Transcript Atomic 1

Quantum
Physics
2002
Atomic Physics
Recommended Reading:
Harris Chapter 7
Angular Momentum and Magnetic Dipole Moment.
Imagine an electron orbiting a proton counter-clockwise in an orbit of
radius r. Being negatively charged, it represents a conventional
clockwise current of magnitude I = e/T, where e is the fundamental
charge and T is the period of revolution.
If we have a current I flowing around a loop of area A then the magnetic
dipole moment  is defined as  = IA. Thus
e 2
e
evr
e
e
2
me vr   
μ  IA   πr  
πr  

L
T
2 πr v
2
2me
2me
where me is the mass of the electron, v is
its velocity and we have used the
definition of angular momentum L = mevr.
Both the angular momentum and
magnetic moment are vectors so
e
μ L 
L
2m e
(2)
(1)
Current
L
r
m
v

Result: any orbiting charged particle will have a magnetic dipole moment
We can exploit this relationship to reveal angular momentum quantisation
In the absence of an external force to align them the magnetic moments
 of atoms point in random directions in space. However if we place the
atoms in an external magnetic field B then they gain potential energy
V  μ  B
(3)
And they will experience a force given by
F  V   μ  B  μ x B x  μ z Bz  μ z Bz 
dB y
dB x
dBz
 μx
 μy
 μz
dx
dy
dz
(4)
If the magnetic field is constant everywhere in space then the
derivatives w.r.t x, y and z are all zero and there is no force acting on the
atoms  they have a constant potential energy.
 if we apply a nonuniform magnetic field then we can exert a force on
an atom if it has a magnetic dipole moment.
In 1922 O. Stern and W. Gerlach carried out an experiment to see if they
could detect this force by applying a nonuniform magnetic field to a
beam of atoms
Stern - Gerlach Experiment.
Stern - Gerlach Experiment.
The pole pieces of the magnetic are situated such that there is a
channel in which there is a non-uniform field in the z-direction. The
magnetic field in the x and y (horizontal) is uniform so there is no force
in the x and y direction.
Atoms produced from a metal (Ag) heated in an oven are sent down the
channel in the magnet. Because of the non-uniform field they will
experience a force that will deflect them in the vertical direction.
• If the angular momentum is quantised then it can only point in certain
directions relative to the z-component of the magnetic field (space
quantisation) and the amount of deflection will depend on the zcomponent of the angular momentum
• However. if the angular momentum behaves classically then it can
point in any direction in space, it will still experience a force but this
force will be in a random direction and therefore the magnitude of the
deflection will also be random.
The pattern observed on the screen will then distinguish between
‘classical’ and ‘quantum ‘ behaviour of the atoms in the magnetic field.
Since only the z-component of the force remains in equation (4)
dB z
F  μz
ẑ
dz
(5)
We can relate this force to the angular momentum using equation (2)

 dB
e
F   
L z  z ẑ
 2m e
 dz
(6)
l=0
If the angular momentum L is zero then it will
have no z-component and there will be no force
acting on the atom  it should be undeflected by
the magnetic field
classical, l  0
If L  0, then the classical expectation is that Lz
could be any of the continuous range of values
from -L to +L leading to a continuous band of
deflected atoms on the screen
If the angular momentum L, is quantised then its z-component will be
quantised and therefore the force exerted on the atom will be
quantised. Using the quantisation condition Lz =mL 

 dB
e
F   
m L   z ẑ (mL  - l ....  l)
 2m e
 dz
Suppose we passed hydrogen atoms in the l = 1
l=1
state through the magnet, in this case the
quantum number mL can take on three values -1,
0 and +1 and we would expect to see three lines
on the screen corresponding to to an upward
force (mL = -1), a downward force (mL = +1) and
no force (mL = 0). In general we would expect to
see 2l+1 lines on the screen
Stern and Gerlach used Ag atoms which has l = 0
and therefore they should have seen only one
undeflected line on the screen. However when
they did the experiment they observed two
deflected lines and no undeflected line in the
centre of the screen
mL = -1
mL = 0
mL = +1
Ag: l = 0
This result was baffling, since there should be 2l + 1 lines on the
screen and this result implies that the angular momentum quantum
number l should be 1/2, but we know that l can only be an integer.
Clearly a problem here. What is going on?
in the early 1920’sW. Pauli was the first to suggest
that a fourth quantum number, ( in addition to
n,l,mL) assigned to the electron was needed to
solve this problem.
In 1925 S. Goudsmit and G. Uhlenbeck, proposed
that the electron must have an intrinsic (built in)
angular momentum. This proposal runs into
problems if it is considered classically, so we must
regard this additional angular momentum an a
purely quantum mechanical effect
To explain the experimental data Goudsmit and Uhlenbeck proposed
that the electron has an intrinsic spin quantum number s = 1/2 and
therefore it also has an associated magnetic moment.
We can treat the spin quantum number, in exactly the same was as we
treated the orbital angular momentum.
By analogy, there should be 2s+1 = 2(1/2)+1 = 2 components of the spin
angular momentum vector s.
The magnetic spin quantum number ms (I.e. the z-component of the spin
vector) should have have only two values +1/2 and -1/2, so the electron’s
spin components will be either ‘up’ or ‘down’ in an applied magnetic
field.
The magnitude of the spin vector is
The wavefunction of the hydrogen
atom now has four quantum
numbers to describe each state:
Ψ n,l,m L ,m S
ss  1  
1 1 
3

