How many electrons can be spin-up in all the orbitals in the #n = 4# shell?

1 Answer
Truong-Son N.
Sep 26, 2017

Well, if the spin is specified to be #+1/2#, then there are #2l+1# electrons having this configuration in each subshell. And there are four such subshells, #s, p, d, f#, for #n = 4#.

Thus, there are:

#overbrace(2(0) + 1)^(4s) + overbrace(2(1) + 1)^(4p) + overbrace(2(2) + 1)^(4d) + overbrace(2(3) + 1)^(4f)#

#= 0 + 1 + 2 + 1 + 4 + 1 + 6 + 1#

#=# #ulbb"16 electrons"#

Recall that the angular momentum quantum number #l# maxes out at #n - 1#.

Therefore, for #n = 4#,

  • the maximum angular momentum is #l = 3#, or an #f# orbital.
  • the minimum angular momentum is #l = 0#, an #s# orbital.

This gives you the range of #s, p, d, f# orbitals, each of which is to contain an electron of spin #+1/2# in each orbital. The number of orbitals per subshell is given by its degeneracy, #2l+1#.

Lastly, the Pauli Exclusion Principle allows only one spin direction in each quantum state, and only two different spins per orbital. Therefore, if you specify #m_s = +1/2#, you only allow one electron per orbital.

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