PROGRAM OF “PHYSICS2B”

Lecturer: Dr. DO Xuan Hoi

Room 413

E-mail : dxhoi@hcmiu.edu.vn

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.phy-

astr.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

From a modern viewpoint, the light has both wave and particle characteristics

Diffraction, Interference phenomena light has wave nature

Photoelectric, Compton’s effects light has particle nature

What is the nature of light?

ANSWER:

“… the wave and corpuscular descriptions are only to be regarded as complementary ways of viewing one and the same objective process…” (Pauli, physicist)

The Wave-Particle Duality of Light That is

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

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)

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.

1. 3 The photoelectric effect

Einstein's Photon Explanation: a beam of light consists

of small packages of energy called photons or quanta.

The photoelectric effect:

light strikes a metal surface emission of electrons

light

_ _ _ _ + + + +

electron

metal

The energy of a photon: hcE hf

f, : frequency, wavelength of the light

Planck's constant h = 6.626069310-34J.s

The photoelectric effect:

: work function (minimum energy needed to remove an

electron from the surface)

Maximum kinetic energy :

max

hcE hf K

V0: The stopping potential

2max max

1K mv

2

2max max max 0

1K' 0 K mv eV

2

Theorem of variation of kinetic energy:

max max 0K K ' eV

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

pc 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

h

' 1 cosmc

(Compton shift)

Energy conservation:

Momentum conservation:

2epc mc p 'c E

ep p ' P

2 2 2eP p p ' 2pp 'cos

hp

hp '

'

With and :

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

Particle could act like a wave;

both X rays and electrons are

WAVE

The pattern of the two experiments are the same

Discussion

Similar diffraction and interference experiments have been realized with protons, neutrons, and various atoms

Small objects like electrons, protons, neutrons, atoms, and molecules travel as waves - we call these waves “matter waves”

Wave interference patterns of 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

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, ):

fhE

1hp

particle

wave

De Broglie’s hypothesis:

To a particle (E, p) is associated a matter wave

which has a frequency f and a wavelength

h

Ef

p

h

h

Ef

p

hand From

if we put: 2h fE 2

E

2p

Kp

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

SOLUTION

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:

mv

h

p

h

smkg

sJ

/100.61011.9

.10625.6631

34

m10102.1 For the dust speck:

dd

dmv

h

p

h

smkg

sJ

/020.00.6100.1

.10625.69

34

md

23103.3 DISCUSSION: The de Broglie wavelength of the dust speck is so

small that we do not observe its wavelike behavior

SOLUTION

PROBLEM 2 An electron microscope uses 40-keV electrons.

Find the wavelength of this electron.

The velocity of this electron:

mKv /2

31

193

101.9

106.110402

v sm /102.1 8

The wavelength of this electron:

mv

h

831

34

102.1101.9

1063.6

o

A1.6m10101.6

2 The Schrödinger’s Equation

Matter waves:

A moving particle (electron, photon) with momentum p is described

by a matter wave; its wavelength is ph /

2.1 Wave Function and Probability Density

A matter wave is described by a wave function: );,,( tzyx(called uppercase psi)

);,,( tzyx is a complex number (a + ib with i2 = -1, a, b: real numbers)

);,,( tzyx depends on the space (x, y, z) and on the time (t)

The space and the time can be grouped separately: tiezyxtzyx ),,();,,(

frequency) angular(part dependent -time

psi)case (lowerpart dependent -space

::

:),,(

tie

zyx

• The meaning of the wave function: ),,( zyx

The function ),,( zyx has no meaning

Only 2

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 wave is

proportional to the value at that point of 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 N.B.: with

* is the complex conjugate of

If we write iba iba* (a, b: real numbers)

2 greater it is easier to find the particle

For the case of one-dimensional motion,

when a particle with the mass m has a potential energy U(x) Schrödinger’s equation is

0)]([8

2

2

2

2

xUEh

m

dx

d

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

0)( xU and 2

2

1mvE

Schrödinger’s equation becomes: 02

18 22

2

2

2

mv

h

m

dx

d

2.2 The Schrödinger’s equation

By replacing: pmv 02

2

2

2

h

p

dx

d

With the de Broglie wavelength: h

p

1

and the wave number:

2K

we have the Schrödinger’s equation for free particle:

022

2

Kdx

d

This differential equation has the most general solution: iKxiKx BeAex )(

The time-dependent wave function:

(A and B are arbitrary constants)

tiiKxiKxti eBeAeextx )()(),(

)()(),( tKxitKxi BeAetx

Probability density:

)()(),( tKxitKxi BeAetx : traveling waves

:)( tKxiAe wave traveling in the direction of increasing x

:)( tkxiBe wave traveling in the negative direction of x

Assume that the free particle travels only in the positive direction

Relabel the constant A as :0iKxex 0)(

The probability density is: 22

0

2

02

)( iKxiKx ee Because:

1))(())(( 0*2 eeeeee iKxiKxiKxiKxiKx

we have:

const 20

2)(

const 20

2)(What is the meaning of a constant probability?

