Why Do g Factors Vary Among Different Atoms?

  • Context: Graduate 
  • Thread starter Thread starter mimocs
  • Start date Start date
  • Tags Tags
    Atoms G factor
mimocs
Messages
13
Reaction score
0
Hi, last week I read Rabi's paper "The Molcular Beam Resonance Method".

This paper contains the basic idea of the oscillation which we call "Rabi Oscillation" as many of you guys know.

However, at the end of this paper, Rabi calculates nuclear magnetic moments of Li (atomic mass 6), Li (atomic mass 7), F (atomic mass 19) by measuring the g factor of the atoms above.

Here's my Question.
g factor for Li (atomic mass 6) is 0.820,
Li (atomic mass 7) is 2.167
F (atomic mass 19) is 5.243
These values differ greatly. Are there any logical reasons to explain the differences of g factor?
Or is it just an intrinsic property of each atoms (just like the spin of electron is 1/2)?

Have a nice day
 
Physics news on Phys.org
The g factor will depend on the energy differences of each state...so why not?
 
ChrisVer said:
The g factor will depend on the energy differences of each state...so why not?
I don't get it...
Isn't the g factor something very similar to gyromagnetic ratio?
And, I would like to know what you mean by 'energy differences of each state'.
Are you talking about the electric potential energy? Or is it something else?
 
It's part of a definition. In general the g-factor is given by the energy difference between the transitions and also the angular momenta: So because of that it also depends on the angular momenta, something that the gyromagnetic ratio doesn't...

Suppose that your atom has a total angular momentum [itex]F=I+J[/itex] where [itex]I[/itex] and [itex]J[/itex] the nuclear and electronic angular momenta respectively. The Hamiltonian is given by:
[itex]H= g~ I \cdot J = \frac{g}{2} [ F^2 - I^2 - J^2 ][/itex]
passing into energies ([itex]\hbar=1[/itex]):
[itex]E= \frac{g}{2} [ F(F+1) - I (I+1) - J (J+1) ][/itex]

That's the energy for a given state. Now it depends... after some transition, the energy difference will be:: [itex]\Delta E[/itex]
and the [itex]g[/itex] will be that energy difference divided by some factor that comes from the angular momenta of those states...
 
geeeee...

Never knew something like that
Thanks a lot for your help

ChrisVer said:
It's part of a definition. In general the g-factor is given by the energy difference between the transitions and also the angular momenta: So because of that it also depends on the angular momenta, something that the gyromagnetic ratio doesn't...

Suppose that your atom has a total angular momentum [itex]F=I+J[/itex] where [itex]I[/itex] and [itex]J[/itex] the nuclear and electronic angular momenta respectively. The Hamiltonian is given by:
[itex]H= g~ I \cdot J = \frac{g}{2} [ F^2 - I^2 - J^2 ][/itex]
passing into energies ([itex]\hbar=1[/itex]):
[itex]E= \frac{g}{2} [ F(F+1) - I (I+1) - J (J+1) ][/itex]

That's the energy for a given state. Now it depends... after some transition, the energy difference will be:: [itex]\Delta E[/itex]
and the [itex]g[/itex] will be that energy difference divided by some factor that comes from the angular momenta of those states...
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 6 ·
Replies
6
Views
5K
  • · Replies 50 ·
2
Replies
50
Views
13K
  • · Replies 13 ·
Replies
13
Views
8K
Replies
2
Views
7K
  • · Replies 4 ·
Replies
4
Views
3K