Theoretical Analysis of the Hyperfine Structure of NaK

Angela Wilkins

Advisors : Dr. Hickman of Lehigh U.

& Dr. Semak of UNC

Outline

• Molecular Spectroscopy

• Energy levels of NaK

• Angular Momentum Coupling

• Conclusions

Alkali Molecular Structure• Each successive

orbital has a higher energy and lower energy orbitals are filled first

• Alkali atoms have 1 valence electron

• NaK acts like a 2 electron molecule

1s

2p2s

3s3p

4s

3d4p

Energy Electron orbitals of an atom

(Na)

(K)

Molecular Spectroscopy

spectroscopy allows study of different

energy levels

Na K

(R)

Internuclear separation

Experimental setup

M

MM

Moveable Mirror

LL

NaK Heat Pipe Oven

Ti-Sapphire Laser

Dye Laser

Green Fluor. PMT

Red fluor. PMT

M- Mirror

L- Lens

Electronic State Notation13nS+1

•Numeric label

•S-electron spin:

2 electron molecules have parallel

(S=1, triplet) or anti-parallel (S=0,

singlet) spins

•-orbital angular momentum along internuclear axis:

Whole integer numbers (

Different Electronic States

Energy Levels of a Diatomic Molecule

Electronic State (i.e. 13)

Vibrational levels (v)

Rotational levels (N)

Fine Structure

Hyperfine Structure

Energy Levels of NaKEnergy levels are labeled by the angular momentum quantum

numbers: R,L,S, and I.

rotation of nuclei

R is the nuclear orbital angular momentum

L is the electronic orbital angular momentum

S is the electron spin momentum

I is the nuclear spin momentum

Na

K

Fine Structure

L precesses rapidly about the inter- nuclear axis, is a component of L.

N=+R

J=N+S

J=|N-S|,…, N+S

For the triplet NaK cases, S=1,

So J= N-1, N, N+1

LNa K

Fine Structure Levels

N = rotational angular momentum

J = total angular momentum (excluding the nuclear spin)

N=17 J=16

J=17

J=18

N=16 J=15

J=16

J=17

N=15 J=14

J=15

J=16

Hyperfine Structure (Includes Nuclear Spin)

N=+R J=N+SF=J+I

F=|J-I|,…,J+IFor 13 of NaK, I=3/2 soF = J-3/2, J-1/2, J+1/2, J+3/2

Hyperfine structure

N = rotational angular momentum

J = total angular momentum (excluding the nuclear spin)

F=total angular momentum (including nuclear spin)

F=18.5

N=16

J=15

J=16

J=17

F=14.5F=15.5F=16.5F=17.5

F=15.5F=16.5F=17.5

F=13.5F=14.5F=15.5F=16.5

N=17

J=16

J=17

J=18

F=15.5F=16.5F=17.5F=18.5

F=16.5F=17.5F=18.5F=19.5

F=14.5F=15.5F=16.5F=17.5

Experimental Data

N=15

N=45

N=38N=26

As N becomes larger, the spacing between the groups of peaks becomes less.

N=86

More Angular Momentum Coupling

F= N+S+I

Case 1 Case 2 F= [N+S] + I F=N + [S+I] J=N+S G=S+I F=J+I F=N+G

Recall: For 13 of NaK, S=1 and I=3/2

G=|S-I|,…,S+I

G=1/2, 3/2, 5/2

Energy Levels for Limiting Cases

F=15.5F=16.5F=17.5F=18.5

F=17.5F=18.5

N=17

G=5/2

G=3/2

G=1/2

F=14.5F=15.5F=16.5F=17.5

F=13.5

F=18.5N=17

J=16

J=17

J=18

F=15.5F=16.5F=17.5F=18.5

F=16.5F=17.5F=18.5F=19.5

F=14.5F=15.5F=16.5F=17.5

Case 1 Case 2

Model Hamiltonian for NaK (3)H = Hspin-orbit + Hrotation + Hhyperfine + Hspin-rotation

Hspin-orbit = AvLS

Hrotation= Bv [(N(N+1) - 2 ] - Dv [(N(N+1) - 2 ] 2

Hhyperfine= bIS

Hspin-rotation= RS

The 12 energy levels are the eigenvalues of this Hamiltonian.

We adjusted Av, b, and to fit the experimental energies.

Case 1 Case 2

BvN >> Av >> b BvN >> b >>Av

Intermediate CaseF=J+I F=N+G

J=N-1

J=N

J=N+1

G=5/2

G=3/2

G=1/2

Hyperfine coupling strength

Rel

ativ

e E

nerg

y

0.0

-1.0

2.0

1.0

-2.0Case 1 limit Case 2 limit

N=15F=J+I F=N+G

Hyperfine coupling strength

Rel

ativ

e E

nerg

y

Case 1 limit Case 2 limit

0.0

-1.0

2.0

1.0

-2.0

N=38F=J+I F=N+G

Hyperfine coupling strength

Rel

ativ

e E

nerg

y

0.0

-1.0

2.0

1.0

-2.0Case 1 limit Case 2 limit

N=86F=J+I F=N+G

Hyperfine coupling strength

Rel

ativ

e E

nerg

y

0.0

-1.0

2.0

1.0

-2.0Case 2 limitCase 1 limit

Comparison of Experiment and Theory

Hyperfine coupling strength

N=15

N=26

N=38 & 45 N=86 & 87

Red

uced

Ene

rgy

Case 1 limit Case 2 limit

Conclusions

1) The intermediate angular momentum coupling cases explain data.

2) The coupling scheme changes with N.

3) Plan to work further and continue analysis on data at N values > 86 to check agreement with limiting cases and include other electronic states.

Acknowledgements

• Dr A. Peet Hickman

• Dr Matthew Semak

• Dr. Huennekens

• Laurie Sibbach & Catherine Deibel

• NSF for funding

Transition from LS to jj coupling

Light atoms tend to exhibit LS coupling, and heavy atoms tend to exhibit jj coupling. The transition from one to the other can be seen as one goes down a column in the periodic table.

Diagram adapted from Condon and Shortley

Electron Transition