The plot of const 20

2)(

is a straight line parallel to the x axis:

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

x

2

O

2

0

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:

0p (because: )021 ppp

and when the position x is completely unspecified we write:

x

• In general, we always have: a constantx p

This constant is known as:

2

h

h is the Planck’s constant

(called h-bar)

).10625.6( 34 sJh

So we can write: px

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

xpx ypy zpz

• With the definition of the constant : 2// hKhp Kp

• Uncertainty for energy :

E t

SOLUTION

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

pxFrom the uncertainty principle:

if we want to have the minimum uncertainty: px

We evaluate the momentum: mvp )1005.2()101.9( 631

smkgp /.1035.9 27The uncertainty of the momentum is:

241087.1100/5.0%5.0 pp smkg /.1035.9 27

p

x

27

34

1035.9

2/10635.6

nmm 111013.1 8

SOLUTION

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?

xmm

pv

2100.1101.9

.1063.6431

34

mkg

sJv

smv /2.1

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

SOLUTION

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?

xmm

pv

2100.1101

.1063.656

34

mkg

sJv

smv /101.1 23

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

SOLUTION

PROBLEM 6 An electron is confined within a region of width

1.010- 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 1.610-8 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?

The fractional uncertainty of the photon energy is

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

The electric potential energy is : r

e

r

eeU

0

2

0 4

))((

4

1

We can demonstrate that the energies of the quantum electron is

2220

4 1

8 nh

meEn

2

6.13

n

eVEn

where n is an integer, that is the principal quantum number

The energies of the hydrogen atom is quantized

4 The Bohr Atom

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:

nmmn

mn EEhc

hf

• If En E1 : Lyman series • If En E2 : Balmer series

• If En E3 : Paschen series

• If En E4 : Backett series .

.

. E1

E2

E3 E4

Lyman

Balmer

Paschen

E

PROBLEM 8

SOLUTION

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?

1/ For the Lyman series: 11

1 EEhc

hf mm

m

The least energetic photon is the transition between E1 and the level

immediately above it; that is E2 . The energy difference is:

22121

6.13

2

6.13 eVeVEEE eV2.10

E

hc

19

834

106.12.10

)103()1063.6(

nmmm 122122.01022.1 7

(in the ultraviolet range)

2/ The line H in the Balmer series corresponds to the transition

between E3 and E2 .

The energy difference is:

22232

6.13

3

6.13 eVeVEEE eV89.1

E

hc

19

834

106.189.1

)103()1063.6(

nmmm 658658.01058.6 7

(red color)

• The line H : E4 E2

• The line H : E5 E2

• The line H : E6 E2

• The line H : E3 E2 ( = 658 nm)

( = 486 nm)

( = 434 nm)

( = 410 nm)

PROBLEM 9

SOLUTION

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.

22

2

8n

ma

hEn

2192931

234

1106.1)102.0(101.98

)10625.6(

22

2

2 28

ma

hE 14E

22

2

1 18

ma

hE

eV40.9

eV6.37 Energy required:

eVEEE 2.2840.96.3712

The wavelength that can cause this transition: nmEhc 0.44/

PROBLEM 10

SOLUTION

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 :

; 1,2,3,...mvr n n

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.

2

21

2

eE KE PE mv k

r

2

22

keKE

r

;F ma

The energy of the atom:

Newton’s second law: 2 2

2

e vk m

rr

2

2

eE k

r

1/

With: 2 2

2

2 2

:n

v

m r

2 2 2 2 2

2

2 2 2

;ke m n n

ke

r mrr m r

We have: 2 2

2n

nr

mke

(the electronic orbits are quantized)

With: 2

0 20.0529a nm

mke

(Bohr radius), 2

0nr a n

2 4

2 2

1

2

mk eE

n

2/ We have : 2

13.6

n

eVE

n

The frequency of the emitted photon is given by: i fE E

f

h

2 4

3 2 2

1 1 1

4f i

f mk e

c c n n

With:

