# Normalizing a radial wave function

1. Nov 2, 2009

### atarr3

1. The problem statement, all variables and given/known data
Show that the radial function R$$_{31}$$ is normalized.

2. Relevant equations
$$\frac{1}{a_{0}^{3/2}}\frac{4}{81\sqrt{6}}\left(6-\frac{r}{a_{0}}\right)\frac{r}{a_{0}}e^{-r/3a_{0}}$$

$$\int^{\infty}_{0}r^{2}R_{31}*R_{31}dr=1$$

3. The attempt at a solution
So I plugged that radial function in and got $$\int^{\infty}_{0}a_{0}^{2}u^{2}\left(6u-u^{2}\right)^{2}e^{-2u/3}du=1$$ all multiplied by some constant and $$u=\frac{r}{a_{0}}$$

I'm getting $$\frac{243}{4}$$ times the constant, and that does not equal one. So I feel like I'm not using the right equation for this one.

Last edited: Nov 2, 2009
2. Nov 2, 2009

Remember what coordinate system you are working in.

3. Nov 2, 2009

### atarr3

Spherical coordinates. So there's an r^2 thrown in there. Did I not account for everything in my integral?

4. Nov 3, 2009

Well, it looks like you didn't in your equation with the u's.

5. Nov 3, 2009

### dotman

I'm seeing an extra $a_o^2$, since there's an $(\frac{1}{a_0^{3/2}})^2 = \frac{1}{a_0^3}$ term, an $r^2 = a_0^2u^2$ term, and a $dr = a_0 du$ term:

$$\frac{1}{a_0^3}a_0^2u^2a_0 du = u^2 du$$

I'm curious as to how you're going about integrating

$$C\int_0^\infty (u^6 - 12u^5 + 36u^4)e^{-2u/3} du$$ ?

Does this have to be done by repeated parts (after you split it up of course), or is there a trick?

6. Nov 4, 2009

### kuruman

A very useful definite integral that was given to me when I learned about the hydrogen atom is

$$\int^{\infty}_{0}x^n \; e^{-ax}dx=\frac{n!}{a^{n+1}}; \; (a>0;\; n\; integer \:>0)$$