Quantum Theory of the Atoms

X 射线

准直缝 晶体劳厄斑

electrons

slit crystalfringes

Electrons diffraction

§20-6 The infinite Potential Well 一维无限深的势阱

§20-1 Rutherford’s Experiment and the Nuclear Atom 卢瑟夫实验和原子有核模型

§20-2 Atomic Spectra Bohr Model of Hydrogen Atom 原子光谱 玻尔的氢原子模型

§20-4 The Uncertainty Principle 不确定关系

§20-3 De Broglie’s Postulates and Matter Waves 德布罗意假设和物质波

§20-5 Wave Function and Schrodinger Equation Born’s Interpretation 波函数和薛定谔方程

§20-7 Hydrogen Atom and Electron Spin 氢原子和电子自旋

§20-8 Multielectron Atom and Periodic Table 多电子和周期表

§20-1§20-1 Rutherford’s Experiment and the Nuclear Atom

1. Thomson’s model1. Thomson’s model

+Ze+Ze of an atom distributes of an atom distributes uniformly in sphere.uniformly in sphere.

Atom size (Atom size (1010-10-10mm))

electronselectrons panel in the atom panel in the atom uniformly.uniformly.

electron

Plum pudding model

2. Rutherford’s 2. Rutherford’s –particles scattering experiment–particles scattering experiment (1907)(1907)

Some of the scattering angles are so large Some of the scattering angles are so large that it cannot be interpreted with “that it cannot be interpreted with “Plum pudding model”

atomatom

––particlesparticles

All the positive charge and almost all the mass of All the positive charge and almost all the mass of the atom is concentrated in about the atom is concentrated in about 1010-15-15~10~10-14-14 m m the volume of the atom.the volume of the atom.

3. Rutherford’s “nuclear atom model” (1911)3. Rutherford’s “nuclear atom model” (1911)

4. Limitation Rutherford’s model 4. Limitation Rutherford’s model

Cannot explain why the structure of atom is Cannot explain why the structure of atom is stable .stable .

Cannot explain why the radiating spectrum Cannot explain why the radiating spectrum is discrete.is discrete.

1. The experimental laws of hydrogen atom1. The experimental laws of hydrogen atom

§20-2 Atomic spectra and the Bohr model of hydrogen atom

H H H H H

6563

6563

ÅÅ

4861

4861

ÅÅ

4341

4341

ÅÅ

4102

4102

ÅÅ

3646

Å

visible tultraviole

J.Balmer found their regularities in 1885 J.Balmer found their regularities in 1885 ：：

)1

2

1(~

22 nR -- -- Balmer formulaBalmer formula

Wave numberWave number

J.R.Rydberg proposed a general formula in 1890J.R.Rydberg proposed a general formula in 1890

)11

(~22 nk

R

RR=1.096776=1.096776101077mm-1-1

-- Rydberg formula-- Rydberg formula

-- -- Rydberg constantRydberg constant

,2,1kkn nn are integers with are integers with

1~

(1)(1)k=1,n=2,3,…k=1,n=2,3,… Lyman seriesLyman series, ultraviolet, ultraviolet

(2)(2)k=2,n=3,4,…k=2,n=3,4,… Balmer seriesBalmer series,, visblevisble

(3)(3)k=3,n=4,5,…k=3,n=4,5,… Paschen seriesPaschen series,, infraredinfrared

(4)(4)k=4,n=5,6,…k=4,n=5,6,… Brackeff Brackeff seriesseries, infrared, infrared

(5)(5)k=5,n=6,7,…k=5,n=6,7,… Pfund Pfund seriesseries, infrared, infrared

… … … … … …… … … … … …

)11

(~22 nk

R

Same Same kk, different , different n – n – a series line (a series line ( 线系线系 ))

)11

(~22 nk

R )()( nTkT

----T(k) , T(n) terms of spectrum

)()(~ nTkTkn

)]()([)]()([ nTnTnTkT

nnnk ~~

----Ritz combination principle

2.Bohr’s postulates2.Bohr’s postulates In order to support the nuclear model, Bohr In order to support the nuclear model, Bohr

proposed three postulates for explaining the proposed three postulates for explaining the experimental regularities of hydrogen atom.experimental regularities of hydrogen atom.

