How many electrons have quantum numbers #(n,m_s) = (3,1/2)#?
1 Answer
Nine.
#ul(darr color(white)(uarr))" "ul(darr color(white)(uarr))" "ul(darr color(white)(uarr))" "ul(darr color(white)(uarr))" "ul(darr color(white)(uarr))#
#underbrace(" "" "" "" "" "" "" "" "" "" "" "" "" "" ")#
#" "" "" "" "" "" "3d#
#ul(darr color(white)(uarr))" "ul(darr color(white)(uarr))" "ul(darr color(white)(uarr))#
#underbrace(" "" "" "" "" "" "" "" ")#
#" "" "" "3p#
#ul(darr color(white)(uarr))#
#3s#
Well,
For each

#l = 0# ,#m_l = {0}# ,
number of#s# orbitals =#2(0) + 1 = bb1# 
#l = 1# ,#m_l = {1,0,+1}# ,
number of#p# orbitals =#2(1) + 1 = bb3# 
#l = 2# ,#m_l = {2,1,0,+1,+2}# ,
number of#d# orbitals =#2(2) + 1 = bb5#
(recalling that
#m_l = {l, . . . , 0, . . . , +l}# )
Hence,
The spin quantum number of
#ul(darr color(white)(uarr))# or#ul(darr uarr)#
are allowed (the choice of spin for the first electron is arbitrary), but
#ul(darr darr)# or#ul(uarr uarr)#
is not.
Therefore, the total number of configurations for one spindown electron in each possible orbital is the same number as the total number of orbitals.
In other words, there are