3s Hydrogen Wave Function

  • Thread starter chrisa88
  • Start date
  • #1
chrisa88
23
0
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:
[itex]\Psi[/itex](r,[itex]\theta[/itex],[itex]\phi[/itex])= RnlY[itex]^{m}_{l}[/itex]([itex]\theta[/itex],[itex]\phi[/itex])
and Rnl(r) uses the Laguerre polynomial L[itex]^{1}_{3}[/itex](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!!
 

Answers and Replies

  • #2
Simon Bridge
Science Advisor
Homework Helper
17,874
1,657
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:
  • #3
chrisa88
23
0
No, this is all from the same textbook. Which makes it that much more nerve racking!
 
  • #5
chrisa88
23
0
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.
 
  • #6
chrisa88
23
0
Thank you all though!!
 
  • #7
Simon Bridge
Science Advisor
Homework Helper
17,874
1,657
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:

$$L_2^{(1)}\big(\frac{2r}{3a_0}\big) = \frac{1}{2}\big(\frac{2r}{3a_0}\big)^2-3\big(\frac{2r}{3a_0}\big)+3$$
 
  • #8
DrDu
Science Advisor
6,253
905
and Rnl(r) uses the Laguerre polynomial L[itex]^{1}_{3}[/itex](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.
 

Suggested for: 3s Hydrogen Wave Function

  • Last Post
Replies
4
Views
2K
Replies
1
Views
1K
Replies
4
Views
1K
Replies
3
Views
2K
Replies
3
Views
2K
Replies
7
Views
4K
Replies
1
Views
1K
Replies
4
Views
420
  • Last Post
Replies
2
Views
1K
Replies
5
Views
2K
Top