Transcript Document

 PROGRAM OF “PHYSICS2B”
Lecturer: Dr. DO Xuan Hoi
Room A1.413
E-mail : [email protected]
ANALYTICAL PHYSICS 2B
03 credits (45 periods)
Chapter 1 Geometric Optics
Chapter 2 Wave Optics
Chapter 3 Relativity
Chapter 4 Quantum Physics
Chapter 5 Nuclear Physics
Chapter 6 The Standard Model of Particle Physics
References :
Young and Freedman, University Physics, Volume 2, 12th Edition,
Pearson/Addison Wesley, San Francisco, 2007
Halliday D., Resnick R. and Merrill, J. (1988), Fundamentals of
Physics, Extended third edition. John Willey and Sons, Inc.
Alonso M. and Finn E.J. (1992), Physics, Addison-Wesley
Publishing Company
Hecht, E. (2000), Physics. Calculus, Second Edition.
Brooks/Cole.
Faughn/Serway (2006), Serway’s College Physics, Brooks/Cole.
Roger Muncaster (1994), A-Level Physics, Stanley Thornes.
http://ocw.mit.edu/OcwWeb/Physics/index.htm
http://www.opensourcephysics.org/index.html
http://hyperphysics.phyastr.gsu.edu/hbase/HFrame.html
http://www.practicalphysics.org/go/Default.ht
ml
http://www.msm.cam.ac.uk/
http://www.iop.org/index.html
.
.
.
PHYSICS 2B
Chapter 4
Quantum Physics & Quantum Mechanics
Nature of light - Matter Wave
The Schrödinger’s Equation
The Heisenberg’s uncertainty principle
The Bohr Atom
The Schrödinger Equation for the Hydrogen Atom
The Zeeman Effect
The Dirac Equation and antimatter
1. Nature of light - Matter Wave
Diffraction, Interference phenomena  light has wave nature
Photoelectric, Compton’s effects  light has particle nature
What is the nature of light?
ANSWER:
From a modern viewpoint,
the light has both wave and particle characteristics
That is
The Wave-Particle Duality of Light
“… the wave and corpuscular descriptions are only to be
regarded as complementary ways of viewing one and the same
objective process…” (Pauli, physicist)
1. 1 Continuous and line spectra
Prism (or a diffraction grating) to separate the various
wavelengths in a beam of light into a spectrum
Light source: Hot solid (such as a light-bulb filament) or liquid
 continuous spectrum
Source: Gas  line spectrum (isolated sharp parallel lines)
Each line: a definite wavelength and frequency.
Continuous spectrum
White light: 380 nm (violet)  760 nm (red)
Line spectrum
Explain?
1. 2 Rutherford's gold foil experiment
The scattering of alpha particles () by a thin metal foil:
The source of alpha particles: a radioactive element (Ra)
 alpha particles are scattered by the gold foil.
Result: The atom has a nucleus; a very small, very dense
structure (10-14 m in diameter; 99.95% of the total mass of the
atom)
1. 3 The photoelectric effect
The photoelectric effect:
light strikes a metal surface  emission of electrons
electron
light
_
_
_
_
++ + +
metal
Einstein's Photon Explanation: a beam of light consists
of small packages of energy called photons or quanta.
The energy of a photon:
hc
E  hf 

f,  : frequency, wavelength of the light
Planck's constant h = 6.626069310-34J.s
The photoelectric effect:
hc
E  hf 
 K max  

