Lecture36.pptx

Using socalled spectroscopic notabon we have nlm r

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ) × (↑) or (1s ) 1 1 22 (1s )2 × (↑↓ − ↓↑) or (1s )2 00 Also note: each of these one ­electron orbitals has principal quantum number n=1, and orbital angular momentum l=0. Using so ­called “spectroscopic” notaBon we have: ψ nlm (r ) → (nl ), so ψ 100 (r ) = (1s ) Using this notaBon, we can parBally describe the ground states as: Hydrogen: (1s ) Helium: Instead, we “encode” the spin state of the atom in this way: 2 S +1 LJ L = total orbital angular momentum of the atom S = total intrinsic spin angular momentum J = grand total (spin + orbital) angular momentum 3 4 (1s ) (1s ) → (1s )2 For hydrogen: L=0, S=½, J=½, so: Why? Because. L value 0 1 2 3 4 5 6 7 symbol S P D F G H I K 2 S +1 LJ = 2 S1/ 2 Now, let’s look at the first four elements: Z 1 2 3 4 Element H He Li Be configura0on (1s) (1s)2 (1s)2(2s) (1s)2(2s)2 2S+1L J 2S 1/2 0 1S 2S 1/2 0 1S Note: 1. All orbitals are 1s or 2s, with no orbital angular momentum, so total L=0 (S). 2. ...
View Full Document

This note was uploaded on 04/23/2011 for the course PHYS 360 taught by Professor Durbin,stephen during the Spring '11 term at Purdue.

Ask a homework question - tutors are online