Download Chapter 41: Quantization of Angular Momentum and of Energy Values

Chapter 41: Quantization of
Angular Momentum and of Energy
Values
Classical mechanics: Angular
momentum and energy are in a
continuous range
1 2
p = mv ; E = mv ; L = mr × v
2
v
Quantum mechanics: Angular
momentum and energy quantized in
discrete values.
Similar to classical waves confined in a
cavity
A(x) or Ψ(x)
p or E
x
P ( x) = Ψ
2
1
Classical “picture” of an atom
What’s wrong with this picture?
+
-
Quantum mechanical “picture”
of an atom
P ( x) = Ψ
2
2
Atomic spectral lines
http://www.colorado.edu/physics/2000/quantumzone/index.html
3
Quantization of energy
(frequency) in standing waves
Ψ (x)
note : Ψ = 0 at boundaries
P(x)= Ψ (x)
2
probability
λ for
electron?
4
Quantization of energy
(frequency) in standing waves
2b
λn =
n = 1, 2 ,3…
n
h nh
=
pn =
λn 2b
2
2
2
Energy α n
2
pn
nh
=
En =
2m 8mb 2
0
5
Bohr model of hydrogen
since m p >> me we treat nucleus as stationary
→ m = me
v2
p2
p2
e2
F=
= me a = me =
→
=
2
r me r
4πε 0 r
2me 8πε 0 r
p2
e2
e2
e2
e2
E = K +U =
−
=
−
=−
2m 4πε 0 r 8πε 0 r 4πε 0 r
8πε 0 r
L = me vr , v =
e
4πε 0 me r
→ L = e me r 4πε 0
Classical
e2
L2
r=
me e 2 4πε 0
Angular momentum quantized: Ln = n
n2 2
2
=
rn =
n
a0
2
me e 4πε 0
n = 1, 2, 3…
2
, a0 =
me e 2 4πε 0
=0.53 ×10-10 m
e2
1 e2
13.6 eV
=- 2
En = −
=8πε 0 rn
n 8πε 0 a0
n2
The circumference of the orbit is a integral multiple of λ :
2π rn
de Broglie:
= n , λn = h pn (de Broglie)
λn
2π rn rn pn Ln
→
=
=
=n
h pn h 2π
6
Bohr model of hydrogen
−13.6 eV
En =
n2
E >0 ?
E∞ = ?
http://www.colorado.edu/physics/2000/quantumzone/bohr.html
Note radiation is emitted
ONLY when electron
changes energy levels.
E1 = ?7
Energy of radiation emitted or
absorbed by atoms
Population in nth state ∝ e − En kT
HW # 38 & 39
(Boltzmann distribution
is introduced on p. 560
521,
Fishbane et al, Vol. 1)
hf = Ei − E f
E f < Ei energy released (emission)
E f > Ei energy absorbed (absorption)
What if T is very small?
8
Ex. 41-1: A certain laser
emits light λ=3391 nm, what is
∆E=Ei-Ef in eV?
9
Hydrogen emission/absorption
wavelengths
hf = h
→
1
λ
c
λ
=
= Ei − E f
Ei − E f
hc
 1 1 
= R∞  2 − 2 
 n nf 
 i

R∞ = ( me 2hc ) ( e 4πε 0
2
)
2
= 1.0974 ×107 m -1
Rydberg constant
10
Ex. 41-2: ∆E between lowest
energy of hydrogen (n=1) and
first excited state (n=2)? At
which T is substantial fraction
of a gas of H atoms in first
excited state?
11
The true spectrum of hydrogenHeisenberg and Schroedinger
1. Classical Orbits → wavefunction Ψ (x)
P(x)= Ψ (x)
2
→ states
→ quantum numbers (e.g., n, )
2. L = , is a vector, so L can have different
projections on the z-axis → Lz = m where
m = , − 1, − 2…1, 0, −1, − ( − 1) ,→ 2 + 1 orientations
z
Lz
z
m=+1
L
x
L
m=0
x
m=-1
Example: if = 1 ( L = ), then Lz can have three
values corresponding to
m = 1, 0, or -1
(Lz = , 0, or - )
12
Hydrogen orbits
n = 0,1,2, … (principal quantum number)
l = 0,1,2,…, (n − 1) (orbital momentum)
m = 0,±1,±2, …,±l
l =1
l =0
l =1
l =1
l=2
l=2
l=2
l=2
l=2
13
Projection of L onto z axis: Lz
(
L=
L2 =
(
+ 1) ≈
+ 1)
2
Lz = m
≤ n −1
m ≤
n → En
principal
quantum
number
14
Stern-Gerlach experiment
Beam of atoms (silver) sent through an inhomogeneous
magnetic field. The Ag atoms have a magnetic
momentum µ (like a little bar magnet)
Classically: µ can point in any direction
Reality (and Quantum mechanicly): only certain 15
z-components allowed
Potential energy of the atoms in magn. field:
U mag = − µ ⋅ B = − µ z B
B along zˆ
Force on Ag atoms in inhomogeneous magn. Field:
Fz = −
dU mag
dz
µ z ∝ Lz
dB
= µz
dz
Quantified
⇒only certain values occur
⇒No continuous deflection
but separate beams
16
3. Energy values for hydrogen atom
13.6 eV
E=−
n2
for each n, there are n values ,
as long as ≤ n − 1
for each , there are 2 + 1 of values m ,
as long as m ≤
Examples
n =1→ = 0 → m = 0
1 state total
n=2→ =0→m=0
or = 1 → m = 1, 0, or − 1
4 states total
n =3→ =0→m =0
or = 1 → m = 1, 0, or − 1
or = 2 → m = 2,1, 0, −1, or − 2
9 states total
n=4→?
The total number of states labeled by n is n 2
The energy of the state depends only on n
17
n states total (note that energy only depends on n)
2
Transitions in a H atom
s
p
d
f
g
L photon = → = 1
photon absorption/emission
→ ∆ = ±1
Angular momentum and energy
are conserved in QM
Angular momentum of a photon: Lphoton =
18
Ex. 41-4: Balmer series are
transitions in a H atom that
end at n=2. Longest and
shortest λ in series?
Paschen
Lyman
19
Zeeman Effect: splitting the energies of the
different l quantum states with a magnetic field
Place an atom into a magnetic field B => the different
orientation of the magn. momenta (from the angular
momentum) have different pot. Energies in the
magnetic field
mag
z
U
=> Levels with different
= −µ ⋅ B = − µ B
Lz (or ml
) are split
Energy levels in hydrogen
For an electron in a circular orbit
 e L
where µ B ≡ Bohr magneton
µz = µB = 

