2006-07-11 03:32:27 · 3 answers · asked by eshaghi_2006 3 in Science & Mathematics ➔ Chemistry
For the record, both answers are wrong.
Austin is referring to ORBITAL angular momentum, not the angular momentum of an electron. Furthermore, the angular momentum of a 5s orbital is zero, but it's energy is much greater than a 3d (which has orbital angular momentum of 2).
A better answer is the one I gave here:
http://answers.yahoo.com/question/index;_ylt=AokhAdwvkn2yWFlD.WLeRrHsy6IX?qid=20060711073728AASQumu
2006-07-11 05:10:54 · answer #1 · answered by Iridium190 5 · 0⤊ 0⤋
This is a lot more complicated question than it sounds, but in a very simplified view the angular momentum of an electron will define what orbital (s,p,d or f) it is in. Obviously, the greater the angular momentum, the greater the energy, the further it will be from the nucleus (on average) and which orbital it falls in to.
2006-07-11 10:38:28 · answer #2 · answered by austin.simonson 2 · 0⤊ 0⤋
An electron travels in a curved path.
It has a momentum to continue traveling in a straight line, but it also is bound to the nucleus by electric/magnetic attraction. It then has a momentum toward the nucleus. The net result is a momentum along a curve. That is its angular momentum.
2006-07-11 10:38:19 · answer #3 · answered by bequalming 5 · 0⤊ 0⤋