(1) (1) Stable states postulateStable states postulate ：：

The motion of electron in The motion of electron in a circular orbit is stablea circular orbit is stable

(2) (2) Postulate about the quantization of orbit aPostulate about the quantization of orbit angular momentum ngular momentum ::

2

hnL

2

h --simplified Plank’s constant--simplified Plank’s constant

,2,1n

nn—quantum number—quantum number

n

The orbit angular momentum The orbit angular momentum L=mvrL=mvr of an of an electron must satisfy electron must satisfy ,,

(3) (3) radiation postulateradiation postulate::

h

EE nkkn

EEnn>>EEkk--- a photon is emitted--- a photon is emitted

EEnn<<EEkk---a photon is absorbed---a photon is absorbed

Bohr got Noble Prize of physics in 1922.Bohr got Noble Prize of physics in 1922.

When an electron charges its orbit, a photon When an electron charges its orbit, a photon with frequency is emitted or absorbed.with frequency is emitted or absorbed.

3. The results of Bohr’s model3. The results of Bohr’s model(1) The radius of orbit(1) The radius of orbit

According to According to Newton’s lawNewton’s law r

vm

r

e 2

20

2

4

mvrL 2

hnandand

mr

nhv

2

2

202

me

hnrn

So So ,2,1n

--quantization--quantization

rr11=0.53=0.531010-10-10 m mFor For nn=1=1,,

----first Bohr’s radius----first Bohr’s radius

12rnrn

(2) energy(2) energyThe energy of hydrogen atom=potential+kineticThe energy of hydrogen atom=potential+kinetic

nnn r

emvE

0

22

42

1

2

202

me

hnrn

r

vm

r

e 2

20

2

4

r

emvn

0

22

82

1

usingusing We getWe get

nn r

eE

0

2

8

220

4

2 8

1

h

me

n

---quantization---quantizationThe quantized energy ---- energy-leverlThe quantized energy ---- energy-leverl

nn----quantum number----quantum number

eVE 6.131 ----ground state----ground state

discussiondiscussion ：：

ForFor nn=1=1,,(1)(1)

Lowest energy, most stable stateLowest energy, most stable state

ForFor nn>1>1(2)(2) ----excited state----excited state

(3)(3) nFor For 0,, nn Er

---The ionization energy of hydrogen atom in ---The ionization energy of hydrogen atom in ground state is ground state is 13.6eV13.6eV

220

4

2 8

1

h

me

nEn

2

202

me

hnrn

Electronic orbits Energy-level

ground1E

1n2n

3n

4n

2E

3E4E

ExcitedExcited

Lyman

Balmer

Paschen

-13.6-13.6

-3.39-3.39

-1.51-1.51

00

EEnn/eV/eVn

1

2

34-0.85-0.85

If an electron jumps If an electron jumps EEnnEEk k , the H-atom emits a p, the H-atom emits a photon.hoton.

h

EE kn

22320

4 11

8 nkh

me

Wave number isWave number is

c

~

22

11

nkR

22320

4 11

8 nkch

me

220

4

2 8

1

h

me

nEn

Here Here ch

meR

320

4

8

---agree perfectly with the experiment.---agree perfectly with the experiment.

17 m100973731.1

RRexpexp=1.096776=1.096776101077mm-1-1

Bohr’s theory made great success in hydrogen Bohr’s theory made great success in hydrogen atoms and atoms and hydrogenlike ionshydrogenlike ions ( ( 类氢离子类氢离子 ) )

But this theory cannot explain the experimental But this theory cannot explain the experimental results of more complex particles.results of more complex particles.

More than one electron outside the nuclear

4. The limitation of Bohr’s model4. The limitation of Bohr’s model

(1) using classical physics theory, but the electr(1) using classical physics theory, but the electron has acceleration and no radiation is inexon has acceleration and no radiation is inexplicable. plicable.

