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
Solutions:
The Lande g-factor is:
g = 1 +
[J (J + 1) + S (S + 1) - L (L + 1)/ 2 J(J+1)]
For 3 F 3: J = 3, S
= 1, L = 3
g = 1 +
[3 (3 + 1) + 1 (1 + 1) - 3 (3 + 1)/ 2 x3(3+1)]
g = 1 +
[12 + 2 - 12 / 24] = 1
+ 1/12 = 1.08
For 3 F 2: J = 2, S
= 1, L = 3
g = 1 +
[2 (2 + 1) + 1 (1 + 1) - 3 (3 + 1)/ 2 x2(2+1)]
g = 1 +
[6 + 2 - 12/ 12] = 1 +
(-4/12) = 8/12= 2/3
For 3 F 4: J = 4, S
= 1, L = 3
g = 1 +
[4 (4 + 1) + 1 (1 + 1) - 3 (3 + 1)/ 2 x3(3+1)]
g = 1 +
[20 + 2 – 12 / 24] = 1 +
5/12 = 1.41
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:
L, S, J,
g
, M J , D E (in eV x 10 -5 )
Solution:
From the
example problem :
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.
Then: D E = e ħ /2m g B M J
= (5.79 x 10-5 eV/T ) g (0.05T) M
J
So, calculate the value of g for each energy state, i.e.
g = 1 + [J (J + 1) + S (S + 1) - L (L + 1)/ 2 J(J+1)]
Then substitute into the equation for D E with the correct value of M J
We obtain for the results table:
g = 1 + [J (J + 1) + S (S + 1) - L (L + 1)/ 2 J(J+1)]
Then substitute into the equation for D E with the correct value of M J
We obtain for the results table:
m L = (-e (L)/ 2m)
Show in a vector diagram that m and J are not parallel.
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