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- Thread starter albertsmith
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It looks like you don't have the entire wave function. It seems as if you're looking only at the radial portion of the wave function "R(r)". You can look at the "effective radial wave function" by multiplying by r^2. This acts just the the r^2 necessary in the spherical coordinate integral. Someone else can better explain why you would do this.

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jtbell

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First note that the wave function gives you (by way of [itex]|\psi|^2[/itex]) the

To simplify the discussion, suppose we have a uniform volume probability density [itex]\rho[/itex].

Now ask the question, what is the probability that the particle is located a distance r from some point (e.g. the center of a sphere)? Loosely speaking, if r is large, there are more points at that distance; and if r is small, there are fewer points at that distance. So the total probability of being at some point at distance r increases as r increases, and decreases as r decreases.

To make this more precise, consider the probability that the particle is located in a thin spherical shell of thickness dr and radius r. For a uniform volume probability density, the probability of being in the shell is approximately (thickness of shell)(area of shell)(probability density) = [itex]4 \pi r^2 dr \rho[/itex]. Factoring out the thickness of the shell we have the

The hydrogen 1s wave function and volume probability density are not uniform, but the same idea applies, when we switch from volume probability density to radial probability density.

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