(2) No any theory can explain the angular mom(2) No any theory can explain the angular momentum quantization. entum quantization.

(3) Cannot get the intensity of the spectra from t(3) Cannot get the intensity of the spectra from this theory.his theory.

1. De Broglie’s postulates(1924)1. De Broglie’s postulates(1924)

§20-3 De Broglie’s postulates §20-3 De Broglie’s postulates and matter wavesand matter waves

Just as radiation has particle-like properties, Just as radiation has particle-like properties, electron and other material particles possess electron and other material particles possess wave-like properties.wave-like properties.

hE

h

p

The waves related to the material The waves related to the material particles----particles----matter wavesmatter waves

De Broglie won Noble prize in physics in 1929.De Broglie won Noble prize in physics in 1929.

---De Broglie’s formulae---De Broglie’s formulae

The wave property of eleThe wave property of electrons was confirmed by ctrons was confirmed by Davisson and Germer in Davisson and Germer in experiment in 1927.experiment in 1927.

2. Testing De Broglie hypothesis2. Testing De Broglie hypothesis

They observed that the They observed that the diffraction patterns of the diffraction patterns of the electrons electrons are similar to with are similar to with the ones of the ones of xx-ray.-ray.

E-gunE-gun detector

Nickle crystal

Acce.anode

--electron have --electron have wave wave property.property.

Meanwhile, Thomson’s Meanwhile, Thomson’s experiment gave another experiment gave another method to confirm the method to confirm the wave property of wave property of electrons.electrons.

They won Noble prize in They won Noble prize in physics in 1937physics in 1937

xx-ray-ray electronselectrons

It was confirmed by great amount of experimentIt was confirmed by great amount of experiments that s that all micro-particlesall micro-particles such as electrons, neutr such as electrons, neutrons, proton, as well as atoms and molecules ons, proton, as well as atoms and molecules havhave wave propertiese wave properties. And their wave properties ag. And their wave properties agree with De Broglie’s formulae.ree with De Broglie’s formulae.

singlesingle doubledouble three four

The results of The results of slits diffraction experiments in slits diffraction experiments in 1960 1960

[[ExampleExample] an electron is accelerated by electric ] an electron is accelerated by electric field. If the Acc. potential is field. If the Acc. potential is UU, the electron is at , the electron is at rest before it is accelerated and its speed rest before it is accelerated and its speed V<<CV<<C after it is accelerated. Find its wavelength after it is accelerated. Find its wavelength

Solution Solution 2

02

1vmeU

0

2

m

eUv

vm

h

0

Uem

h

02

o

A2.12

U

Scanning electron microscope---SEM

---observe the microscopic morphology.

Mosquito Head with X 1,000

A shell of a radiolarian — a single-celled animal with X 2,000

Transmission electron microscope --TEM

---study the microstructure.

a leaf of a green plant.

§20-4 The Uncertainty Principle§20-4 The Uncertainty Principle For microscopic particles For microscopic particles ：： their position , their position ,

velocity and others cannot be determined by velocity and others cannot be determined by actual experiment actual experiment at same instantat same instant because of because of wave properties.wave properties.

The single slit diffraction of electrons The single slit diffraction of electrons

A beams of electrons is incident on a slit A beams of electrons is incident on a slit normally,normally,

x1 y

x pyp

xp

considering the central bright fringe only,considering the central bright fringe only,

1sin0 ppx 1sin ppx

The first dark fringe The first dark fringe satisfies, satisfies,

1sinx

xppx

ka sin

xhpx hpx x

Considering all diffraction fringes,Considering all diffraction fringes,

hpx x --estimated result--estimated result

Hesbreg deduced the precise resulHesbreg deduced the precise results in 1927.ts in 1927.

2,

2,

2

zyx pzpypx

ph xppx

Hesbreg won Noble prize in physics in 1932.Hesbreg won Noble prize in physics in 1932.

