Why does the possible orbitals equal n squared?

[1MileCrash]

If i take any integer, x

write out all integers from 0 to x - 1

write out all numbers between negative and positive values of those numbers

I get x squared total values. why? what is the mathematical logoc behind this?

x = 3
0 to x - 3 = 0, 1, 2

0
-1, 0, 1
-2, -1, 0, 1, 2

9 total values, x squared total values, but why does this work? how is that process related to squaring?

 

[Mute]

For a given value of the quantum number \ell, the magnetic quantum number m can run from -\ell to \ell, which means there are 2\ell + 1 values of m for each \ell. The quantum number \ell can run from 0 to n-1, where n is the principal quantum number. You can fill up orbitals for each \ell until you hit this max, so the total number of orbitals is

\sum_{\ell = 0}^{n-1} (2\ell + 1) = 2\left(\sum_{\ell = 0}^{n-1}\ell\right) + n

It is a know result that

\sum_{k=1}^{N} k = \frac{N(N+1)}{2}
hence, we get

2\left(\sum_{\ell = 0}^{n-1}\ell\right) + n-1 = 2\frac{(n-1)n}{2} + n = n(n-1 + 1) = n^2.

 

[1MileCrash]

Thank you, I just couldn't find that connection and needed to know out of pure curiosity..

 

[gb7nash]

Thank you, I just couldn't find that connection and needed to know out of pure curiosity..

You may not realize it, but that's a great thing that you discovered that relationship without knowing about the equation. There's a lot of cool relationships you'll discover in math.