Moderators: Chem_Mod, Chem_Admin

Joanne Kang 3I
Posts: 50
Joined: Thu Jul 25, 2019 12:17 am


Postby Joanne Kang 3I » Wed Oct 23, 2019 11:09 pm

What are the principal and orbital angular momentum quantum numbers for each of the following orbitals: (a) 6p; (b) 3d; (c) 2p; (d) 5f?

Can someone please explain how to solve this

Ryan Chang 1C
Posts: 105
Joined: Sat Aug 24, 2019 12:17 am

Re: 1D.15

Postby Ryan Chang 1C » Wed Oct 23, 2019 11:24 pm

The principle quantum number (n) is the number before the letter in the given orbital. To find the angular momentum quantum number (l) you use to following:

s; l=0
p; l=1
d; l=2
f; l=3

For part a, n=6 since the orbital is 6p, and l=1, since it is a p orbital.
For part b, n=3 since the orbital is 3d, and l=2, since it is a d orbital
Part C: n=2, l=1
Part D: n=5, l=3

Posts: 66
Joined: Fri Aug 30, 2019 12:17 am

Re: 1D.15

Postby LReedy_3C » Thu Oct 24, 2019 9:31 pm

The number is the principle quantum number, and the letter corresponds to the orbital angular momentum quantum number. If l=0 its s, l=1 its p, if l=2 its d, and if l=3 its f. From there you can find the answers for all of them.

Victor James 4I
Posts: 50
Joined: Wed Sep 18, 2019 12:20 am

Re: 1D.15

Postby Victor James 4I » Thu Oct 24, 2019 11:50 pm

also, the l value can always only be as high as the n value minus 1

Return to “Quantum Numbers and The H-Atom”

Who is online

Users browsing this forum: No registered users and 3 guests