The associated Laguerre polynomial required for the Schrodinger equation for orbitals is L(2l+1,n+l-1), where the polynomial is a function of p. Would you agree? p=2*Z*r/(n*a

_{0}), where Z is the atomic number, r is the distance from nucleus to electron, and n is the first quantum number. Let us take l as 0 (so we are looking at the ns orbital case).

The problem is then L(1,2), which should pertain to n=3. According to Wikipedia's sum for n=3 (in terms of a, which we know to be 1), which is L(1,2)=(1/2)p

^{2}-3p+3, I am getting, after I substitute in p=2*Z*r/(n*a

_{0}), which is now p=2*Z*r/(3*a

_{0}), ((2/9)*Z

^{2}*(r/a

_{0})

^{2})-(2*Z*(r/a

_{0}))+3. When I put Z=1 then I get ((2/9)*(r/a

_{0})

^{2})-(2*(r/a

_{0}))+3. Whereas I should be finding (1-(2/3)*(r/a

_{0})+(2/27)*(r/a

_{0})^2) acording to this website (

http://panda.unm.edu/Courses/Finley/P262/Hydrogen/WaveFcns.html). Anyone know the problem?

Strangely enough, I can reach the right answer by dividing by 3. But why is this that I have to? I can't see what I did wrong so that diving by 3 should rectify it. Because I did use the correct expression for p and I've checked the polynomial simplification with WolframAlpha. So something must be going on outside the polynomial which reduces it down by 1/3, but I can't see what.