How many electrons have #n = 3#, #m_l = 0#, and #m_s = +1/2#?

1 Answer
Oct 4, 2015

Answer:

Three electrons.

Explanation:

The idea here is that each electron in an atom has an unique set of quantum numbers, which describes exclusively that electron.

The four quantum numbers are

http://socratic.org/questions/what-are-quantum-numbers

In your case, the only quantum number that is not accounted for is the angular momentum quantum number, #l#, which can take values from #0# to #(n-1)#, where #n# is the principal quantum number.

So, the question wants you to find the number of electrons that

  • can be located on the third energy level, since #n=3#;
  • can be located in an orbital that has a specific orientation, since #m_l=0#
  • have spin-up, since #m_s = +1/2#

Start by finding how many orbitals can accomodate these electrons. SInce you know what the relationship of #l# is to #n#, you can say that

#l = 0, 1, ..., (n-1) implies l = {0; 1; 2}#

For #l=0#, which corresponds to the s-orbital, you can have one electron located in the 3s-orbital that has spin-up.

For #l=1#, which corresponds to the 3p-orbitals, you can have an electron located in the #3p_y# orbital, for which #m_l=0#, that has spin-up.

Finally, for #l=2#, which corresponds to the 3d-orbitals, you can have an electron located in the #3d_(yx)#, which would correspond to #m_l=0#, that has spin-up.

Therefore, these quantum numbers can correspond to a total of three electrons

  • #n=3" " l=0" "m_; = 0" "m_s = +1/2#
  • #n=3" "l=1" "m_l = 0" "m_s = +1/2#
  • #n=3" "l=2" "m_l = 0" "m_s = +1/2#