Any experiment cannot determine Any experiment cannot determine simultaneously the exact values of a simultaneously the exact values of a component of momentum and its component of momentum and its corresponding coordinate.corresponding coordinate.

Similarly, measuring the energy of a Similarly, measuring the energy of a particle in a time interval particle in a time interval tt, the , the uncertainty uncertainty EE of the energy is of the energy is

hpx x ---Coordinate-momentum uncertainty relation---Coordinate-momentum uncertainty relation

hEt

[[ExampleExample] according to classical ] according to classical electromagnetism and mechanics, the speed of electromagnetism and mechanics, the speed of the electron moving around its nuclear is about the electron moving around its nuclear is about 10106 6 m/sm/s. the size of an atom is . the size of an atom is 1010-10-10mm. Estimate . Estimate the uncertainty of the speed.the uncertainty of the speed.

Solution Solution the uncertainty of the position of the uncertainty of the position of the electron is the electron is m10 10x

m

pv x

x

xm

h

1031

34

10101.9

1063.6

m/s103.7 6v v vv. the uncertainty of the speed of the . the uncertainty of the speed of the electron in atoms is very apparent.electron in atoms is very apparent.

1.wave function1.wave function A classical plane wave function travels in A classical plane wave function travels in ++xx axis axis,,

)(2cos0 xtyy

§20-5 The wave function and §20-5 The wave function and Schrodinger equation Schrodinger equation

Born’s interpretation of wave functionBorn’s interpretation of wave function

Write it in complex form,Write it in complex form,)(2

0 xtieyy

hE h

p

Similarly, for a free particle with energy Similarly, for a free particle with energy E E and and momentum momentum pp, its wave function, , its wave function,

)(2

0,

x

tietx

pxEti

e

0

pxEti

etx

0,

2. Schrodinger equation2. Schrodinger equation

Free particleFree particle ：： moves in +moves in +x x axisaxis

2

2

2

2

p

x

m

p

xm 22

2

2

22

E

i

t

Et

i

ti

xm

2

22

2m

pE

2

2

---S-Equation of a free particle

The particle has potential energy The particle has potential energy VV((xx,t), ,t), its total energy isits total energy is

),( txVEE k ),(22 txVmp

E

i

t

)],(

2[

2

txVm

pi

)],(

2[

2

txVm

p

ti

i.e.i.e.

titxV

xm

),(2 2

22

m

p

xm 22

2

2

22

t

tritrVtr

m

),(

),(),(2

22

Introducing energy operator,Introducing energy operator,

),(2

ˆ 22

trVm

H

--Hamiltonian operator--Hamiltonian operator

thenthent

iH

ˆ

If a particle moves in three dimension space,If a particle moves in three dimension space,

2

2

2

2

2

2

2

2

zyxx

2

---S-Equation of a particle in a potential field

Stationary state: Stationary state: potential is independent on tipotential is independent on timeme

)(xVV

Let Let )()(),( tfxtx

)()()()(

)(2 2

22

tfxxVx

xtf

m

t

tfxi

)(

)(

titxV

xm

),(2 2

22

t

tf

tfixV

x

x

xm

)(

)(

1)(

)(

)(

1

2 2

22

Divided by Divided by )()( tfx , we get, we get

Left side of the equation is the function of Left side of the equation is the function of xx, , and the right side is the function of and the right side is the function of tt, ,

If it is useful for any If it is useful for any tt or any or any xx, it must , it must satisfies, satisfies,

left side = right side = a constantleft side = right side = a constant

(1)(1)Et

tf

tfi

)(

)(

1

(2)(2)ExVx

x

xm

)()(

)(

1

2 2

22

i.e.i.e.

the solution of Eq. (1) is the solution of Eq. (1) is Et

i

etf

~)(

E E has the demission of energy.has the demission of energy.

