# The expectation value for the radial part of the wavefunction of Hydrogen.

1. Sep 29, 2011

### mjordan2nd

The wavefunction of hydrogen is given by

$$\psi_{nlm}(r, \theta, \phi) = R_{nl}(r)Y_{lm}(\theta, \phi)$$

If I am only given the radial part, and asked to find the expectation value of the radial part I integrate the square of the wavefunction multiplied by r cubed allowing r to range from 0 to infinity. I don't understand where the extra factor of r squared comes from? I suspect it has something to do with multiplying by a volume element, but it is unclear to me why the factor of 4 pi that would normally come with spherical integration that depends on r alone disappears. I missed this on a test, recently, and was hoping someone could explain.

Thanks.

2. Sep 29, 2011

### G01

You're integrating in spherical coordinates so the volume element is $dV=r^2sin\theta dr d\theta d\phi$

Last edited: Sep 29, 2011
3. Sep 29, 2011

### Matterwave

The 4pi is normalized away by the spherical harmonics. You essentially already integrated over a spherical shell, and over that entire spherical shell, you should get the integral to be 1.