# 3s Hydrogen Wave Function

## Main Question or Discussion Point

Hi, I am wondering why the associated Laguerre Polynomial for the 3s hydrogen wave function is (27-18σ+2σ2).
My physical chemistry book tells me that the complete hydrogen wave function is given by:
$\Psi$(r,$\theta$,$\phi$)= RnlY$^{m}_{l}$($\theta$,$\phi$)
and Rnl(r) uses the Laguerre polynomial L$^{1}_{3}$(x)=-3!(3-3x+(1/2)x2).

How does this become (27-18σ+2σ2)? I understand that the σ = Zr/a0, but how does the 27 and 2 come about? According to my calculations it should be (18-18σ+3σ2).

Please help me!! I've been wracking my brain about this for way too long and cannot seem to find how this difference from the wave function given in my book and the wave function I produced using the complete hydrogen atomic wave function form comes about.

Thank you!!

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Simon Bridge
Homework Helper
Off the top of my head - there are several representations of the Laguerre Polynomial.
Are you using two different sources? What are they?

Last edited:
No, this is all from the same textbook. Which makes it that much more nerve racking!

DrClaude
Mentor
σ = Zr/a0
Try with σ = 2Zr/a instead.

I'm pretty sure that still does not work. From a quick look at it all that would give is (18-36σ+6σ2). I will discuss this with my teacher tomorrow, time is not permitting me to continue with this issue anymore.

Thank you all though!!

Simon Bridge
Homework Helper
I have: $$L_3^{(1)}=\frac{-x^3}{6}+2x^2-6x+4$$

But I see the polynomial in the general wavefunction given as $$L_{n-l-1}^{2l+1}\big(\frac{2r}{na_0}\big)$$

The 3s states would ne n=3, and l=0 ... so the polynomial is:

$$L_2^{(1)}\big(\frac{2r}{3a_0}\big) = \frac{1}{2}\big(\frac{2r}{3a_0}\big)^2-3\big(\frac{2r}{3a_0}\big)+3$$

DrDu
and Rnl(r) uses the Laguerre polynomial L$^{1}_{3}$(x)=-3!(3-3x+(1/2)x2).
If I set $\sigma=\frac{2}{3}\sigma'$ i get $18-18\sigma +3 \sigma^2=(18-18\frac{2}{3}\sigma'+3(\frac{2}{3}\sigma')^2)=2/3(27-18\sigma' +2\sigma'^2)$