Rewriting (2),Rewriting (2),

oror

----Schordinger’s equation in stationary state ----Schordinger’s equation in stationary state

The wave function of the particle is The wave function of the particle is

Eti

extx

,

)()()()(

2 2

22

xExxVx

x

m

0)()(2)(

22

2

xVE

m

x

x

3.The physical meaning of wave function 3.The physical meaning of wave function

--- --- complex functioncomplex function

M.Born postulated M.Born postulated

),(),(),(2* trtrtr

---Born’s interpretation of wave function ---Born’s interpretation of wave function

no physical meaningno physical meaning

represents represents probability densityprobability density, ,

i.e. the probability of finding the particle per i.e. the probability of finding the particle per unit volume about the point at time unit volume about the point at time tt..r

dVdw2

The probability of finding the particle within thThe probability of finding the particle within the volume element e volume element dvdv ==dxdydzdxdydz about the point about the point at time at time tt is is

r

The wave function should satisfy ),( tr

continuous, single-valued and finite.

And

12

dVV

The space of the particle moving in

---Standard conditions

---Normalized condition

The superposition principle

i.e

then

is also the solutions of the S- equation.

If the and are two solutions of the Schrodinger equation,

1 2

2211 cc

Here, c1 and c2 are normalized factor.

The conservation of probability2

),( tr ------probability densityprobability density

][2

** mi

j

------probability current densityprobability current density

From Sch. Eq., we can get From Sch. Eq., we can get

0

t

j

------probability is conservative.probability is conservative.

4. Operators and physical observables4. Operators and physical observables

The value of a physical observable=The value of a physical observable=

V* (its operator dV)

Such as

Energy operator is ),(2

ˆ 22

trVm

H

So dVHEV ˆ*

momentum operator is

iP̂

So dVPPV ˆ*

If the particle is in stationary state,2

2)(),(

Eti

ertr

2)(r

-- probability density has nothing to do with -- probability density has nothing to do with tt..

Schrodinger won NoSchrodinger won Noble prize in 1933.ble prize in 1933.

A particle with mass A particle with mass mm moves along moves along xx axis. axis. It potential isIt potential is

§ 20-6 The infinite potential well§ 20-6 The infinite potential well

x0 a

Out wellOut well )(xV

0

)()(

2 2

2

xEdx

xd

m

In well:In well: 0)( xV

)(xV )(00 wellinax )(,0 welloutaxx

Using the continuous condition of wave Using the continuous condition of wave function,function,0)0(

0)( a

0sin C 0

0sin kaC nka

a

nk

,2,1n

x0 a

0 0

xa

nCx

sin)(

0)()( 2

2

2

xkdx

xd Let Let 22 2 mEk

Its general solution is Its general solution is kxCx sin)(

CC 、、 ---integral constants---integral constants

Using normalized condition of wave function,Using normalized condition of wave function,

dxtxa 2

0),( dxx

a 2

0)( dxx

a

nC

a2

0sin

1

We getWe get aC 2

0,00

)(xx

xn axx

a

na 0sin2

22 2

mEk

2

222

2ma

nEn

,2,1n

a

nk

--quantized energy--quantized energy

nn ：： quantum numberquantum number

NotesNotes ：：

122 2

22

nma

En

nn=1=1 ：：

-- zero point energy-- zero point energy

2

22

1 2maE

--- the interval of energies between two --- the interval of energies between two

adjacent states is not uniform. adjacent states is not uniform.

2nEn

-- the lowest energy of a particle cannot be z-- the lowest energy of a particle cannot be zero in quantum mechanics.ero in quantum mechanics.

For an eigen state For an eigen state nn with quantum number with quantum number nn, th, th

ere are ere are n+1n+1 nodes nodes and and nn antinodes antinodes in the well. in the well.

x0 a

)(xn

x0 a

2)(xn

1n2n

3n

4n

Minimum probabilityMinimum probability Maximum probabilityMaximum probability

[[ExampleExample] A particle with mass ] A particle with mass mm locates in a locates in a infinite potential well with length infinite potential well with length aa. Calculate . Calculate the probability of finding the particle in the the probability of finding the particle in the range of range of 00xxaa/4/4 for the two different states for the two different states with with nn=1=1 and and nn==. . the the positions ofpositions of maximum maximum probability for the state probability for the state n=n=22..

