Download atom

Quantum Theory of the Atoms
slit
准直缝
electrons
X 射线
crystal
晶体
fringes
劳厄斑
Electrons diffraction
§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
§20-3 De Broglie’s Postulates and Matter Waves
德布罗意假设和物质波
§20-4 The Uncertainty Principle 不确定关系
§20-5 Wave Function and Schrodinger Equation
Born’s Interpretation 波函数和薛定谔方程
§20-6 The infinite Potential Well 一维无限深的势阱
§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 Rutherford’s Experiment
and the Nuclear Atom
1. Thomson’s model
electron
+Ze of an atom distributes
uniformly in sphere.
Atom size (10-10m)
electrons panel in the atom
uniformly.
Plum pudding model
2. Rutherford’s –particles scattering experiment
(1907)
–
particles
atom
Some of the scattering angles are so large
that it cannot be interpreted with “Plum
pudding model”
3. Rutherford’s “nuclear atom model” (1911)
All the positive charge and almost all the mass of
the atom is concentrated in about 10-15~10-14 m
the volume of the atom.
4. Limitation Rutherford’s model
Cannot explain why the structure of atom is
stable .
 Cannot explain why the radiating spectrum
is discrete.
§20-2 Atomic spectra and
the Bohr model of hydrogen atom
1. The experimental laws of hydrogen atom


 
ultraviolet
4102Å
4341Å
H  H H
4861Å
6563Å
H
H
3646Å
visible
found their regularities in 1885 ：
1 1
~
  R( 2  2 ) -- Balmer formula
2 n
1
~
Wave number  
J.Balmer

 J.R.Rydberg
proposed a general formula in 1890
1
1
~
  R( 2  2 )
k
n
k  1,2,
n are integers with n  k
-- Rydberg formula
R=1.096776107m-1 -- Rydberg constant
1
1
~
  R( 2  2 )
k
n
Same k, different n – a series line (线系)
(1)k=1,n=2,3,… Lyman series, ultraviolet
(2)k=2,n=3,4,… Balmer series, visble
(3)k=3,n=4,5,… Paschen series, infrared
(4)k=4,n=5,6,… Brackeff series, infrared
(5)k=5,n=6,7,… Pfund series, infrared
………………
1
1
~
  R( 2  2 )  T ( k )  T ( n )
k
n
----T(k) , T(n) terms of spectrum
~
 kn  T (k )  T (n)
 [T ( k )  T ( n)]  [T ( n)  T ( n)]
 ~kn  ~nn
----Ritz combination principle
2.Bohr’s postulates
In
order to support the nuclear model, Bohr
proposed three postulates for explaining the
experimental regularities of hydrogen atom.
(1) Stable states postulate：
The motion of electron in
a circular orbit is stable
(2) Postulate about the quantization of orbit
angular momentum :
The orbit angular momentum L=mvr of an
electron must satisfy ,
h
L  n  n
2
n  1,2,
n—quantum number
h

--simplified Plank’s constant
2
(3) radiation postulate:
When an electron charges its orbit, a photon
with frequency is emitted or absorbed.
 kn 
Ek  En
h
En>Ek--- a photon is emitted
En<Ek---a photon is absorbed
Bohr
got Noble Prize of physics in 1922.
3. The results of Bohr’s model
(1) The radius of orbit
According to
Newton’s law
e
2
2
v

m
2
4 0 r
r
nh
v
2mr
h
and L  mvr  n
2
  0h 

n

1
,
2
,

So rn  n 
2
 me 

 --quantization
2
2
For n=1,
  0h 

rn  n 
2 
me 

----first Bohr’s radius
2
r1=0.5310-10 m
2
rn  n r1
2
(2) energy
The energy of hydrogen atom=potential+kinetic
2
1
e
2
E n  mv n 
2
4 0 rn
2
2
e
v
using
m
2
4 0 r
r
2
1
e
2
We get mv n 
2
8 0 r
4

e
1  me 
 En  
 2 2 2
8 0 rn
n  8 0 h 
2
---quantization
The
quantized energy ---- energy-leverl
n----quantum
number
discussion：
4

