Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

HELP! Can you find an electron in the exact middle of an S orbital?

  1. Sep 15, 2008 #1
    I have been fighting about this with my inorganic chemistry professor for the past week. As far as I can tell when you take the radial wave function for the 1 s orbital, square it, then times it by r^2, you get a zero probability to find the electron in the middle of the 1s orbital, and a local maximum at the bohr radius.

    He told me that the radial function squared represents probability or some non-sense (instead of the radial wave function squared, times r^2 (I thought you can only square the entire wave function to get probability not just the radial part)). He also stated that there are a few concepts in chemistry that require the electron to be at the exact center of the atom to work. He also said the 1 S orbital has no nodes, so the electron can therefore exist anywhere in the universe.

    I can't find any other way to describe electron density online (or in any text book), besides the radial and angular wave functions, which suggest that there is a 0 chance, to find the electron at the exact center of an atom (ie s orbital).

    I need someone to tell me that I'm wrong or right, then explain why. If I'm right, tell me something that can shut my prof up, and prove i'm right, so i can get sleep at night. If I'm wrong tell me why please.

    This problem is taking up waaaaay tooo much of my free time.

    My understanding of quatum mechanics, is about the level of solving the schrodinger equation, and applying the 1d ideas to 3d space.

    Thank You
  2. jcsd
  3. Sep 16, 2008 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    What is confusing you is the following. It is something that is not particularly related to quantum mechanics, but rather to densities in 3 dimensions.

    Imagine you have a constant density throughout a sphere. Constant density in *space*. That means, if you have a sphere of radius R, then the volume is 4/3 pi R^3, so the density is everywhere equal to rho(x,y,z) = 3/(4 pi R^3) as long as the point (x,y,z) is within the sphere.
    So the space density is constant.

    However, what would be the radial density here ? That is, if you pick a point with the above distribution, what would be its r-value, and what would be the distribution of these r-values ?

    Well, if you think it is constant, you're wrong. In fact, the radial density distribution goes like r^2 (up to R). The probability to be on the outer shell is higher than the probability to be half-way. The reason is simply that the outer shell has more points to it than the shell half way.

    The probability density on the r-axis takes on the form rho(r) = 3 r^2/R^3 (because the integral from r=0 to r=R must be 1).

    So although the probability density in *space* is constant, the radial probability density goes as r^2. Although you have the same probability to be in a small cube in the center than to be in a small cube near the outer surface, the probability density to have r=0 is 0, while the probability density to have r = R is 3/R.

    Back to the s-orbital. This time, the probability density in space is not uniform, but goes as something like exp(- r/a). But that's the probability density in space. It means that the highest probability density in space is actually in the center. However, if we are going to calculate the probability density on the r-axis (to find a specific value of r), then this will go as r^2 exp(- r/a). It is 0 for r=0, and it has a maximum at r = 2a.

    So of all r-values, r = 2a is the most probable, and r=0 is not probable at all. However, of all cubes in space, the cube at r=0 is the most probable (but there's only a single small one, while at r = 2a there are many more).
  4. Sep 16, 2008 #3
    Yea I get all of that. I understand how there is no surface area when r = 0, such that the radial probability will be 0 at 0 (summed over all angles).

    You still havent directly answered my question though. Can you find an electron in the exact middle of an S orbital?

    I read some where yesterday that the wave function has a max at the bohr radius, exactly where the radial probability has a maximum. Does this mean that the wave function itself goes to zero at r = zero?

    Over all I just want to know if you can find an electron in the exact center of an s orbital, because my prof is convinced you can, and I can't find any proof to suggest you can...
  5. Sep 16, 2008 #4
    You should precise your question because it can have different menings, for example:

    1. you mean the exact point (0,0,0); then the probability to find the electron there is 0, as well as the probability to find it in any other points (because they are points and psi is not a Dirac delta).
    2. you mean the probability to find the electron in a small cube with fixed side delta_r and centered in (0,0,0); such prob. is not 0, you can find the electron there.
    3. you mean the prob. to find the electron in a spherical layer of radius r, fixed thickness delta_r and centered in (0,0,0); the prob. is not zero as well, but it goes to zero when r-->0 because, fixed delta_r, the volume of the layer is 4(pi)r^2*delta_r and so goes to 0 when r-->0, while the cube's volume is (delta_r)^3 and so is fixed.
  6. Sep 17, 2008 #5


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    :approve: Exactly.