2 4

7 1

31.097 10

4H

mk eR m

c

2 2

1 1 1

H

f i

R

n n

(Rydberg constant)

2 2

2

e vk m

rr

PROBLEM 11

SOLUTION

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

2 2 4

2 2

1

2

mk Z eE

n

For He+: Z = 2

2

2

(13.6 )

n

Z eVE

n

14(13.6 ) 54.4E eV eV The ground state energy:

• The radius of the ground state : 2

0n

nr a

Z

2 2

2n

nr

mkZe

2

1 0

10.0265

2

r a nm (The atom is smaller; the electron is more

tightly bound than in hydrogen atom)

5 The Schrödinger Equation for the Hydrogen Atom

In spherical coordinates (r, , ), the hydrogen-atom problem

is formulated as :

2

0

1( )

4

eU r

r

Electron cloud

( ) ( , , ) ( ) ( , ) ;r r R r Y

The potential energy is :

( , ) ( ) ( )Y

5.1 Radial function of the hydrogen atom

If depends only on r : (r)

2 2 2 2 ;r x y z 2 2r

r xx

r x

x r

;d r x d

x dr x r dr

2

2

x d

x x x r drx

2 2

2 2 2

1 d x r d x d r

r dr x dr r xx r dr

2 2 2

3 2 2

1 d x d x d

r dr drr r dr

The Schrödinger equation in three dimensions :

2 2 2

2 2 2 2

2[ ( )] 0

mE U r

x y z

2 2 2 2

2 2 2 20

2 10

4

m eE

rx y z

2 2 2 2

2 3 2 2

1 d x d x d

r dr drx r r dr

The same result for y and z : 2 2 2 2

2 3 2 2

1 d y d y d

r dr dry r r dr

2 2 2 2

2 3 2 2

1 df z d z d

r dr drz r r dr

2 2 2 2

2 2 2 2

2d d

r drx y z dr

2 2

2 20

2 2 10

4

d d m eE

r dr rdr

The Schrödinger equation :

Wave function for the ground state

The wave function for the ground state of the hydrogen atom:

Where a is the Bohr radius: nmma 9.5210529.0 10

/

3/2

1( ) r ar e

a

Wave function for the first excited state

/2( ) (2 )arr Ce ar

PROBLEM 12

SOLUTION

Knowing that the wave function for the ground state of the

hydrogen atom is arAer /)(

Where a is the Borh radius: nmma 9.5210529.0 10

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?

1/ Because the electron moves in the three dimensional space, the

probability of finding the electron in a volume dV is written:

dVeAdV ar /222

where: drrdV 24

Normalization condition: 1)(0

2

dVr

)1(140

2/22

drreA ar

By changing variable: arz /2

We put:

0

2/2 drreI ar

0

23

2dzze

aI z

By integration by parts, we can demonstrate the general formula:

nnndzze nz

)1(...21!0

4/2/22/ 33

0

23 aadzzeaI z

Substituting the value of I into (1): 14/4 32 aA2/3

1

aA

area

r /2/3

1)(

2/ Because:

drrea

drrea

dVeAdV ararar 2/23

2/23

/222 44

1

When P (r) has a maximum:

the radial probability density is

2/23

4)( re

arP ar

0)(

dr

rdP

2/23

4)(re

dr

d

adr

rdP ar

0)(8 /24

arerara

ar

(The value r = 0 corresponds to a minimum of P( r) )

arar ree

ar

a

/2/223

2)2

(4 r a

P(r)

O

Physical meaning: The position r = a is the most probable for the

electron electronic clouds

PROBLEM 13

SOLUTION

Calculate the probability that the electron in the ground state of

the hydrogen atom will be found outside the Bohr radius

area

r /2/3

1)(

With:

a

dVrP2

)(The probability is found by: drrdV 24

a

ar drera

P /223

4

where:

By changing variables: ,/2 arz

2

2

2

2 }22{2

1

2

1 zz ezzdzezP 677.05 2 e %7.67 P

PROBLEM 14

SOLUTION

The wave function of a particle is given as: axCer /)(

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

1/ With: 1/22

dxeC ax 120

/22

dxeC ax

12

2 2

aC

aC

1

2/

aa

a

dxxdxxP0

22)(2)(

aax dxeC

0

/222

)1)(2/(2 22 eaCP %5.86)1)(2/()/1(2 22 eaa

5.2 Quantization of Orbital Angular Momentum

Torque and angular momentum in rotational motion

Force and linear momentum in translational motion :

dvF ma m

dt

( )d mv dp

dt dt

Fd L pr

F r L r p

NOTES : In classical mechanics :

2( ) ( )L mvr m r r mr I

The requirement that the () function must be finite at = 0 and = gives the result : L can take some possible values :

In the wave function of electron in hydrogen atom :

( 1)L l l

( 0 ;1 ; 2 ; 3 ;...; 1 )l n

In quantum mechanics :

( ) ( , , ) ( ) ( , ) ;r r R r Y ( , ) ( ) ( )Y

The number l is called : the orbital angular-momentum quantum number or the orbital quantum number for short.