1  me 
En   2  2 2 
n  8 0 h 
2



h
2
0

rn  n 
2 
 me 
(1) For n=1, E1  13.6eV ----ground state
Lowest energy, most stable state
(2) For n>1 ----excited state
(3) For n  
, rn  , En  0
---The ionization energy of hydrogen atom in
ground state is 13.6eV

E4 
E3

E 2
n 1
n2
Excited
n3
n4
Electronic orbits
E1
Energy-level
ground
En/eV
0
-0.85
-1.51

Paschen
-3.39

Balmer
-13.6 
Lyman

n
4
3
2
1
1  me 4 
En   2  2 2 
n  8 0 h 
an electron jumps EnEk , the H-atom emits
a photon.
 If
En  Ek
me  1
1 


 2
2 3  2
h
n 
8 0 h  k
4
Wave number is

me  1 1 
 1 1
~
   2 3  2  2   R 2  2 
c 8 0 h c  k n 
k n 
4
4
me
7
1
Here R 
2 3  1.0973731  10 m
8 0 h c
---agree perfectly with the experiment.
Rexp=1.096776107m-1
Bohr’s theory made great success in hydrogen
atoms and hydrogenlike ions (类氢离子)
But this theory cannot explain the experimental
results of more complex particles.
More than one electron outside the nuclear
4. The limitation of Bohr’s model
(1) using classical physics theory, but the
electron has acceleration and no radiation is
inexplicable.
(2) No any theory can explain the angular
momentum quantization.
(3) Cannot get the intensity of the spectra from
this theory.
§20-3
De Broglie’s postulates
and matter waves
1. De Broglie’s postulates(1924)
Just as radiation has particle-like properties,
electron and other material particles possess
wave-like properties.
E  h
h
p

---De Broglie’s formulae

The waves related to the material
particles----matter waves
De
Broglie won Noble prize in physics in 1929.
2. Testing De Broglie hypothesis
 The
wave property of
electrons was confirmed
by Davisson and Germer
in experiment in 1927.
observed that the
diffraction patterns of the
electrons are similar to with
the ones of x-ray.
E-gun
detector
 They
--electron have wave
property.

Acce.
anode
Nickle crystal
Thomson’s
experiment gave another
method to confirm the
wave
property
of
electrons.
Meanwhile,
They
won Noble prize in
physics in 1937
x-ray
electrons
The
results of slits diffraction experiments in
1960
single
It
double
three
four
was confirmed by great amount of
experiments that all micro-particles such as
electrons, neutrons, proton, as well as atoms and
molecules have wave properties. And their wave
properties agree with De Broglie’s formulae.
[Example] an electron is accelerated by electric
field. If the Acc. potential is U, the electron is at
rest before it is accelerated and its speed V<<C
after it is accelerated. Find its wavelength 
Solution
1
2
eU  m0 v
2
h

m0v
2eU
v 
m0
h

2em0U
12.2 o

A
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

For microscopic particles ：their position ,
velocity and others cannot be determined by
actual experiment at same instant because of
wave properties.
a sin   k
 The
single slit diffraction of electrons
A beams of electrons is incident on a slit normally,
considering the central bright fringe only,
0  px  p sin 1  px  p sin1
The first dark fringe
satisfies,
x px
py
x sin 1  

 p x  p
x
p
x
1
y
  h p
 px  h x
xpx  h

p x  p
x
Considering all diffraction fringes,
xpx  h --estimated result
Hesbreg
deduced the precise
results in 1927.



xp x  , yp y  , zpz 
2
2
2
Hesbreg
won Noble prize in physics in 1932.