SolutionSolution AsAs xa

n

ax

sin2

)(

the probability of finding the particle in the the probability of finding the particle in the range of range of 00xxaa/4/4 is is

dxxwa 2

4

0)( dx

a

n

a

a

4

0

2sin2

1nFor For 2

1

4

1w %9

nForFor4

1w

1n

n

2sin

2

1

4

1

n

n

xaa

x 2

sin2

)(2

xaa

x 2

sin2

)( 22

2

Let Let

0x

0)(2

2 xdx

d

We get We get 2

,a

4

3,

a a,

minimumminimum

4,

a

maximummaximum

The potential distribution isThe potential distribution is

x

U

1 2 3A particle move along x axis,

0UE If , the particle can pass through area If , the particle can pass through area 22 and arrive to area and arrive to area 33..

0UE If , the particle cannot pass through If , the particle cannot pass through area area 22 and cannot be found in area and cannot be found in area 3.3.

)(xU0

0U ax 0

axx ,0

Tunneling effectTunneling effect

0U

For classical particle,

121

22

2

Edx

d

m

22022

22

2

EUdx

d

m

323

22

2

Edx

d

m

Area 2Area 2

Using Schrodinger equation,Using Schrodinger equation,

Area 3Area 3

For macroscopic particle,

Area 1Area 1

The solutions of S.Eq. areThe solutions of S.Eq. areArea Area 11 ：： incident waves incident waves ++reflected wavesreflected waves

Area Area 22 ：： transmitted waves transmitted waves + + reflected reflected waveswavesArea Area 33 ： ： transmitted waves transmitted waves

E

0 a x

No matter No matter EE>>UU00 or or EE<<UU00

---Tunneling effect---Tunneling effect

癌细胞表面图像癌细胞表面图像硅表面图像硅表面图像

扫描隧道显扫描隧道显微镜微镜 (STM)(STM)

ApplicationApplication Scanning Tunnel Microscope Scanning Tunnel Microscope (STM)(STM)

一氧化碳“分子人”

“ 原子和分子的观察与操纵” -- 白春礼

“ 扫描隧道绘画”

量子围栏量子围栏-------- 实现波函数的测实现波函数的测量量

19931993 年人们用年人们用 STMSTM 所做的量子围栏所做的量子围栏第一次看到第一次看到波函数波函数

操纵操纵 4848 个铁原子在铜表个铁原子在铜表面围成半径为 面围成半径为 71.371.3ÅÅ 的的圆圈。表面电子在铁原圆圈。表面电子在铁原子上强烈反射，被禁锢子上强烈反射，被禁锢在该围栏中，它们的波在该围栏中，它们的波函数形成同心圆形驻波函数形成同心圆形驻波

1.The Schrodinger equation of hydrogen atom1.The Schrodinger equation of hydrogen atom

r

erU

0

2

4)(

§20-7 The Hydrogen atom and Electron spin§20-7 The Hydrogen atom and Electron spin

Schrodinger equation is Schrodinger equation is

0)4

(2

0

2

22

2

2

2

2

2

r

eE

m

zyx

In the hydrogen atom system, the potential In the hydrogen atom system, the potential energy isenergy is

introducing the transitions,introducing the transitions,

cossinrx sinsinry

cosrz S.Eq. changes intoS.Eq. changes into

)(sinsin

1)(

12

22

rrr

rr

0)4

(2

sin

1

0

2

22

2

22

r

eE

m

r

Let Let )()()(),,( rRr

We getWe get 02

2

2

lmd

d(1)(1)

0)sin

()(sinsin

12

2

lm

d

d

d

d(2)(2)

0)4

(2

)(1

20

2

22

2

R

rr

eE

m

dr

dRr

dr

d

r

(3)(3)

2.the results of the solution of Schrodinger equat2.the results of the solution of Schrodinger equation for H-atomion for H-atom

2220

2

4 1

32 n

meEn

It is agrees with Bohr’s result.It is agrees with Bohr’s result.