( 1)L l l

( 0 ;1 ; 2 ; 3 ;...; 1 )l n

In quantum mechanics :

z lL m

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

( 0 ; 1 ; 2 ; 3 ;...; )lm l

( ; 1 ; 2 ; ... 1 ; 0 ;1 ;...; 1 ; )lm l l l l l

We call ml the orbital magnetic quantum number

( 1)L l l ( 0 ;1 ; 2 ; 3 ;...; 1 )l n

EXAMPLE :

z lL m

For n = 3; the possible values of l are : 0 ; 1 ; 2

( 0 ; 1 ; 2 ; 3 ;...; )lm l

( ; 1 ; 2 ; ... 1 ; 0 ;1 ;...; 1 ; )lm l l l l l

With l = 2; the possible values of ml are : -2 ; -1 ; 0 ; +1 ; +2

2(2 1) 6 2.45L

0 ; 1 ; 2zL

PROBLEM 15

SOLUTION

How many distinct (n, l, ml ) states of the hydrogen atom with n = 3 are there? Find the energy of these states.

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.

2

6.13

n

eVEn 3 2

13.61.51

3

eVE eV

PROBLEM 16

SOLUTION

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?

When n = 4, the maximum value of the orbital angular-momentum quantum number l is ( 1) (4 1) 3n

( 1)L l l 3(3 1) 12

(a)

(b) For l = 3 the maximum value of the magnetic quantum number ml is 3 :

z lL m 3

PROBLEM 17

SOLUTION

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?

(c) The minimum allowed angle between L and the z-axis corresponds to the maximum allowed values of Lz and ml

maxmin

( )cos zL

L

0

min30

3

12

SOLUTION

Represent all the possible orientations of the

angular momentum with the value l = 0 ; 1 ; 2 ; 3

l = 0 : L = 0 ; ml = 0

l = 1

l = 2

ml = -1 ; 0 ; +1

( 1)L l l 1(1 1) 2

0 ; 1z lL m

ml = -2 ; -1 ; 0 ; +1 ; +2

( 1)L l l 2(2 1) 6

0 ; 1 ; 2z lL m

0

2

2

zL

PROBLEM 18

0

0

l

L

1

2

l

L

2

6

l

L

SOLUTION

2 2 2 2

x y zL L L L 2 2 2( 1) ll l m

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 . Write an expression for this quantity in terms

of l and ml .

(b) What is the meaning of ?

(c) For a state of nonzero orbital angular momentum, find the

maximum and minimum values of . Explain your results.

2 2

x yL L

2 2

x yL L

2 2

x yL L

2( 1) ll l m (a)

This is the magnitude of the component of angular momentum perpendicular to the z-axis (b)

The maximum value : (c) 2 2 ( 1)x yMAX

L L l l

when ml = 0 The minimum value : 2 2

x yMIN

L L 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.

1( 1) ;

2S s s s

1 1 31

2 2 2S

Like orbital angular momentum. the spin angular momentum of an electron (denoted by S) is found to be quantized.

The projection of the spin on z-axis is called Sz

2z sS m

1( : )

2s magnetic spin numm ber

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

and "spin down" with a z-component of 2

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 :

, , , 1 /2, 1 /2l sn l m s m

SOLUTION 1

0

2 (2 1)n

l

N l

PROBLEM 20

(a)

The n = 5 shell (O - shell) has 50 states (b)

(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 ?

1 1

0 0

4 2 1n n

l l

l

1 1

0 0

4 2 1n n

l l

l

( 1)4 2( )

2

n nn

22 2 2n n n

22N n

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 : `

2

B

e

m

6. The Zeeman Effect

: the Bohr magneton

( 0; 1; 2; 3;...)

l B

l

U m B

m

SOLUTION

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.

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

7. The Dirac Equation and antimatter

Antiparticle: positron (e+), the electron's antiparticle,

which is identical to the electron but with positive charge