6th Edition HW 2.29
Moderators: Chem_Mod, Chem_Admin
-
- Posts: 61
- Joined: Fri Sep 29, 2017 7:03 am
6th Edition HW 2.29
I don't really understand 2.29, the question asks to identify how many electrons can have the following quantum numbers in an atom. I don't know how to figure out the possible number of electrons that can have a specific quantum number. Can someone explain how to do this?
Re: 6th Edition HW 2.29
in a) n=2, l=1, that means that ml=-1,0,1, but there are also different spins in electrons, so since -1,0,1 (3 options), and there's a possibility of either electron being 1/2 or -1/2 (2 options), 3*2=6. So there are 6 different electrons that can have the following quantum numbers.
b) n=4, l=2, ml=-2 (1 options), then there's a possibility of either electron being 1/2 or -1/2 (2 options) 1*2=2. So there are 2 different electrons that can have the following quantum numbers.
And the same goes for the rest of 2.29
b) n=4, l=2, ml=-2 (1 options), then there's a possibility of either electron being 1/2 or -1/2 (2 options) 1*2=2. So there are 2 different electrons that can have the following quantum numbers.
And the same goes for the rest of 2.29
-
- Posts: 61
- Joined: Fri Sep 29, 2017 7:03 am
Return to “Quantum Numbers and The H-Atom”
Who is online
Users browsing this forum: No registered users and 9 guests