1





22
4

 z
ms  
2

ms  
2
3
s

4
and the following eigenvalue
equations apply.
ˆ 2 Ψ n,l,m ,m  l l  1 2 Ψ n,l,m ,m
L
L S
L S
ˆ z Ψ n,l,m ,m  m L Ψ n,l,m ,m
L
L S
L S
Sˆ 2 Ψ n,l,m L ,m S  ss  1 2 Ψ n,l,m L ,m S
Sˆ z Ψ n,l,m L ,m S  ms Ψ n,l,m L ,m S
For each state described by the quantum numbers (n, l, mL) discussed
previously there are now two distinct states one with ms =+1/2 (spin
up) and a second with ms = -1/2 (spin-down) . These states will be
degenerate unless the atom is placed in a magnetic field. In a magnetic
field these states will have different energies and this degeneracy is
removed.
n l mL mS state Degeneracy
In a magnetic field these
1 0 0
1s
2
½
states will have different
2 0 0
2s
2
½
energies and this
-1
2p
½
degeneracy is removed.
1 0
2p
6
½
+1 ½
2p
The degeneracy of a state
3 0 0
3s
2
½
with principle quantum
-1
3p
½
number n is now
1 1
3p
6
½
n1
+1 ½
3p
2  2l  1  2n2
l 0
-2
3d
½
-1
3d
½
where the factor of 2
2 0
3d
10
½
comes from the spin
+1 ½
3d
quantum number.
+2 ½
3d
we now have two magnetic moments associated with the electron.
The first is due to the orbital motion of the electron about the nucleus.
μ L 
μ
e
L BL
2m e

