Normalizing a radial wave function

[atarr3]

Homework Statement

Show that the radial function R[tex]_{31}[/tex] is normalized.

Homework Equations

[tex]\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}}[/tex]

[tex]\int^{\infty}_{0}r^{2}R_{31}*R_{31}dr=1[/tex]

The Attempt at a Solution

So I plugged that radial function in and got [tex]\int^{\infty}_{0}a_{0}^{2}u^{2}\left(6u-u^{2}\right)^{2}e^{-2u/3}du=1[/tex] all multiplied by some constant and [tex]u=\frac{r}{a_{0}}[/tex]

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

 

[badphysicist]

Remember what coordinate system you are working in.

 

[atarr3]

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

 

[badphysicist]

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

 

[dotman]

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

[tex]\frac{1}{a_0^3}a_0^2u^2a_0 du = u^2 du[/tex]

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

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

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

 

[kuruman]

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

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

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

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

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