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Clearly the probability density to find an electron at point x is
(x) = R(r)*R(r), |
(11.22) |
but what is the probability to find the electron at a distance r from
the proton? The key point to realise is that for each value of r the electron
can be anywhere on the surface of a sphere of radius r, so that for larger
r more points contribute than for smaller r. This is exactly the source of
the factor 4
r2 in the normalisation integral. The probability to find
a certain value of r is thus
(r) = 4 r2R(r)*R(r). |
(11.23) |
Figure 11.2:
The probability to find a certain value of r for the first
two Harmonic oscillator wave functions.
|
|
These probabilities are sketched in Fig. 11.2. The peaks are of some
interest, since they show where the electrons are most likely to be found.
Let's investigate this mathematically:
1 = 4r2/a03e-2r/a0. |
(11.24) |
if we differentiate with respect to r, we get
 1 = (2re-2r/a0 - 2r2/a0e-2r/a0). |
(11.25) |
This is zero at r = a0. For the first excited state this gets a little
more complicated, and we will have to work harder to find the answer.
Next: 11.6 Spherical harmonics
Up: 11. 3D Schrödinger equation
Previous: 11.4 The hydrogen atom