Derive formula for #2n^2#?

1 Answer
Dec 9, 2017

I assume you are referring to #2n^2# electrons per energy level...

Refer to this answer for info on quantum numbers.


Well, for a given #n#, we have the following properties:

  • #l_max = n-1#
  • #2l+1# orbitals per subshell

From this, we find that for a given energy level #n#, there exist

#sum_(l=0)^(n-1)(2l+1)#

#= (2(0) + 1) + (2(1) + 1) + . . . + (2l_max + 1)#

orbitals in the energy level. Each term contains a #1#, for a total of #n# times, since we sum from #0# to #n-1# inclusive. Therefore, the sum becomes:

#= 2(0 + 1 + 2 + . . . + l_max) + n#

The sum from #0# to #l_max# adds up to #(l_max cdot (l_(max) + 1))/2#, or #(n(n-1))/2#.

As a result, the sum becomes:

#= cancel2 cdot 1/cancel2 n(n-1) + n#

#= n(n-1) + n#

#= n^2# orbitals per energy level #n#.

There exist two electron spins possible for a single electron: #m_s = pm1/2#. From the Pauli Exclusion Principle, there can only be two electrons in the same orbital, without sharing all four atomic quantum numbers.

Therefore, the maximum number of electrons per energy level is #color(blue)(2n^2)#.