Rotation
   Contents
   Applications
   Worked examples
© The scientific sentence. 2010
Formulas
θ = (1/2) α t2 + ωot + θo
ω = α t + ωo
ω2 - ωo2 = 2 α (θ - θo)
ar = ω2(t) r
at = α r
Circ. Unif. motion:
1 rev = 2Ï€ rad
θ = ωt
v = ω r
T = 2π/ω = 1/ƒ
ar = ω2r = v2/r
at = 0
|
| spin orbit
For the hydrogen atom, Schrodinger equation gives energy levels in function of
the principal quantum number n. This equation also
quantify the orbital angular momentum. A state of atom
is given by n, l, and ml. This description of
the state is not sufficient.Using the related orbital quantum number
l and the associated magnetic quantum number ml explains
normal Zeeman effect. The real situation is the
anomalous Zeeman effect.
Experiments show that split of energy
levels by an external magnetic field are not complete in Zeeman effect
described using only orbital angular momentum and its associated magnetic
quantum number l. Sometimes, the split levels are even enaqually spaced.
The numer of split levels are more than those given by
orbital angular momentum alone, that is normal Zeeman effect. By
using high-resolution spectrographs, Some energy levels exhibit
splitting even when there is no external magnetic field. Some lines
of the hydrogen spectrum, for example, are sets of several closely
spaced lines called multiplets. An other example is the sodium
that has a doublet at the 4p level.
1.Electron spin
The well known experiment is the Stern-Gerlach experiment.
The physicists Otto Stern and Walter Gerlach performed in 1922 an
experiment where passed a beam of neutral atoms through a non uniform
magneticfield. With respect to the used magnetic field, the atoms
are deflected according the orientation of their magnetic moments.
The results show some atomic levels are split into an even number
of cmponents, which not comply with the odd (2l+1) number of levels
given by the orbital angular momentum alone. Finding an
even number of levels suggests replacing the orbital quatum number
l by another that has is a half-integer, say j. Therefore, the number (2j+1) fits the
results. Hence, we must add another angular momemtum, say S,
to the orbital angular momentum L to find J. The total angular momentum
vector J is the sum vectors of the L and the S, J = L + S.
All these results cannot be predicted by neither by Bohr model nor by
Schrodinger equation.
As the earth around the sun, the earth spins about itself
and about the sun. This analogy leads to consider that
the electron spnins about itself, that is its axis, that is an electron
is a sphere of charge that has a spin angular momentum S and
a spin magnetic moment μ quantized the same way as the orbital amgular
momentum. Linking L and S, precisely by the dot product L.S, is
called Spin-Orbit coupling. Like L, S and J are quantified:
J = [j(j + 1)]1/2 ħ and Jz = mj ħ
j = |l + s| = |l +- 1/2| that is |l - s| <= j <= |l + s|
- j < mj < +j
S = [s(s + 1)]1/2 ħ and Sz = ms ħ
For an electron (called fermion = spin-1/2 particle):
S = 1/2 and ms = - 1/2 or + 1/2
The spin magnetic quantum numbers ms = -1/2 (spin down) or + 1/2 (spin up) are
the only possible values for ms. S is always equal to 1/2. Therefeore,
to label completely the state of an electron in an atom, we need four quantum
numbers: n, l, ml, and ms.
Example:
For l = 1, we have j = |l + s| = |1 +- 1/2| = 3/2 or 1/2. Then
For j = 1/2: mj = -1/2 and + 1/2
For j = 3/2: mj = -3/2, -1/2, + 1/2, and + 3/2.
In an external magnetic field, the electron interacts with it
by two contributions, the one as discussed in Zeeman effect, that is
the interaction orbit-field; and the second is the interaction spin-field.
These total interaction energy will give an additional Zeeman shifts
due to the spin magnetic moment.
2. Spin magnetic moment
The z-component of the spin magnetic moment μz is
gvien by
μz = - 2.0023 (e/2m) Sz
The gyromagnetic ratio μz/Sz = - 2.0023 (e/2m). This
value is predicted by the Quantum electrodynamics theory.
It is approximately twice as great as the value e/2m for orbital angular momentum
and associated magnetic dipole moment. Equation of Dirac (1928) that takes into account
relativistic effects and spin-orbit coupling, gives a spin gyromagnetic
ratio of 2(e/2m).
When an atom is placed in an external magnetic field, the interaction
energy U = - μ.B of the spin magnetic moment with the field produces
further splittings in energy levels and in the corresponding spectrum lines.
If the field points along the z-axis:
U = - μz.B = (+/-) 0.93 x 10-23B (J/T)
Note that this splitting given by the the spin magnetic dipole moment
exists even there is no external magnetic field. Placing an atom
in an external filed to observe these splittings is just a technique
as a spectroscopic analysis.
3. Landé Factor
The contributions for the energy of interaction of the elctron in an atom with an external magnetif field B are:
U(L) = (e/2m)L.B, due to the orbital motion, and
U(S) = 2(e/2m)S.B due to the spin.
Therefore
the totatl energy of interaction of the magnetic moments ( spin and orbit) is:
U = U(L) + U(S) = μtotal. B = (e/2m) (L + 2S).B
Projecting the vectors L an S in the J direction
gives:
U = (e/2m) (L + 2S).J J.B/J2 =
(e/2m) (L + 2S).(L + S) Jz.B/J2 =
(e/2m) (L2 + 3L.S + 2S2) mjħ B/J2
Since J2 = (L + S)2 =
L2 + S2 + 2L.S, we have then:
3L.S = (3/2) [J2 - L2 - S2]
Therefore:
U = (e/2m) (L2 + (3/2) [J2 - L2 - S2] + 2S2) mjħ B/J2 =
(eħ/2m) (3J2 - L2 + S2) mj B/2J2
= μB g mj B
Where:
μB = eħ/2m: Bohr magneton, and
g = [3j(j + 1) - l(l + 1) + s(s + 1)] /2j(j + 1), the Landé g-factor.
U = g mj μBB
g = [3j(j+1) - l(l+1) + s(s+1)]/2j(j+1),
the Landé factor
|
  |