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Radial probability density / quantum numbers.

  1. Apr 24, 2010 #1
    In my notes for a module on atomic and molecular physics it has this statement:

    "For a given n the probability density of finding e- near the nucleus decreases as l increases, because the centrifugal barrier pushes the e- out. So the low-l orbitals are called penetrating."

    I just want to clear a few things up to make sure I understand this correctly.
    In a text book I found a good set of graphs comparing different combinations of n and l:

    http://img.photobucket.com/albums/v319/Adwodon/IMG.jpg

    Taking the n=3 set as the example, it appears to me that for l=0 the average distance is actually further from the nucleus than l=1,2.

    However there are 2 other peaks which seem to be roughly the same distance as the most probable distance for n=1,2 (I don't know if this is just coincidental?)

    So would I be totally wrong if I thought of these as sets of orbits (ie n=3 l=0 has 3 sets) and that low-l orbitals can 'penetrate' into sets of orbits which are closer to the nucleus (which are of a similar orbits to lower energy orbitals). As l increases the electron can no longer "penetrate" into lower sets due to increased angular momentum / centrifugal barrier(?), and the most probable position of the individual sets moves closer to the nucleus.

    For example (distances are made up and represent the location of the peaks):

    n=3 l=0 has
    Set 1 @ r=1 , Set 2 @ r=5, Set 3 @ r= 15

    at n=3 l=1, set one is now inaccessible and set 2 /3 have moved closer:
    Set 2 @ r=4.5, Set 3 @ r= 14

    and for n=3 l=2 both set 1/2 are inaccessible
    Set 3 @ r=12

    Although I cant say I really know why the average position would move closer to the nucleus as the angular momentum is increased?

    Ultimately im just trying to put it into a form I can understand rather than have to learn by rote so unless my thinking is completely self defeating I don't mind if it doesn't give a totally accurate picture.
    Also please correct me if i'm using incorrect terminology, this is all fairly new to me and ive always been slow when it comes to buzzwords.
     
  2. jcsd
  3. Apr 24, 2010 #2

    SpectraCat

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    Science Advisor

    First of all, the s-orbitals (l=0) are characterized as "penetrating" simply because they have non-zero probability at the nucleus ... the mathematical form of the wavefunction is just a (normalized) decaying exponential, so it actually has its highest value at r=0 (i.e. at the nucleus). All other l-values have an angular node at the nucleus (due to the centrifugal barrier), and so the wavefunction goes to zero at r=0. Furthermore, the centrifugal barrier is larger for larger l, meaning that the wavefunction is excluded from a larger region around the nucleus.

    Also, you seem to be confusing the peaks in plots of the radial probability density (i.e. [tex]r^2\psi^*\psi[/tex] with the average value of the radius. The average value is given by the expectation value of r,
    [tex]<r>=\int_0^{2\pi}d\phi\int_0^{\pi}sin\theta d\theta\int_0^{\infty}r^2 dr \psi^*\:r\:\psi[/tex]

    which will not necessarily correspond to the peak in the probability density, particularly for wavefunctions with radial nodes. Have you learned about those yet? They are the reason the radial probability density goes to zero at the specific values of r you mentioned.
     
  4. Apr 24, 2010 #3
    Hi, adowodon.
    Let us consider the states that have same n but different l. The state energy is determined by n, not l or m.
    Energy of the states consists of plus rotation energy and minus potential energy.
    The state of high l has more rotation energy than lower l.
    In order energy to be same, the state of higher l must have more minus potential energy than lower l. It means that the former is closer to origin than the latter is.

    PS the graph in the textbook seems to be P(r) the probability density that the distance of electron is between r and r+dr regardless of angle.
    P(r) dr =|R_n,l(r)|^2 r^2 dr, P(0)=0.


    Regards.
     
    Last edited: Apr 24, 2010
  5. Apr 24, 2010 #4

    SpectraCat

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    Science Advisor

    Sorry .. just noticed this got cut off somehow, and my full answer didn't get posted.. here is the bit that got lost:

    The general formula for the average radius for a given orbital is given by:

    [tex]<r> = \frac{a_0}{2}[3n^2-l(l+1)][/tex]

    So, as you can see, the expectation value does get smaller as l increases, as you noticed from the plots. The reason for this is that the radial nodes "push" the density out farther from the nucleus, where the volume element in the radial integral (i.e. [tex]4\pi r^2[/tex]) is larger.

    However, this does not change the explanation I gave earlier of why lower-l orbitals are more "penetrating". That has to do with the behavior near the nucleus, not the average value of the radius.
     
  6. Apr 24, 2010 #5
    Ah i see, thanks!

    About nodes though, looking at this picture:

    http://upload.wikimedia.org/wikipedia/en/thumb/e/e7/Hydrogen_Density_Plots.png/660px-Hydrogen_Density_Plots.png [Broken]

    I see that the (2,0,0) (3,0,0)(4,0,0) all have probability at the nucleus, whereas ones with l>0 dont (and the bigger l is the bigger the 'gap' at the centre), so thats whats meant by penetrating...

    Would the dark circles on these l=0 pictures correspond to radial nodes? Where the radial equation = 0?

    And on the l>0 pictures, do the dark lines going out of the centre represent the angular nodes? Where the angular eq = 0?

    It looks like if l=n-1 there are only angular nodes? Im guessing that has something to do with the (n-l-1)! part of the angular equation?

    And the lack of radial nodes because eg:

    n=2 l=0 there is a term which goes (1-Zr/2a) so if Zr/2a = 1 the radial eq = 0

    but for n=2 l=1 its just (Zr/a) so the radial equation cant = 0 and so no nodes. (except at r=0)?

    I think its all starting to fall into place :)
     
    Last edited by a moderator: May 4, 2017
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