# Probability distribution of an electron - comparing to Bohr

1. May 9, 2014

### Emspak

1. The problem statement, all variables and given/known data

The ground sate of a Hydrogen-like atom is given by the wavefunction:

$$R_{10}(r)= 2\left(\frac{Z}{r_0}\right)^{3/2}e^{Zr/r_0}$$

and the Probability distribution is $$P(r)=4\pi r^2 |R_{10}|^2$$

At what distance is the most probable radius of the electron?

2. Relevant equations

see above

3. The attempt at a solution

OK, so I want the spot where P(r) is a maximum. That's where the derivative of P(r) = 0.

so, $\frac{d}{dr}P(r) = \frac{d}{dr}\left[4\pi r^2 \left(2\left(\frac{Z}{r_0}\right)^{3/2}e^{-Zr/r_0}\right)^2\right]=0$

and doing little algebra and taking a derivative

$\frac{d}{dr}P(r) = \frac{d}{dr}\left[16\pi r^2\left( \frac{Z}{r_0}\right)^3 ( e^{-2Zr/r_0})\right]=0$
$\frac{d}{dr}P(r) = \left[16\pi r^2\left( \frac{Z}{r_0}\right)^3 (-2\frac{Z}{r_0}e^{-2Zr/r_0})+32\pi r \left( \frac{Z}{r_0}\right)^3e^{-2Zr/r_0}\right]=0$
$-32\pi \left( \frac{Z}{r_0}\right)^3 \left[r^2(\frac{Z}{r_0}e^{-2Zr/r_0})+ r e^{-2Zr/r_0}\right]=0$

solve for r:

$\left[r^2(\frac{Z}{r_0}e^{-2Zr/r_0})+ r e^{-2Zr/r_0}\right]=0$
$r^2(\frac{Z}{r_0}e^{-2Zr/r_0})=- r e^{-2Zr/r_0}$
$r^2(\frac{Z}{r_0})=- r$
$r=- (\frac{r_0}{Z})$

My problem is that this seems wrong. I would not expect the radius to get smaller as Z rises. Now, I realize this is for a Hydrogen-like atom and hat does not describe real atoms, and at Z=1 the result is the Bohr radius which is fine, but I want to make sure I am not missing something or making a basic mathematical error. If the idea is to find that the Bohr model doesn't work for anything with Z> 1 that's also fine -- I just wanted to be sure that was what I was seeing.

anyhow, if I did this right -- or wrong -- I'd love to know.

Last edited: May 9, 2014
2. May 9, 2014

### TSny

Hello, Emspak.

You made a sign error when factoring out the constants just before "solve for r". Note that you are getting a negative value for r, which can't be right.

It is natural that the most probable r will decrease as Z increases. The attraction of the electron to the nucleus increases as Z increases, thus the probability distribution is "pulled in". The wave function is proportional to $e^{-Zr/r_0}$. Think about how the graph of this function changes as Z increases.

3. May 9, 2014

### Emspak

Thanks a lot- I saw the sign error and it was one of those cases where I wasn't believing the right answer because I kept thinking that since the nucleus of an atom (Bohr like or not) is getting bigger, the Bohr radius would have to go up with it. Bu then I realized that the space the actual nucleus takes up relative to the Bohr radius is tiny.

4. May 9, 2014

### TSny

Yes, sounds good.