: work function (minimum energy needed to remove an
electron from the surface)
Maximum kinetic energy : K max
1
 mv 2max
2
V0: The stopping potential
Theorem of variation of kinetic energy:
K max  K 'max  eV0
K 'max  0  K max 
1
mv 2max  eV0
2
EXAMPLE While conducting a photoelectric-effect
experiment with light of a certain frequency, you find that a
reverse potential difference of 1.25 V is required to reduce the
current to zero. Find
(a) the maximum kinetic energy and
(b) the maximum speed of the emitted photoelectrons.
(a)
(b)
EXAMPLE
When ultraviolet light with a wavelength of
254 nm falls on a clean copper surface, the stopping potential
necessary to stop emission of photoelectrons is 0.181 V.
What is the photoelectric threshold wavelength for this copper
surface?
Photon Momentum:
E hf h
p 

c
c 
Photons have zero rest mass.
The direction of the photon's momentum
is the direction in which the
electromagnetic wave is.
Compton Scattering
When X rays strike matter,
some of the radiation is
scattered: Some of the
scattered radiation has
smaller frequency (longer
wavelength) than the
incident radiation and that the
change in wavelength depends
on the angle through which
the radiation is scattered.
Compton Scattering
2
Energy conservation: pc  mc  p'c  Ee
Momentum conservation: p  p ' Pe
Pe2  p2  p'2  2pp'cos 
h
h
With p 
and p ' 
:
'

h
   '  
1  cos  
mc
(Compton shift)
EXAMPLE You use the x-ray photons  = 0.124 nm in a
Compton scattering experiment.
(a) At what angle is the wavelength of the scattered X rays 1.0%
longer than that of the incident X rays?
(b) At what angle is it 0.050% longer?
(a)
(b)
1.4 The Davisson - Germer experiment
Electron diffraction experiment (Davisson, Germer, Thomson, 1927)
A beam of either X rays (wave) or
electrons (particle) is directed onto a
target which is a gold foil
The scatter of X rays or electrons by
the gold crystal produces a circular
diffraction pattern on a photographic
film
The pattern of the two experiments
are the same
Particle could act like a wave;
both X rays and electrons are
WAVE
Discussion
 Similar diffraction and interference experiments have been
realized with protons, neutrons, and various atoms
(see “Atomic Interferometer”, Xuan Hoi DO, report of practice
of Bachelor degree of Physics, University Paris-North, 1993)
In 1994, it was demonstrated with iodine molecules I2
Wave interference
patterns of atoms
Small objects like electrons, protons, neutrons, atoms, and
molecules travel as waves - we call these waves “matter waves”
1.5 De Broglie’s Theory - Matter Wave
De Broglie’s relationships
We recall that
for a photon (E, p) associated to an electromagnetic wave (f, ):
 h f
1
p  h

E

particle

wave
De Broglie’s hypothesis:
To a particle (E, p) is associated a matter wave
which has a frequency f and a wavelength 
E
f 
h
h

p
E
h
 From f 
and  
h
p
if we put:
h  2 
E  


p K


E  2f 

2
p


Planck-Einstein’s relation
 is called de Broglie wavelength
Conclusion. Necessity of a new science
 When we deal with a wave, there is always some quantity
that varies (the coordinate u, the electricity field E)
 For a matter wave, what is varying?
ANSWER: That is the WAVE FUNCTION
 New brand of physics: THE QUANTUM MECHANICS
PROBLEM 1 In a research laboratory, electrons are
accelerated to speed of 6.0  106 m/s. Nearby, a 1.0  10-9 kg
speck of dust falls through the air at a speed of 0.020m/s.
Calculate the de Broglie wavelength in both case
 For the electron:
SOLUTION
h
h
6.625 1034 J .s
 

p mv 9.111031 kg  6.0 106 m / s
  1.2 10 10 m
 For the dust speck:
h
h
6.625 1034 J .s
d 


pd mvd 1.0 109 kg  6.0  0.020m / s
d  3.3 10 23 m
DISCUSSION: The de Broglie wavelength of the dust speck is so
small that we do not observe its wavelike behavior
PROBLEM 2
An electron microscope uses 40-keV electrons.
Find the wavelength of this electron.
SOLUTION
The velocity of this electron:
v  2K / m
v
2  40  103  1.6  1019
 1.2  108 m / s
9.1 1031
The wavelength of this electron:
h

