Answer:
Yes, the n for potassium would be 4, and for neon would be 2.
Just count which row of the periodic table you are on.
The "L" tells you whether the highest-energy electron is in
an "s" orbital (L=0) or a "p" orbital (L=1) or a "d" orbital (L=2) or an "f" orbital (L=3).
The manner in which these orbitals are filled is:
for each of the first three rows (up to argon),
two electrons in the "s" orbital are filled first, then 6 electrons in the "p"orbitals.
The potassium row also starts with filling the "s" orbital at the new "n" level (4)
but then goes back to filling up the "d" orbitals of n=3 before it fills up the "p"s for n=4.
OK, so potassium has n=4, L = 0, while neon has n=2, L = 1.
The quantum numbers connected with "an element" are always referring to the
highest-energy electron, i.e., the one that was absent in the predecessor element
of the periodic table. When you go from potassium to calcium,
you still get n=4, L = 0, because there are two positions in the "s" orbital.
But when you go from calcium to scandium, suddenly you go back to n=3, L = 2 ("d" orbital).