,2,1n ---- ---- principle quantum principle quantum number number nn..

2220

4 1

8 nh

me

The energy of H-atom system can be gotten The energy of H-atom system can be gotten from Eq. (3)from Eq. (3)

Quantized energy and principle quantum numQuantized energy and principle quantum number ber nn..

Quantized angular momentum and azimuthal quaQuantized angular momentum and azimuthal quantum number ntum number ll..

)1( llL

l l has has nn possible values for a given value of possible values for a given value of nn..

The magnitude of orbital angular momentum abThe magnitude of orbital angular momentum about the electron moving around the nuclear can out the electron moving around the nuclear can be gotten from Eq.(2)be gotten from Eq.(2)

)1(,2,1,0 nl

--azimuthal quantum number--azimuthal quantum number(( 角量子数角量子数））

The direction of orbital angular momentum aboThe direction of orbital angular momentum about the electron moving around the nuclear can be ut the electron moving around the nuclear can be gotten from Eq.(3)gotten from Eq.(3)

lz mL mmll has has (2(2 ll+1)+1) possible values for a given value of possible values for a given value of ll ..

lml ,2,1,0

The component The component LLzz of of L L along along zz axis is axis is---magnetic quantum number---magnetic quantum number(( 磁量子数磁量子数））---the orientation ---the orientation (( 取向取向 )) of is quantized of is quantizedL

3. The “electron cloud”The distribution of the probability density of finding the electron.

Unlenbeck and Goudsmit arranged a experiUnlenbeck and Goudsmit arranged a experiment to check whether the orientation ment to check whether the orientation (( 取向取向)) of is quantized (1921) of is quantized (1921)

4. Electron spin4. Electron spin

No magnetNo magnet

In In magnetmagnet

Atom Atom sourcesource

N

S

L

The results of the experimentThe results of the experiment ：： No magnetNo magnet ：： there is one track there is one track

of the atoms on the film. of the atoms on the film. With magnetWith magnet ： ： there are two there are two

tracks of the atoms on the film.tracks of the atoms on the film.

无磁场无磁场

有磁场有磁场

contradictioncontradiction ：： when the azimuthal quantum when the azimuthal quantum number isnumber is ll , the orientation of in the space , the orientation of in the space should have should have (2(2ll+1)+1) possible values. possible values.

L

So the number of the tracks of the atoms So the number of the tracks of the atoms should be a odd number.should be a odd number.

. In order to explain the experimental results, . In order to explain the experimental results,

Unlenbeck and Goudsmit proposed a postulation Unlenbeck and Goudsmit proposed a postulation in 1925 :in 1925 :

An electron not only revolves around a nucleus An electron not only revolves around a nucleus but has a spin. This is analogous that the earth but has a spin. This is analogous that the earth revolves around the sun and meanwhile revolves around the sun and meanwhile rotating about its own axis.rotating about its own axis.

2

1sm

----spin quantum number----spin quantum number

2

1s

sz mS

The spin angular momentum of an electron The spin angular momentum of an electron is quantizedis quantized

S

1 ssS

here

5. Four quantum numbers (5. Four quantum numbers (n, l, mn, l, mll, m, mss))

---determine the state of an electron. ---determine the state of an electron. (1) Principle quantum number (1) Principle quantum number nn （（ n=1,2,n=1,2, ）） ------

determine the energy of the atom system.determine the energy of the atom system.(2) azimuthal (2) azimuthal ll （（ ll=0,1,2,=0,1,2,,,nn-1-1 ）） ---determi---determi

ne the magnitude of orbital angular momentum ne the magnitude of orbital angular momentum of an electron.of an electron.

(3) (3) magnetic magnetic mmll （（ mmll=0,=0,1,1,2,2,,,ll ）） ---d---d

etermine the orientation of orbital angular metermine the orientation of orbital angular momentum of an electron.omentum of an electron.

(4) Spin (4) Spin mmss （（ mmss==1/21/2 ）） ---determine the orie---determine the orie

ntation of spin angular momentum of an electrontation of spin angular momentum of an electron.n.

