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11.5 Now where does the probability peak?

Clearly the probability density to find an electron at point x is

$\displaystyle \mathcal {P}$(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$ \pi$r2 in the normalisation integral. The probability to find a certain value of r is thus

$\displaystyle \mathcal {P}$(r) = 4$\displaystyle \pi$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.
\includegraphics[width=6.0cm]{Figures/R12.ps}

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:

$\displaystyle \mathcal {P}$1 = 4r2/a03e-2r/a0. (11.24)

if we differentiate with respect to r, we get

$\displaystyle {\frac{d}{dr}}$$\displaystyle \mathcal {P}$1 = $\displaystyle {\frac{4}{a_0^3}}$(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 up [*]
Next: 11.6 Spherical harmonics Up: 11. 3D Schrödinger equation Previous: 11.4 The hydrogen atom

© 1998 Niels Walet, UMIST
Email Niels.Walet@umist.ac.uk