mv
o
6.63 1034
10

m  6.1 A
31
8  6.1 10
9.110 1.2 10
2 The Schrödinger’s Equation
2.1 Wave Function and Probability Density
Matter waves:
A moving particle (electron, photon) with momentum p is described
by a matter wave; its wavelength is  h / p
A matter wave is described by a wave function: ( x, y, z; t )
(called uppercase psi)
( x, y, z; t ) is a complex number(a + ib with i2 = -1, a, b: real numbers)
( x, y, z; t ) depends on the space (x, y, z) and on the time (t)
The space and the time can be grouped separately:
( x, y, z; t )   ( x, y, z ) e it
 ( x, y, z ) : space- dependentpart (lower casepsi)
 i t
: time - dependentpart ( : angular frequency)
e
• The meaning of the wave function:  ( x, y, z )
The function  ( x, y, z ) has no meaning
2
Only  has a physical meaning. That is:
The probability per unit time of detecting a particle in a small
volume centered on a given point in the matter 2wave is
proportional to the value at that point of 

2
greater  it is easier to find the particle
*
*




N.B.:
with  is the complex conjugate of 
If we write   a  ib   *  a  ib (a, b: real numbers)
2
• How can we find the wave equation?
Like sound waves described by Newtonian mechanics,
or electromagnetic waves by Maxwell’s equation,
matter waves are described by an equation called
Schrödinger’s equation (1926)
2.2 The Schrödinger’s equation
For the case of one-dimensional motion,
when a particle with the mass m has a potential energy U(x)
Schrödinger’s equation is
d 2
dx
2

8 2 m
h
2
[ E  U ( x)]  0
where E is total mechanical energy (potential energy plus kinetic energy)
Schrödinger’s equation is the basic principle
(we cannot derive it from more basic principles)
EXAMPLE: Waves of a free particle
For a free particle, there is no net force acting on it, so
1 2
U ( x)  0 and E  mv
2
d 2 8 2 m  1 2 
Schrödinger’s equation becomes:


2 
2  2 mv   0
dx
h
d 2
 2 p 2
By replacing: mv  p 


2  h    0
dx
1 p
With the de Broglie wavelength:

 h
2
and the wave number: K 

we have the Schrödinger’s equation for free particle:
d 2
 K 2  0
dx 2
This differential equation has the most general solution:
 ( x)  AeiKx  Be iKx
(A and B are arbitrary constants)
The time-dependent wave function:
( x, t )   ( x) ei t  ( AeiKx  Be iKx ) ei t
 ( x, t )  Aei ( Kx t )  Be i( Kx t )
 ( x, t )  Aei ( Kx t )  Be i ( Kx t ) : traveling waves
 Aei ( Kx t ) : wave traveling in the direction of increasing x
 Be i (kx t ) : wave traveling in the negative direction of x
Probability density:
Assume that the free particle travels only in the positive direction
Relabel the constant A as 0 :  ( x)  0eiKx
The probability density is:
2
   0e
iKx 2
2 iKx 2
 (0 ) e
Because:
e
iKx 2
 (eiKx )(eiKx )*  (eiKx )(e iKx )  e0  1
we have:
2
  (0 ) 2  const
What is the meaning of a constant probability? 
2
 (0 ) 2  const
2
The plot of   (0 ) 2  const
is a straight line parallel to the x axis:

2
0
2
x
O
The probability density is the same for all values of x
The particle has equal probabilities of being anywhere
along the x axis: all positions are equally likely expected
3 The Heisenberg’s uncertainty principle
• In the example of a free particle, we see that if its momentum is
completely specified, then its position is completely unspecified
• When the momentum p is completely specified we write:
p  0 (because: p  p1  p2  0)
and when the position x is completely unspecified we write:
x  
• In general, we always have: x  p  a constant
This constant is known as:
(called h-bar)
h