+Ze

Multi-electron:Multi-electron:

§20-8 Multi-electron atoms and §20-8 Multi-electron atoms and The Periodic tableThe Periodic table

+Ze““Average field” theory Average field” theory

+Ze protons+Ze protons++ (Z-1) electrons (Z-1) electrons

forms a uniform field.forms a uniform field.

moving uniform fieldmoving uniform field

Hydrogenlike ( 类氢 ) motion

ConclusionsConclusions

The state of each electron can be specified byThe state of each electron can be specified by n, l , mn, l , mll, m, mss . .

Pauli exclusion principlePauli exclusion principle(( 泡利不相容原理泡利不相容原理 )) ：： nno two electrons in any one atom can have the o two electrons in any one atom can have the same four quantum numbers same four quantum numbers n, l , mn, l , ml l , m, ms s ..

Corresponding Corresponding samesame n, l , m n, l , ml l , ,

2

1sm

i.e. only i.e. only 22 electrons with electrons with samesame n, l , m n, l , ml l ..

mmss has two values has two values

Corresponding Corresponding samesame n, l , n, l ,

i.e. i.e. 2(2l+1)2(2l+1) electrons with electrons with samesame n, l n, l ..

mmll has the values has the values mmll=0,=0,1,1,2,2,,,ll

The electrons which have same The electrons which have same l l are said to beloare said to belong to a subshell ng to a subshell (( 次壳层次壳层 )). .

Corresponding Corresponding ll= 1, 2, 3, 4, 5, 6 = 1, 2, 3, 4, 5, 6 With code lettersWith code letters s s, , p p,, d d,, e e, , f f,, g g, , to specify.to specify.

So the maximum number of the electrons witSo the maximum number of the electrons with same h same nn is is

1

0

122n

le lN 22n

The electrons which have same The electrons which have same n n are said to belare said to belong to a shellong to a shell(( 壳层壳层 )). .

Corresponding Corresponding nn= 1, 2, 3, 4, 5, 6 = 1, 2, 3, 4, 5, 6 With code lettersWith code letters K K,, L L,, M M,, N N,, O O,, P P to specify.to specify.

minimum energy principleminimum energy principle ：： If an atom is in If an atom is in ground stateground state, , the electrons must fill in such wthe electrons must fill in such way as to minimize the total energy of the atomay as to minimize the total energy of the atom..

Minimum energy corresponds the minimum Minimum energy corresponds the minimum values of values of n, l n, l . .

the configuration of an atom is represented the configuration of an atom is represented by symbol by symbol

21 )()( 2211xx lnln

ix -- -- the number of the electrons with same the number of the electrons with same nni i ,, l lii

For example, For example,

H-atom has H-atom has one electronone electron, , its configuration is represented by its configuration is represented by [H][H] 1s 1s11

Helium-atom has Helium-atom has 2 electrons2 electrons, , its configuration is represented byits configuration is represented by [H[Hee]] 1s 1s22

Lithium-atom has Lithium-atom has 3 electrons3 electrons, , its configuration is represented byits configuration is represented by [L[Lii]] 1s 1s2 2 2s2s11

Beryllium-atom has Beryllium-atom has 4 electrons4 electrons, , its configuration is represented byits configuration is represented by [B[Bee]] 1s 1s2 2 2s2s22

Boron-atom has Boron-atom has 5 electrons5 electrons, , its configuration is represented byits configuration is represented by [B][B] 1s 1s2 2 2s2s2 2 2p2p11

Neon-atom has Neon-atom has 10 electrons10 electrons, , its configuration is represented byits configuration is represented by [N[Nee]] 1s 1s2 2 2s2s2 2 2p2p66

Sodium-atom has Sodium-atom has 11 electrons11 electrons, , its configuration is represented byits configuration is represented by [N[Naa]] 1s 1s2 2 2s2s2 2 2p2p66

It is a great triumph for atomic theory to understand The periodic Table of the Elements in terms of atomic levels (or quantum states).