Any experiment cannot determine
simultaneously the exact values of a
component of momentum and its
corresponding coordinate.
xpx  h
---Coordinate-momentum uncertainty relation

Similarly, measuring the energy of a
particle in a time interval t, the
uncertainty E of the energy is
tE  h
[Example] according to classical
electromagnetism and mechanics, the speed of
the electron moving around its nuclear is about
106 m/s. the size of an atom is 10-10m. Estimate
the uncertainty of the speed.
Solution the uncertainty of the position of
the electron is x  10 10 m
34
p x
h
6
.
63

10
 v x 


 31
10
m
m x 9.1  10  10
6
 7.3  10 m/s
v  v. the uncertainty of the speed of the
electron in atoms is very apparent.
§20-5 The wave function and
Schrodinger equation
Born’s interpretation of wave function
h
E  h p 
1.wave function

 A classical plane wave function travels in +x axis,
y  y0 cos 2 (t  x  )
 Write
it in complex form,
y  y0 e
 i 2 (t  x  )
Similarly,
for a free particle with energy E and
momentum p, its wave function,
x
i
 i 2 (t  )
  Et  px 

  x , t   0e
 0e

2. Schrodinger equation
Free
  x , t   0 e

i
 Et  px 

particle：moves in +x axis

p
  2
2
x

2
  p

 
2
2m x
2m
2
2

i
  E
t

2
p
E 
2m
2
2

i
 E
t
2
2
 



i

2
2m x
t
---S-Equation of a free particle
 The
particle has potential energy V(x,t),
its total energy is
E  Ek  V ( x, t )  p 2m  V ( x, t )
2
i p

i
 V ( x , t )]
  E   [
 2m
t

2


p
i.e. i
[
 V ( x , t )]
2
t
2m
 2  2
p2



2
2m x
2m
 


 V ( x , t )  i
2
2m x
t
2
2
 If
a particle moves in three dimension space,




2



2
2
2
2  
x
x
y z
2
2
2
2

 2 
 ( r , t )


  ( r , t )  V ( r , t )  i
2m
t
2
 Introducing
energy operator,
2


2
ˆ
H 
  V ( r , t ) --Hamiltonian operator
2m
---S-Equation of a


then Hˆ   i
particle in a
t
potential field
 


 V ( x , t )  i
2
2m x
t
2
 Stationary
time
Let
2
state: potential is independent on
V  V ( x)
 ( x, t )   ( x ) f (t )

  ( x)

f (t )
 V ( x ) ( x ) f ( t )
2
2m
x
2
2
f ( t )
 i ( x )
t
Divided by  ( x ) f ( t ) , we get
 1   ( x)
1 f (t )

 V ( x )  i
2
2m  ( x ) x
f (t ) t
2
2
Left side of the equation is the function of x,
and the right side is the function of t,
If it is useful for any t or any x, it must
satisfies,
left side = right side = a constant
i.e.
1 f ( t )
 E (1)
i
f ( t ) t
 1   ( x)
(2)


V
(
x
)

E
2
2m  ( x ) x
2
2
the solution of Eq. (1) is f ( t ) ~ e
E has the demission of energy.
i
 Et

Rewriting (2),
   ( x)

 V ( x ) ( x )  E ( x )
2
2m x
2
or
2
  ( x ) 2m
 2 ( E  V ) ( x )  0
2
x

2
----Schordinger’s equation in stationary state
The
wave function of the particle is
  x , t     x e
i
 Et

3.The physical meaning of wave function
---Born’s interpretation of wave function
 --- complex function
no physical meaning
 M.Born
postulated


 ( r , t ) ( r , t )  
*
2

  (r , t )
represents probability density,
i.e. the probability of finding the particle per

unit volume about the point r at time t.
The probability of finding the particle within 
the volume element dv =dxdydz about the point r
at time t is
dw   dV
2
The wave function  ( r , t ) should satisfy
 continuous, single-valued and finite.
---Standard conditions
And 