2
h is the Planck’s constant
(h  6.625 1034 J .s)
So we can write:
x  p  
That is the Heisenberg’s uncertainty principle
“ it is impossible to know simultaneously and with exactness
both the position and the momentum of the fundamental particles”
N.B.: • We also have for the particle moving in three dimensions
x  px  
y  p y  
z  pz  
• With the definition of the constant  :
p  h /   hK / 2
• Uncertainty for energy :
E  t 
p  K
PROBLEM 3 An electron is moving along x axis with the speed
of 2×106 m/s (known with a precision of 0.50%).
What is the minimum uncertainty with which we can
simultaneously measure the position of the electron along the x
axis? Given the mass of an electron 9.1×10-31 kg
SOLUTION
From the uncertainty principle: x  p  
if we want to have the minimum uncertainty: x  p  
We evaluate the momentum: p  mv  (9.11031)  (2.05 106 )
p  9.35 1027 kg.m / s

The uncertainty of the momentum is:
p  0.5% p  0.5 / 100 1.87 1024  9.35 1027 kg.m / s

6.635 1034 / 2
8

1
.
13

10
m  11nm
x 


27
p
9.35 10
PROBLEM 4 In an experiment, an electron is determined to be
within 0.1mm of a particular point. If we try to measure the
electron’s velocity, what will be the minimum uncertainty?
SOLUTION
p

v 

m mx
6.63 1034 J .s
v 
9.11031 kg 1.0 104 m  2
v  1.2m / s
Observation:
We can predict the velocity of the electron to within 1.2m/s.
Locating the electron at one position affects our ability to know
where it will be at later times
PROBLEM 5 A grain of sand with the mass of 1.00 mg appears
to be at rest on a smooth surface. We locate its position to within
0.01mm. What velocity limit is implied by our measurement of its
position?
SOLUTION
p

v 

m mx
6.63  1034 J .s
v 
1 106 kg  1.0 105 m  2
v  1.11023 m / s
Observation:
The uncertainty of velocity of the grain is so small that we do not
observe it: The grain of sand may still be considered at rest, as our
experience says it should
PROBLEM 6 An electron is confined within a region of width
1.010- 10 m. (a) Estimate the minimum uncertainty in the
x-component of the electron's momentum.
(b) If the electron has momentum with magnitude equal to the
uncertainty found in part (a), what is its kinetic energy? Express
the result in jou1es and in electron volts.
(a)
(b)
SOLUTION
PROBLEM 7 A sodium atom is in one of the states labeled
''Lowest excited levels". It remains in that state for an average
time of 1.610-8 s before it makes a transition back to a ground
state, emitting a photon with wavelength 589.0 nm and energy
2.105 eV. What is the uncertainty in energy of that excited state?
What is the wavelength spread of the corresponding spectrum line?
SOLUTION
The fractional uncertainty of the photon energy is
4 The Bohr Atom
4.1 The energy levels
The hydrogen atom consists of a single
electron (charge – e) bound to its central
nucleus, a single proton (charge + e), by
an attractive Coulomb force
1 (e)( e)
e2

The electric potential energy is : U 
40
r
40 r
We can demonstrate that the energies of the quantum electron is
 me4  1
En   2 2  2
 8 0 h  n
En  
13.6eV
n2
where n is an integer, that is the principal quantum number
The energies of the hydrogen atom is quantized
4.2 Spectral emission lines
When the electron jumps down from an energy level Em to
a lower one En , the hydrogen atom emits a photon of energy:
  hf mn 
• If
• If
• If
• If
.
.
.
hc
 mn
 E m  En
E
E4
E3
En  E1 : Lyman series
En  E2 : Balmer series
En  E3 : Paschen series
En  E4 : Backett series
Paschen
E2
Balmer
E1
Lyman
PROBLEM 8
1/ What is the wavelength of light for the least energetic photon
emitted in the Lyman series of the hydrogen atom spectrum lines?
2/ What is the wavelength of the line H in the Balmer series?
SOLUTION
1/ For the Lyman series:   hf m1 
hc
 m1
 Em  E1
The least energetic photon is the transition between E1 and the level
immediately above it; that is E2 . The energy difference is:
 13.6eV   13.6eV 
 10.2eV



E  E2  E1   



22  
12 
hc
(6.63 1034 )  (3 108 )
7

1
.
22

10
m  0.122  m  122nm



19
E
10.2 1.6 10
(in the ultraviolet range)
2/ The line H in the Balmer series corresponds to the transition
between E3 and E2 .
The energy difference is:
 13.6eV   13.6eV 
  1.89eV
  
