have an equation for)?

 

[atarr3]

Ok so $$V =\frac{1}{4\pi\epsilon_{0}}\frac{-e^{2}}{r}$$ and E is just $$\frac{-E_{0}}{n^{2}}$$? And that will all cancel out if I plug everything in?

 

[gabbagabbahey]

Yup.

 

[atarr3]

Wow. Ok. Thank you so much! You've saved me a great deal of work.

 

[Jasso]

Also, try finding $$\frac {2\mu V}{\hbar ^2}$$ and $$\frac {2\mu E_n}{\hbar^2}$$ in terms of $a_0$ and $r$. It might make it easier.

 

[atarr3]

Just to verify that this is correct, I'm getting $$\frac{2\mu V}{\hbar^{2}}=\frac{-2}{a_{0}r}$$ and $$\frac{2\mu E}{\hbar^{2}}=\frac{-1}{9a_{0}^{2}}$$ I'm getting almost everything to cancel out, but not quite everything. There might be an error in my derivatives.

 

[atarr3]

Ok I just got the answer. Thank you all so much for your help!