The second is associated with the spin of the electron
μ s 
μ
e
S  2 B S
me

Where B is the fundamental unit of magnetic moment, called the
Bohr Magneton
e
μ B 
2m e
We can now explain the result of the S-G experiment. If the atom was
in a state with l = 0, there would be now splitting due to mL. However
there is still space quantisation due to the intrinsic spin of the
electron and the magnetic moment associated with this spin can
interact with the inhomogeneous magnetic field to produce two lines,
one due to the +1/2 state and the other due to the -1/2 state.
Total Angular Momentum.
Each electron had both orbital angular momentum (its L quantum
number) and spin angular momentum (its S quantum number). These
are vector quantities and they can combine vectorially to produce a
Total Angular momentum, denoted by the quantum number J.
If we consider one-electron atoms, for example Hydrogen or the
hydrogen like ions, He+, Li2+ ..., or atoms that have a single electron
outside of a filled core for example the alkali metals such as
Li: 1s2 2s1 = [He] 2s1.
Na: 1s2 2s2 2p6 3s1 = [Ne] 3s1.
K: 1s2 2s2 2p6 3s2 3p6 4s1 = [Ar] 4s1
Then for an electron with orbital angular momentum L and spin
angular momentum S, the total angular momentum J is given by
JLS
Since L, Lz, S and Sz are quantised, the total angular momentum J and
its z-component Jz are also quantised. If j and mj are the
corresponding quantum numbers for the single electron then
J 
j j  1 
and Jz  m j
The value of mj ranges from -j to +j in integer steps.
For example if j = 3 then mj = -3, -2, -1, 0, 1, 2, 3.
The total angular momentum quantum number for a single electron
has the values
jl s
and because ms = ½ for a single electron  j = l + ½ or l - ½ .
Example: for an s-state l = 0 and  j = +½ or -½
S
for an p-state l = 1 and  j = 3/2 or 1/2
for an d-state l = 2 and  j = 5/2 or 3/2 etc
The notation commonly used to describe these
states is
nL j
L
Where n is the principal quantum number, j is the
total angular momentum and L is an uppercase
letter representing the orbital angular momentum
Example: an s-state, l = 0, j = +½ or -½  nS1/2 or nS-1/2
a p-state, l = 1, j = 3/2 or 1/2  nP3/2 or nP1/2
a d-state, l = 2, j = 5/2 or 3/2  nD5/2 or nD3/2
J
S
L
J
Russell Saunders Coupling (LS coupling)
The situation becomes more complicated atoms with more than one
electron outside an inert core. In this case we obtain the total angular
momentum J as follows:
1) we couple the orbital angular momenta li of all the electrons to form
a resultant orbital angular momentum L. For example if we had a pelectron (l1 = 1) and a d-electron (l2 = 2) then the total orbital angular
momentum would be L = l1 + l2 = 3. Note that this is the maximum value
L can have.
2) we couple the spin angular momenta si of all the electrons to form a
resultant spin angular momentum S. For example if we had a pelectron (s1 = 1/2) and a d-electron (s2 = 1/2) then the total spin
angular momentum would be S = s1 + s2 = 1. Note that this is the
maximum value L can have.
3) The total L and S vectors then couple to form the total J.
L  l i
and S   s i
then
JLS
i
i
L, S, J, Lz, Sz and Jz are quantised in the same manner as the single
electron case with quantum numbers L, S, J, mL, ms and mj.
Lˆ 2 Ψ  L L  1 2 Ψ Lˆ z Ψ  m L Ψ
Sˆ 2 Ψ  S  S  1 2 Ψ Sˆ z Ψ  m s Ψ
Jˆ 2 Ψ  J J  1 2 Ψ Jˆ z Ψ  m jΨ
We define the multiplicity of a state as 2S+1. For example if a
multielectron state has a total spin S = 1 then it has a multiplicity of 3 (a
triplet state). For a total spin S = 3/2 the multiplicity is 4 (a quartet state).
Example: Find the possible states under LS coupling for a two electron
atom whose electrons are in p and d states.
electron 1: p-state l1 = 1 and s1 = 1/2.
electron 2: d-state l2 = 2 and s2 = 1/2
L  l 1 - l 2  l 1  l 2  1, 2, 3
S  s1 - s2  s1  s2  0, 1
Since S = 0, 1  multiplicities of 1 (singlet states) and 3 (triplet states)
J can have values from L - S  to  L + S in integral steps
 J = 0, 1, 2, 3, 4
Notation The state of a multi-electron atom is denoted by
n(2S 1)LJ
Example: n = 3, L =1, S = 0, J = 1  31P1 state
n = 5, L = 3, S = 1, J = 4  53F4 state etc...
LS coupling of the p and d electrons leads to all of the following possible
states of the atom:
Singlets: S = 0 then J = L and L = 1, 2, 3 so the singlet states are
1P
Triplets:
1,
1D
2
, 1F3 .
S = 1, L =1  J = 0, 1, 2  3P0 , 3P1 , 3P2 .
S = 1, L =2  J = 1, 2, 3  3D1 , 3D2 , 3D3.
S = 1, L =3  J = 2, 3, 4  3F2, 3F3 , 3F4 .
Note that there are 12 possible angular momentum states of a atom in
which a p and d electron couple together.
In general these states will not be degenerate. So how do we know which
state has the lowest energy???  Pauli Exclusion Principle, Afbau
principle and Hund’s Rules
Pauli Exclusion Principle
Early in the development of the quantum mechanics Wolfgang Pauli
(1900 - 1958) recognized that no two electrons, in the same atom, could
have the same set of four quantum numbers. Two electrons may have the
same n, l and ml quantum numbers, placing them in the same atomic
orbital, but then they must have different spins. Thus each orbital can
accommodate two electrons assuming that their spins are "paired”, I.e
one +1/2 and the other -1/2. This rule is called the Pauli Exclusion
Principle.
Starting with hydrogen, the lowest energy state has one electron in the
1s state, 1s1 (n, l, mL, ms ) = (1,0,0,+1/2) or (1,0,0,-1/2).
Helium 1s2 1st electron = (1,0,0,+1/2), 2nd electron = (1,0,0,-1/2).
Litium 1s2 2s1 = He + (2,0,0,+1/2) or (2,0,0,-1/2).
Beryilium 1s2 2s2 = He + (2,0,0,+1/2) or (2,0,0,-1/2)
Boron 1s2 2s2 2p1 = Be + (2, 1, 0,  1/2) or (2, 1, 0, 1/2) or (2, 1, -1, 1/2)
Boron has a single electron in the 2p level, but where does the next
electron go when we move to carbon?
Let's consider the several possibilities here:
Possible electronic configurations of carbon
(a)
(b)
(c)
We can establish a simple rule that electrons will tend to singly occupy
orbitals until forced to begin pairing up. This favours (b) and (c) over (a)
Unpaired electrons in sublevels with degenerate orbitals (p, d, f, etc) will
tend to have their spins parallel. From an energy standpoint, (c) < (b) <
(a). Thus (c) is the correct configuration.
Afbau (building up) Principle
Nitrogen: 1s2 2s2 2p3
Fluorine: 1s2 2s2 2p5
Oxygen: 1s2 2s2 2p4
Neon: 1s2 2s2 2p6
See http://www.chem.uidaho.edu/~honors/aufbau.html
Hund’s Rules
1) The term with maximum multiplicity lies lowest in energy
 ( 3P, 3D and
3F
) are lower in energy than the (1P , 1D and 1F ).
2) For a given multiplicity, the term with the largest value of L lies lowest
in in energy  ( 3F < 3D < 3P ) and ( 1F < 1D < 1P).
3) For atoms with less than half-filled shells, the level with the lowest
value of J lies lowest in energy. Otherwise the level with the largest J
lies lowest in energy. Since we have only one electron in the p and one
in the d shell they are less than half full so  (3F2 < 3F3 < 3F4 )
Conclusion: if we have an atomic state with two electrons, one in a pstate and one in a d-state. Then the atomic state with lowest energy is a
3F state
2
Hund’s rules only apply when L-S coupling holds and this breaks down
for electrons in heavy elements (then we have to use a different
coupling, j-j coupling
See: http://hyperphysics.phy-astr.gsu.edu/hbase/atomic/hund.html