E  E3  E2  
2
2


 
3
2
hc (6.63 1034 )  (3  108 )
7



6
.
58

10
m  0.658  m  658nm

19
E
1.89 1.6 10
(red color)
• The line H : E3
• The line H : E4
• The line H : E5
• The line H : E6
 E2
 E2
 E2
 E2
( = 658 nm)
( = 486 nm)
( = 434 nm)
( = 410 nm)
PROBLEM 9
An atom can be viewed as a numbers of electrons moving around a
positively charged nucleus. Assume that these electrons are in a box
with length that is the diameter of the atom (0.2 nm).
Estimate the energy (in eV) required to raised an electron from the
ground state to the first excited state and the wavelength that can
cause this transition.
SOLUTION
En 
h2
8ma
E2 
2

n
2
h2
8ma
E1 

h2
8ma
2

1
2
(6.625 1034 ) 2
8  9.11031  (0.2 109 ) 2 1.6 10
 9.40eV
2

2
 4E1  37.6eV
2
2

1
19
Energy required:
E  E2  E1  37.6  9.40  28.2eV
The wavelength that can cause this transition:   hc / E  44.0nm
PROBLEM 10
According to the basic assumptions of the Bohr theory applied to the
hydrogen atom, the size of the allowed electron orbits is determined
by a condition imposed on the electron’s orbital angular momentum:
this quantity must be an integral multiple of
:
mvr  n ; n  1,2,3,...
1/ Demonstrate that the electron can exist only in certain allowed
orbit determined by the integer n
2/ Find the formula for the wavelength of the emission spectra.
SOLUTION
1 2
e2
1/ E  KE  PE  mv  k
2
r
e2
v2
Newton’s second law: F  ma; k 2  m
r
r
e2
The energy of the atom: E  k
2r
KE 
ke2
2r 2
n2
2
2
v
With: v2  2 2 : k 2  m
r
mr
r
We have: rn 
With: a0 
e2
n2
mke2
r2
2 2
m n2 2
n

; ke2 
r m2r 2
mr
2
2
mke
2

ke2
(the electronic orbits are quantized)
 0.0529nm (Bohr radius), rn  a0n2
2 4
mk
e  1
2/ We have : E  
 
2 2  n2 
En  
13.6eV
n2
Ei  E f
The frequency of the emitted photon is given by: f 
h
2
4
mk e
1 f mk2e4  1
1
7 1
R


1.097

10
m

 
 2  With: H
3
3 2
4 c
 c 4 c  n f ni 
(Rydberg constant)
 1
1
1
 RH  2  2 
 n f ni 



PROBLEM 11
The result of the Bohr theory of the hydrogen atom can be
extended to hydrogen-like atoms by substituting Ze2 for e2 in the
hydrogen equations.
Find the energy of the singly ionized helium He+ in the ground state
in eV and the radius of the ground-state orbit
SOLUTION
For He+: Z = 2
2
mk2 Z 2e4  1 
Z
(13.6eV )
E
 2  En  
2
2
n 
n2
The ground state energy: E1  4(13.6eV )  54.4eV
• The radius of the ground state : rn 
n2
2
mkZe2
n2
rn  a0
Z
12
r1  a0  0.0265nm (The atom is smaller; the electron is more
2
tightly bound than in hydrogen atom)
5 The Schrödinger Equation for the Hydrogen Atom
Electron cloud
In spherical coordinates (r, , ), the hydrogen-atom problem
is formulated as :
 (r )   (r , , )  R (r )Y ( , ) ; Y ( ,  )  ( )()
e2
The potential energy is : U (r )  
4 0 r
1
The Schrödinger equation in three dimensions :
2
x
2
 2
x
2


