Why Does the 3s Hydrogen Wave Function Use the Polynomial (27-18σ+2σ²)?

[chrisa88]

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,$\theta$,$\phi$)= R_nlY$^{m}_{l}$($\theta$,$\phi$)
and R_nl(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!

 

[Simon Bridge]

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

 

[chrisa88]

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

 

[DrClaude]

σ = Zr/a₀

Try with σ = 2Zr/a instead.

 

[chrisa88]

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.

 

[chrisa88]

Thank you all though!

 

[Simon Bridge]

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 R_nl(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).

If I set $\sigma=\frac{2}{3}\sigma'$ i get $18-18\sigma +3 \sigma^2=(18-18\frac{2}{3}\sigma'+3(\frac{2}{3}\sigma')^2)=2/3(27-18\sigma' +2\sigma'^2)$
So up to a change in normalization, your σ and the σ' in your book the seem to differ by a factor 2/3.