Given that the value of the principle quantum number is n, showthat the total nu

Question

Given that the value of the principle quantum number is n, showthat the total number of possible electron states is 2n2for that energy level .

My attempt: Not sure what i'm thinking wrong, but here are thepossible states you can be in for each quantum number:


Principle: n values
Orbital angular momentum: n values
z-component of angular momentum: 2(n-1)+1 = 2n-1 values
Electron spin: 2 values

Muliplying all those together you get: n*n*(2n-1)*2 which is notwhat we are looking for. The factor I'm most doubtful about is thez-component factor, if we took that out, we get what we're lookingfor but since that's part of the electron state's identity, i don'tknow why you wouldn't include it. Any ideas?

Explanation / Answer

    n = 1 ,2,3,4....n              l   = 0,1,2,3.........(n-1)                  = (n-1) The total number of electrons in the nth - shell isgiven by :    2 ( 2n-1) which follows anArthematic progression having first term 'a' as 2 and acommon difference 'd' as 4. The sum of this series is given by :                    Sn = n /2 [ 2 a + (n -1) d] (the given formula)
                         =   n /2 [ 2*2 + ( n-1) 4 ]                          =   n /2 [ 4 + 4 n - 4]                          =   2 n2  
so the total number of possible electron states is 2n2for that energy level .

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