2
y
2
 2
y
2


2
z
2
 2
z 2

2m
2
[E  U (r )]  0
2m 
1 e2 
 2 E 
  0
4 0 r 

5.1 Radial function of the hydrogen atom
If  depends only on r :   (r)
r
r
x
 2x
r  x  y  z ; 2r

x
x
r
2
 d  r
x d          x d  


;
x  r dr 
x 2 x x
x
dr x r dr
2
2
2
2
2
x r d  x d 2 r
1 d  x 2 d  x 2 d 2

 2


 3
 2
2
2
r dr r x dr
r dr x r dr
x
r dr
r dr 2
1 d
2
x 2 d  x 2 d 2

 3
 2
2
r dr
x
r dr
r dr 2
1 d
The same result for y and z :
 2 1 d  y 2 d  y 2 d 2

 3
 2
2
r dr
y
r dr
r dr 2
2
z 2 d  z 2 d 2

 3
 2
2
r dr
z
r dr r dr 2
 2
x
2

1 df 
 2
y
2

 2
z 2
d 2 2 d 


2
r dr
dr
The Schrödinger equation :
d 2 2 d  2m

 2
2
r dr
dr

1 e2 
E 
  0
4 0 r 

 Wave function for the ground state
The wave function for the ground state of the hydrogen atom:
 (r ) 
1
a 3/2 
e r / a
10
a

0
.
529

10
m  52.9nm
Where a is the Bohr radius:
 Wave function for the first excited state
 (r )  Ce ar / 2 (2  ar )
PROBLEM 12
Knowing that the wave function for the ground state of the
hydrogen atom is
r / a
 (r )  Ae
Where a is the Borh radius: a  0.529 1010 m  52.9nm
1/ What is the value of the normalization constant A ?
2/ What is the value of x at which the radial probability density
has a maximum?
SOLUTION
1/ Because the electron moves in the three dimensional space, the
probability of finding the electron in a volume dV is written:
 dV  A2e2r / a dV
2
where: dV  4 r 2dr

  (r )
Normalization condition:

(1)

We put: I 
dV  1
0
4 A2  e  2 r / a r 2 dr  1
0
2
 e2r / a r 2dr

 a 3  z 2
I     e z dz
By changing variable: z  2r / a
2 0
By integration by parts, we can demonstrate the general formula:
0

 e z z n dz  n ! 1 2  ...  (n  1)  n
0

I   a / 2 3  e  z z 2 dz  2 a / 2 3  a 3 / 4
0
Substituting the value of I into (1): 4 A a / 4  1
2 3
 (r ) 
1
a3 / 2 
er / a
A
1
a3 / 2 
2/ Because:
2  2r / a
2
 dV  A e
dV 
1
a
3e
 2r / a
4 r dr 
2
4
a
 2r / a 2
e
r dr
3
the radial probability density is
4  2r / a 2
P(r)
P( r )  3 e
r
a
dP (r )
When P (r) has a maximum:
0
dr


dP(r ) 4 d 2r / a 2
 3
e
r
dr
a dr
O

4  2 2 2r / a
 3  r ( )e
 2re  2 r / a 

a
a 

8
 2r / a
r
(
a

r
)
e
0
4

a
ra
a
(The value r = 0 corresponds to a minimum of P( r) )
Physical meaning: The position r = a is the most probable for the
electron  electronic clouds
r
PROBLEM 13
Calculate the probability that the electron in the ground state of
the hydrogen atom will be found outside the Bohr radius
SOLUTION
1
With:  (r )  3 / 2
er / a
a


The probability is found by: P 
  (r ) dV where: dV  4 r 2dr
2
a
P
4
a
1
P
2
3


 r 2e2r / a dr
a
By changing variables: z  2r / a ,

1
 z 2e z dz   2 {z 2  2 z  2}e z  5e2  0.677  P  67.7%
2
2
PROBLEM 14
The wave function of a particle is given as:  (r )  Ce x / a
1/ Find C in terms of a such that the wave function is normalized in
all space
2/ Calculate the probability that the particle will be found in the
interval -a  x  a
SOLUTION


