In the case of certain spectral lines, the Lande g-factor figures prominently as a means to calculate the relative splitting of different energy levels, i.e. in weak magnetic fields. In general we refer to 'g' as the g-factor (generic) which depends on the values of the quantum numbers L, S and J, see e.g.
http://brane-space.blogspot.com/2014/08/an-introduction-to-quantum-mechanics-2.html
Note that both the orbital and spin angular momentum contribute to the magnetic moment of an atomic electron so that:
m = - (e/ 2m) L and m s = - g S (e/ 2m)
So the interaction energy of an atomic electron can then be written:
D(E) = (e/ 2m) [L + 2S ] · B
Basically, the splitting of energy levels in an atom increases the number of spectral lines. If each energy level is split into 2j + 1 components - i.e. one for each of the values of m J (see preceding link)- then the magnitude of the splitting will be different for levels with different Lande g factors.
Note that for the cases where spectral lines are split into three components (normal Zeeman effect as shown in top left image) the transitions Δ m J = 0, + 1 lead to only 3 spectral lines because there are only 3 possible energy differences for these transitions. (Because these cases correspond to transitions between states for which s = 0)
Thus, in the image shown the spectral line appears in classic "triplet" form. That is, there exists a normal (unaffected) line of wavelength lo on either side of which are lesser and greater wavelength lines, hence "splits". The triplet wavelengths are as follows:
lo + D l H
lo
lo - D l H
George Ellery Hale was the first to apply the Zeeman splitting of a solar spectral line to the problem of quantifying the strength of the magnetic field associated with a sunspot. He thereby arrived at the following cgs version of the equation:
D l H = (lo)2 e H / 4 π me c2
Here: H is the intensity of the sunspot magnetic field (to be found), e is the electron charge in electrostatic units (e.s.u.), me is the electron mass in grams and c is the speed of light in cm/sec.
The formula for the Lande g factor is:
g = 1 + [J (J + 1) + S (S + 1) - L (L + 1)/ 2 J(J+1)]
And is present in the Zeeman splitting formula above except "disguised" since g = 1. This is for the particular case s = 0 applied to the spin orbital angular momentum, whence:
S = [s(s + 1)] 1/2 ħ = [0 (0 + 1)] 1/2 ħ = ħ
Then the electron magnetic dipole moment would be:
m s = - g(s) m B S / ħ = - m B
The more general case in which this doesn't apply is the anomalous Zeeman effect.
To obtain a derivation for the Lande g we consider again the magnetic moment (m ) but this time in terms of two components – for the spin and angular momentum such that:
g = 1 + [J (J + 1) + S (S + 1) - L (L + 1)/ 2 J(J+1)]
And is present in the Zeeman splitting formula above except "disguised" since g = 1. This is for the particular case s = 0 applied to the spin orbital angular momentum, whence:
S = [s(s + 1)] 1/2 ħ = [0 (0 + 1)] 1/2 ħ = ħ
Then the electron magnetic dipole moment would be:
m s = - g(s) m B S / ħ = - m B
The more general case in which this doesn't apply is the anomalous Zeeman effect.
To obtain a derivation for the Lande g we consider again the magnetic moment (m ) but this time in terms of two components – for the spin and angular momentum such that:
m
= m s + m L = - e (2S)/ 2m + (-e (L)/ 2m)
= -e (J
+ S)/ 2m
Since: L
+ S = J
The
vector projection of m on J is:
m · J
/ |
J
|
= (-e/ 2m) [J·J
+ J·S ] / | J
|
L
+ S = J
So: L = J – S
Then:
L·L = (J
– S) (J – S) = J·J + S·S -
2 J·S
Re-arranging:
J·S =
½ ( J·J + S·S - L·L)
Then we
can write:
m · J
/ |
J
| = (-e/2m) [J·J + ½ ( J·J + S·S - L·L)/ | J |]
After
substituting specific values for the J·J, etc. one finds:
m · J
/ |
J
|
= g
(-e ħ /2m) Ö J (J + 1)
Where:
g = 1 + [J (J +
1) + S (S + 1) - L (L + 1)/ 2 J(J+1)]
Example
Problem (1):
Find the Lande
g-factor for an atom in the state 1D 3/2.
Solution:
The 1D 3/2 state implies: L = 2, S = 1 with J =
L + S
= 2 + 1 = 3
then:
g = [3(3 + 1) + 1(1 + 1) - 2( 2+ 1)/ 2(3)(3 + 1)] + 1 = 1.3
then:
g = [3(3 + 1) + 1(1 + 1) - 2( 2+ 1)/ 2(3)(3 + 1)] + 1 = 1.3
Is the
Lande g-factor
Anomalous Zeeman Effect:
In
the normal or semi-classical case, the Zeeman effect appears as triplet line
splitting as we saw at the top. This is associated with the precession of the
magnetic moment m about some external magnetic field B. The stronger the field the faster
the precession and the greater the separation between the three spectral lines.
When
the L·S coupling is
strong compared to the interaction of either vector with B, then S and L precess rapidly about J producing
a rapid precession of m about J. (Previous section). This system then precesses slowly about B.
In this
way the anomalous Zeeman effect arises, given it depends upon the component
of m along J
Example Problem (2):
Determine
the value of the energy splitting of an atom in a magnetic field B if it is
assumed that the splitting depends only on the component of m along J .
Solution:
The
component of m along J
is:
m J
= g (-e ħ /2m) Ö J (J +
1)
Also
expressed in vector notation:
m J
= m J J / | J | = g (-e ħ
/2m) Ö J (J +
1) [ J/ Ö J (J +
1) ħ =
-e ħ /2m (g J)
The energy splitting is then given by:
D E =
- m J B
= e ħ /2m g J·B
= e ħ /2m g B J
z = e ħ /2m g B M J
And we know already from quantum mechanics that:
M J = J, J – 1,……, -J + 1, - J
So that
for a given field intensity B each energy level will split into 2J + 1
sublevels with the amount of splitting determined by the g –factor.
Problems for energetic readers:
1) Find the Lande g-factor for an atom in
each of the following states: 3
F 3 , 3 F 2 and 3 F 4
2)
a) Assuming
the L·S interaction to be much stronger than the
interaction with an external magnetic field, calculate the anomalous Zeeman
splitting of the lowest energy states:
2 S 1/2 , 2 P 1/2 and 2 P 3/2
In the hydrogen atom for a field of 0.05T
Present a table with the results of the calculations showing the energy states in the extreme left side column under ‘State’, with the headers of the other columns, viz:
L, S, J, g , M J , D E (in eV x 10 -5 )
2(b) Given that: m s = - e (2S)/ 2m and:
m L = (-e (L)/ 2m)
Show in a vector diagram that m and J are not parallel.
