Determining <r> for the hydrogen atom

[Potatochip911]

Homework Statement

The equation for the normalized $n=3$, $l=2$, $m=0$ wavefunction is given by $$\psi_{320}=\frac{1}{81\sqrt{6\pi}}\left(\frac{1}{a_0}\right)^{3/2}\left(\frac{1}{a_0^2}\right)r^2e^{-\frac{r}{3a_0}}(3cos^2\theta-1)e^{i\phi}$$
Determine the expectation value $<r>$.

Homework Equations

3. The Attempt at a Solution [/B]
$<r>$ can be found using the equation $<r>=\int_0^{\infty} rP(r)dr$ where $P(r)=r^2R(r)R^*(r)$, now I know that $R(r)$ will contain the parts of $\psi$ that are functions of $r$ but I'm not sure as to how I can find the normalization constant for it. In other words I have $R(r)=Cr^2e^{-\frac{r}{3a_0}}$ where C is the constant I must somehow determine. I'm not quite sure where to go from here and I can't seem to find a textbook where they go over calculating these (checked Griffiths and a few others).

 

[Brian T]

There are two ways to go about doing that.
1) if you want to determine the normalization constant, you have the fact that the integral of the distribution over all possible values must 1, i.e.
$$ \int_0^\infty P(r) dr = 1 $$
Use this integral to solve for C.

2) Again, using the fact that the integral of the distribution is one, you can calculate
$$ <r> = \int_0^\infty rP(r)dr = \frac{\int_0^\infty rP(r)dr }{\int_0^\infty P(r)dr} $$
and the normalization constant will cancel out completely.

 

[Potatochip911]

There are two ways to go about doing that.
1) if you want to determine the normalization constant, you have the fact that the integral of the distribution over all possible values must 1, i.e.
$$ \int_0^\infty P(r) dr = 1 $$
Use this integral to solve for C.

2) Again, using the fact that the integral of the distribution is one, you can calculate
$$ <r> = \int_0^\infty rP(r)dr = \frac{\int_0^\infty rP(r)dr }{\int_0^\infty P(r)dr} $$
and the normalization constant will cancel out completely.

Thanks I managed to get $<r>=\frac{21}{2}a_0$ which matched the answer I saw on a website.