2C 2  e  2 x / a dx  1
1/ With: C 2  e  2 x / a dx  1
2 a 

1
C
a
2C    1
2
a
2/ P 
  ( x)
a
0
2
a
a
dx  2   ( x) dx  2C 2  e  2 x / a dx
0
2
2
0
P  2C 2 (a / 2)(1  e )  2(1 / a )2 (a / 2)(1  e2 )  86.5%
5.2 Quantization of Orbital Angular Momentum
NOTES : In classical mechanics :
 Force and linear momentum in translational motion :
dv
d (mv ) dp
F  ma  m


dt
dt
dt
 Torque and angular momentum in rotational motion
  Fd
  F r
L  pr
L r p
L  mvr  m (r )r  (mr 2 )  I 
In quantum mechanics :
 In the wave function of electron in hydrogen atom :
 (r )   (r , , )  R (r )Y ( , ) ; Y ( ,  )  ( )()
The requirement that the () function must be finite at
= 0 and  =  gives the result : L can take some possible
values :
L  l (l  1)
( l  0 ; 1 ; 2 ; 3 ;...; n  1 )
The number l is called : the orbital angular-momentum
quantum number or the orbital quantum number for
short.
In quantum mechanics :
L  l (l  1)
( l  0 ; 1 ; 2 ; 3 ;...; n  1 )
 On the other hand, the permitted values of the component
of L in a given direction, say the z-component Lz are
determined by the requirement that the () function must
equal ( + 2).
The possible values of Lz are
Lz  ml
( ml  0 ;  1 ;  2 ;  3 ;...;  l )

( ml  l ;  l  1 ;  l  2 ; ...  1 ; 0 ; 1 ;...; l  1 ; l )
We call ml the orbital magnetic quantum number
( l  0 ; 1 ; 2 ; 3 ;...; n  1 )
L  l (l  1)
Lz  ml
( ml  0 ;  1 ;  2 ;  3 ;...;  l )
( ml  l ;  l  1 ;  l  2 ; ...  1 ; 0 ; 1 ;...; l  1 ; l )
EXAMPLE : For n = 3; the possible values of l are : 0 ; 1 ; 2
With l = 2; the possible values of ml are : -2 ; -1 ; 0 ; +1 ; +2
L  2(2  1)
 6
Lz  0 ;  1 ;  2
 2.45
PROBLEM 15
How many distinct (n, l, ml ) states of the hydrogen atom
with n = 3 are there? Find the energy of these states.
SOLUTION
For n = 3; the possible values of l are : 0 ; 1 ; 2
With l = 0 ; the possible value of ml is : 0
With l = 1 ; the possible values of ml are : -1 ; 0 ; +1
With l = 2 ; the possible values of ml are : -2 ; -1 ; 0 ; +1 ; +2
The total number of (n , l , m, ) states with n = 3 is therefore
1 + 3 + 5 = 9.
En  
13.6eV
n
2
E3  
13.6 eV
2
3
 1.51 eV
PROBLEM 16
Consider the n = 4 states of hydrogen.
(a) What is the maximum magnitude L of the orbital angular
momentum?
(b) What is the maximum value of Lz?
(c) What is the minimum angle between L and the z-axis?
SOLUTION
(a)
When n = 4, the maximum value of the orbital angularmomentum quantum number l is (n  1)  (4  1)  3
L  l (l  1)
 3(3  1)
 12
(b) For l = 3 the maximum value of the magnetic quantum
number ml is 3 :
Lz  ml  3
PROBLEM 17
Consider the n = 4 states of hydrogen.
(a) What is the maximum magnitude L of the orbital angular
momentum?
(b) What is the maximum value of Lz?
(c) What is the minimum angle between L and the z-axis?
SOLUTION
(c) The minimum allowed angle between L
and the z-axis corresponds to the
maximum allowed values of Lz and ml
cos min 
(Lz )max
L

min  300
3
12
PROBLEM 18
Represent all the possible orientations of the
angular momentum with the value l = 0 ; 1 ; 2 ; 3
SOLUTION
 l = 0 : L = 0 ; ml = 0
2