20
 2me 
(HW 41-52 and next HW 42 − 30)
L
Zeeman splitting to measure the magnitude of
the magnetic fields on the sun’s surface:
Strong magnetic fields at “sun spots” split
absorbtion lines in two.
21
The Spin of the Electron
Energy levels in silver, l =0 is expected to be
a single state, but gets split into two states!
This means that there is another quantum
number for angular momentum that we have
not included, that is electronic intrinsic
angular momentum or spin:
1
µ=s , s=
2
m =0
1
1
1
→ ms ≤ s → ms = , or 2
2
2
2
number of states = 2s + 1 = 2
(2 + 1 = ?)
≡s
s=
1
ms = spin up;
2
1
ms = − spin down
2
22
Multi-electron atoms
Each electron moves in the attractive Coulomb
potential of the nucleus plus a repulsive potential from
the other electrons
11 electron
1 electron
Note energy now depends on !
23
For a fixed value of n, the energy increases with l
24
Pauli exclusion principle
Due to spin, each ml state is actually
two states (a spin up and a spin down
state). Therefore the total number of
states for the nth level is 2n2.
The Pauli exclusion principle states
the no two electrons may have the
same quantum numbers (unless they
are in different atoms, where one can
distinguish them!). Therefore each
atomic state in a given atom can only
accommodate two electrons (a spin up
and a spin down). To get the lowest
energy of an atom containing more
than one electron, start filling from the
lowest energy state, allowing no more
25
than two electron per state
26
Science Trek 2000/The Periodic Table
Ex. 41-7: Z=37 electrons, what
values of n and for the
electron that is least tightly
bound?
27
Only fermions, which are spin 1/2
particles such as electrons, obey the
Pauli exclusion principle.
Bosons, which are particles with
integer spin (s=0,1,2…), on the other
hand are not only allowed to share the
same quantum state, but prefer to all
be in the same quantum state.
Examples of bosons:
Photons (s=0)
4He (s=n, where n is an integer)
28
Formation of molecules
H2 molecule
+
+
29
Conditions for formation of
molecules
1. Each atom contributes one electron
from outside closed shells to form a
spin up-spin down pair. Since they
have opposite spin, they can move
closer to neighboring nuclei and
attract the atoms together. Each pair
forms a bond, the more pairs the
stronger the binding.
2. Only electrons not in closed shells
can form bonds with other atoms
(electrons in closed shells already
have partners).
3. Similarly, electrons that are already
paired in a shell cannot form bonds
with other atoms.
30
Ionic bond: atom A (alkali metal) has
1 electron outside closed shells, atom
B (halogen) is 1 electron short of
filling outermost shell, the electron is
transferred leaving A+ and B-, ions
attract, strong bond (e.g., NaCl, NaF)
Valence bond: atoms share electrons
to “fill” their outermost shells (e.g.,
H2, GaAs)
31
Science Trek 2000/The Periodic Table
Van der Waals forces
Sometimes inert atoms (all shells
closed) do form molecules. Charge of
one atom induces dipole moment in
other, causes attraction. Remember
the grass seed in electric field field
demo in Physics 108? The van der
Waals force is always attractive but
falls of as 1/r5
E
32
Molecular spectra
H2 (transitions among electronic,
vibrational, and rotational
states)
H (purely electronic transitions)
33
Vibrational motion
Evib = n ω
n is an integer
k
ω=
is the characteristic frequency
M
M 1M 2
M=
is the reduced mass,
M1 + M 2
(M 1 and M 2 are the two nuclear masses)
See Chapter 7 (SHO)
M hydrogen = ?
34
Rotational states
L2
P2
Erot =
Etrans =
2I
2M
Moment of inertia for two masses connected by
a rigid, massless rod of length r0 is: I = Mr02
M=
M 1M 2
M1 + M 2
is the reduced mass
quantize L → L2 =
2
L
=
Erot =
2I
=
(
(
+ 1)
2I
(
2
+ 1)
2
+ 1)
(
=
2Mr02
2
2
me  e  me
+ 1) 

2  4πε 0  4 M
2
8πε 0 2
where we have substituted r0 = 2a0 =
me e 2
Erot =
(
+ 1) E0 me
4M
E0 =13.6 eV
35
Molecular energy states
1 eV
0.0001 eV
0.01 eV
Eelect
− E0
= 2
n
Evib = nE0
me
Z1Z 2
M
E0 = 13.6 eV
me
−4
≈ 10
M
Erot =
(
+ 1) E0 me
4M
36