- #1
Emspak
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Homework Statement
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?
Homework Equations
see above
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, [itex]\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[/itex]
and doing little algebra and taking a derivative
[itex]\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[/itex]
[itex]\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[/itex]
[itex]-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[/itex]
solve for r:
[itex]\left[r^2(\frac{Z}{r_0}e^{-2Zr/r_0})+ r e^{-2Zr/r_0}\right]=0[/itex]
[itex]r^2(\frac{Z}{r_0}e^{-2Zr/r_0})=- r e^{-2Zr/r_0}[/itex]
[itex]r^2(\frac{Z}{r_0})=- r [/itex]
[itex]r=- (\frac{r_0}{Z}) [/itex]
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.
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