Lz
l=1
L  l (l  1)
l 1
l 2
L 2
L 6
0

2
 1(1  1)
ml = -1 ; 0 ; +1
l 0
L 0
 2
Lz  ml  0 ;  1
l=2
L  l (l  1)
 2(2  1)
ml = -2 ; -1 ; 0 ; +1 ; +2
 6
Lz  ml  0 ;  1 ;  2
PROBLEM 19 (a) If the value of Lz is known, we cannot
know either Lx or Ly precisely. But we can know the value of the
quantity L2x  L2y . Write an expression for this quantity in terms
of l and ml .
(b) What is the meaning of L2x  L2y ?
(c) For a state of nonzero orbital angular momentum, find the
maximum and minimum values of L2x  L2y . Explain your results.
SOLUTION
(a) L2x  L2y  L2  L2z  l (l  1)
 ml2
2
2

l (l  1)  ml2
(b) This is the magnitude of the component of angular
momentum perpendicular to the z-axis
(c) The maximum value :
L2x  L2y
 l (l  1)
when ml = 0
The minimum value :



L2x  L2y
MAX

MIN

l
when ml =  l
5.3 The spectroscopic notation and
the shell notation
The existence of more than one distinct state with the same
energy is called degeneracy
Example : n = 2  4 states : degeneracy g = 4
5.4 Electron Spin
Analogy : The earth travels in a nearly circular orbit around
the sun, and at the same time it rotates on its axis. Each
motion has its associated angular momentum. which we call
the orbital and spin angular momentum, respectively.
L
S
Each electron possesses an intrinsic
angular momentum called its spin.
Like orbital angular momentum. the spin
angular momentum of an electron
(denoted by S) is found to be quantized.
1
S  s (s  1) ; s 
2
1 1

S 
 1

22

3

2
The projection of the spin on z-axis is called Sz
S z  ms  
2
1
( ms   : magnetic spin number )
2
The spin angular momentum vector S can have only two orientations
in space relative to the z-axis: "spin up" with a z-component of 
2
and "spin down" with a z-component of 
2
CONCLUSION :
State of an electron is defined by 5 quantum numbers :
n : the principal quantum number
l : the orbital quantum number
ml : the orbital magnetic quantum number
s : the spin number
ms : the magnetic spin number
Wave function of an electron is denoted as :
 n ,l ,ml ,s 1 / 2,ms 1 / 2
PROBLEM 20
(a) Show that the total number of atomic states (including
different spin states) in a shell of principal quantum number n
is 2n2 .
(b) Which shell has 50 states ?
SOLUTION
n 1
n 1
n 1
n 1
n 1
l 0
l 0
l 0
l 0
l 0
(a) N  2 (2l  1)  4  l  2 1  4  l  2 1
(n  1)n
4
 2(n )  2n 2  2n  2n
2
N  2n 2
(b) The n = 5 shell (O - shell) has 50 states
6. The Zeeman Effect
The Zeeman effect is the splitting of atomic energy levels and
the associated spectrum lines when the atoms are placed in a
magnetic field
The interaction energy of an
electron (mass m) with
magnetic quantum number ml ,
in a magnetic field B along the
+ z-direction : `
U  ml B B
(ml  0;  1;  2;  3;...)
e
B 
: the Bohr magneton
2m
PROBLEM 21
An atom in a state with l = 1 emits a photon with wavelength
600.000 nm as it decays to a state with l = 0. If the atom is
placed in a magnetic field with magnitude B = 2.00 T, determine
the shifts in the energy levels and in the wavelength resulting
from the interaction of the magnetic field and the atom's orbital
magnetic moment.
SOLUTION
7. The Dirac Equation and antimatter
Paul Dirac had developed a relativistic generalization of
the Schrödinger equation for the electron (1928). The
Dirac Equation
 explain the spin magnetic moment of the electron
 mechanism for the creation of positrons
 like photons, electrons can be created and destroyed
Antiparticle: positron (e+), the electron's antiparticle,
which is identical to the electron but with positive charge