    I would like to add - as I said in my previous post - that for the 1S orbital (hydrogen), the probability density *in space* is actually MAXIMUM in the center.
    However, for all higher orbitals (p,d,f...), the probability density in space is zero in the center (if I'm not mistaken...).
  7. Sep 17, 2008 #6


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The 3-dim wavefunction of the 1S orbital doesn't have a maximum at the Bohr radius.
    Look at the hydrogen atom wavefunctions at http://en.wikipedia.org/wiki/Hydrogen_atom

    Put l = 0 (s-orbital). The wavefunction goes in exp(-rho/2) (in a normalized unit system).
  8. Sep 18, 2008 #7
    I looked into it quite a bit more. I have concluded, when using the radial coordinate system, that the density increases per unit volume toward the center of the 1s orbital, but the probability goes to zero. The probability goes to zero because the volume goes to zero.

    the squared wave function (when both the radial and the angular components are considered together, and not just the radial portion) for a 1s orbital has a local probability maximum at r = bohr radius.

    The reason there is zero probability at the center to find an electron in a 1s orbital, is because the volume decreases faster than the density increases.

    (I use surface area and volume interchangeably, though this is probably incorrect).
  9. Sep 18, 2008 #8
    Perhaps I can offer some insight on this problem from a different perspective, as discussing probability densities at single points vs in infinitessimal volumes gives me headaches.

    When we solve the Schrodinger equation in spherical coordinates we end up with a radial equation that looks something like
    [tex] \left(-\frac{1}{2m}\nabla^2 + \frac{l\left(l+1\right)}{mr^2} - \frac{e}{r}\right)u\left(r\right) = Eu\left(r\right)[/tex]
    and [tex]u\left(r\right) = rR\left(R\right)[/tex] and R is the radial part of the wavefunction. Thus the solution for u(r) is simply the solution for a particle moving in 1-d with the effective potential
    [tex] V_\textrm{eff}\left(r\right) = \frac{l\left(l + 1\right)}{mr^2} - \frac{e}{r}[/tex]

    The angular part of the wavefunction should have no affect on the radius of the electron (I believe that the bohr radius comes as the expectation value of "r" of the electron, not of a maximum of probability density.. e.g. it is a max of [tex]\psi^* r\psi[/tex] not [tex]|\psi|^2[/tex]), except for that presence of angular momentum.

    For values of l > 0 you can see that Veff goes to +infinity as r goes to zero, this is because the l(l+1)/r^2 term, known as the angular barrier, prevents the electron from colliding with the proton.

    Classically the angular barrier can be explained this way: Angular momentum can be expressed as [tex]\hat{r}x\hat{p}[/tex]. So if the radius of a circular/elliptic orbit decreases then the linear momentum, p, has to increase by the same amount in order to conserve angular momentum. Thus as the electron moves closer and closer to the proton it must orbit the proton faster and faster, requiring more and more energy. So for it to collapse into the proton it has to orbit infinitely fast with infinite energy, which is impossible. However, for l = 0, e.g. no angular momentum, the electron isn't required to orbit the proton and can take a direct path to it. There is no angular barrier when l = 0.

    I hope that explanation helped.
  10. Sep 18, 2008 #9


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yes, that is because you look at the density in r. But if you look at the density *in space*, then, by definition, the volume doesn't go to 0 (it is dV !), and then it is maximal at the center. In r, the corresponding volume density is r^2 dr, and then for r = 0 the volume is 0 too. But not if you work in dV.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: HELP! Can you find an electron in the exact middle of an S orbital?
  1. S orbitals (Replies: 2)

  2. Electron Orbitals (Replies: 2)