V
2
dV  1 ---Normalized condition
The space of the particle moving in
 The superposition principle
i.e
If the  1 and  2 are two solutions of the
Schrodinger equation,
then
  c11  c22
is also the solutions of the S- equation.
Here, c1 and c2 are normalized factor.
 The conservation of probability
2

 ( r , t )   ---probability density


*
*
j
[      ]
2mi
---probability current density
From Sch. Eq., we can get
 
 j 
0
t
---probability is conservative.
4. Operators and physical observables
The value of a physical observable=
(its operator )dV


*
V
Such as
2
 2

ˆ
Energy operator is H  
  V (r , t )
2m
ˆ  dV
So E   * H

V
ˆ
momentum operator is P  i


* ˆ
So P   P dV
V
If the particle is in stationary state,
 2

 ( r , t )   ( r )e
i
 Et

2
 2
  (r )
-- probability density has nothing to do with t.

Schrodinger won
Noble prize in 1933.
§ 20-6 The infinite potential well
 A particle
with mass m moves along x axis.
It potential is
V ( x)  
0
0  x  a( in well )
 x  0, x  a(out well )
Out well V ( x )  
  0


In well: V ( x )  0
 d  ( x)

 E ( x )
2
2m dx
2
0
a
x
Let k  2mE 
2

2
d  ( x)
2

 k  ( x)  0
2
dx
2
Its general solution is
 ( x )  C sinkx   
 0
C、 ---integral constants
0

 0
a
Using the continuous condition of wave function,
 ( 0)  0
 ( a)  0
C sin  0   0
C sin ka  0
n
k 
n  1,2, 
a
ka  n
n
 ( x )  C sin
x
a
x
Using normalized condition of wave function,
2
2
2
a
a
a
n
 ( x, t ) dx    ( x ) dx   C sin
x dx  1
0
0
0
a
We get C  2 a

0  x  0, x  0 
 n ( x )  
 2 a sin n x 0  x  a 
a
2mE
k  2

2
n 
 En 
2ma 2
2
2 2
n
k
a
n  1,2,  --quantized energy
n：quantum number
Notes：
 n=1： E1 
 
2
2
2ma
2
-- zero point energy
-- the lowest energy of a particle cannot be
zero in quantum mechanics.
 En  n2 --- the interval of energies between
two adjacent states is not uniform.
E n 
 
2
2
2ma
2
2n  1
For an eigen state n with quantum number n,
there are n+1 nodes and n antinodes in the well.
Minimum probability
 n (x )
Maximum probability
n4
 n (x )
n3
n2
0
n 1
a x 0
a x
2
[Example] A particle with mass m locates in a
infinite potential well with length a. Calculate 
the probability of finding the particle in the
range of 0xa/4 for the two different states
with n=1 and n=.  the positions of maximum
probability for the state n=2.
2
n
sin
x
Solution  As  ( x ) 
a
a
 the probability of finding the particle in the
range of 0xa/4 is
a
2
4
2 n
4
dx
w    ( x ) dx   sin
0
a 0
a
a
2
1 1
n
 
sin
4 2n
2
n 1
1 1
 9%
For n  1 w  
4 2
For
n
1
w
4
n
2
2
  2 ( x )  sin
x
a
a
2 2 2
2
  2 ( x )  sin
x
a
a
d
2
 2 ( x)  0
Let
maximum
dx
We get
a a 3a
,a
, ,
x0 ,
4 2 4
minimum
Tunneling effect
The potential distribution is
 U0
U ( x)  
0
0 xa
x  0, x  a
A particle move along x axis,
U
U0
1
2
3
For classical particle,
If E  U 0 , the particle can pass through area
2 and arrive to area 3.
If E  U0 , the particle cannot pass through
area 2 and cannot be found in area 3.
x
For macroscopic particle,
Using Schrodinger equation,
 d 1
Area 1 
 E 1
2
2m dx
2
2
 d 2
Area 2 
 U0 2  E 2
2
2m dx
2
2

d

3
Area 3 
 E 3
2
2m dx
2
2
The solutions of S.Eq. are
Area 1：incident waves +reflected waves
Area 2：transmitted waves + reflected waves
Area 3： transmitted waves
No matter E>U0 or E<U0
E
---Tunneling effect
0
a
x
Application Scanning Tunnel Microscope (STM)
扫描隧道显
微镜(STM)
硅表面图像
癌细胞表面图像
“扫描隧道绘画”
一氧化碳“分子人”
“原子和分子的观察与操纵” -- 白春礼
1993年人们用STM所做的量子围栏第一次看到波
函数
操纵48个铁原子在铜表
面围成半径为 71.3Å的
圆圈。表面电子在铁原
子上强烈反射，被禁锢
在该围栏中，它们的波
函数形成同心圆形驻波
量子围栏
----实现波函数的测量
§20-7 The Hydrogen atom and Electron spin
1.The Schrodinger equation of hydrogen atom
In the hydrogen atom system, the potential
energy is
U (r )  
 Schrodinger equation
e
2
4 0 r
is
      2m
e



(
E

)


0
2
2
2
2
x y z

4 0 r
2
2
2
2

introducing the transitions,
x  r sin cos 
y  r sin sin 
z  r cos
S.Eq.
changes into
1  2 
1


(r
) 2
(sin
)
2
r r
r
r sin 

1   2m
e
 2 2
 2 (E 
)  0
2
r sin  

4 0 r
2
2
Let
 (r , ,  )  R(r ) ( ) ( )
d
2
 ml   0
2
d
2
We get
(1)
ml
1 d
d
(sin
)  (  2 )  0 (2)
sin d
d
sin 
2
1 d 2 dR  2m
e

(3)
(
r
)

(
E

)

R

0
 2
2
2
r dr dr  
40 r r 
2
2.the results of the solution of Schrodinger
equation for H-atom
Quantized energy and principle quantum
number n.
The energy of H-atom system can be gotten
from Eq. (3)
4
4
me
1
me 1
En  
 2 2 2
2 2 2 2
32  0  n
8 0 h n
n  1,2,---- principle quantum
number n.
It is agrees with Bohr’s result.
 Quantized angular momentum and azimuthal
quantum number l.
The magnitude of orbital angular momentum
about the electron moving around the nuclear
can be gotten from Eq.(2)
L  l (l  1)
l  0,1,2,( n  1)
--azimuthal quantum number(角量子数）
l has n possible values for a given value of n.
 The direction of orbital angular momentum
about the electron moving around the nuclear
can be gotten from Eq.(3)
ml  0,1,2,  l

---the orientation (取向) of L is quantized
---magnetic quantum number(磁量子数）
The component Lz of L along z axis is
Lz  ml 
ml has (2 l+1) possible values for a given value of l .
3. The “electron cloud”
The distribution of the probability density of
finding the electron.
4. Electron spin
Unlenbeck and Goudsmit arranged a
experiment to check whether the orientation
(取向) of L is quantized (1921)
N
Atom
source
No magnet
S
In
magnet
 The
results of the experiment：
 No magnet：there is one track of
the atoms on the film.
 With magnet： there are two
tracks of the atoms on the film.
无磁场
有磁场
contradiction：when
the azimuthal
 quantum
number is l , the orientation of L in the space
should have (2l+1) possible values.
So the number of the tracks of the atoms
should be a odd number.
. In order to explain the experimental results,
Unlenbeck and Goudsmit proposed a postulation
in 1925 :
An electron not only revolves around a nucleus
but has a spin. This is analogous that the earth
revolves around the sun and meanwhile
rotating about its own axis.

The spin angular momentum S of an electron
is quantized
S  ss  1
S z  ms 
here
1
s
2
1
ms  
2
----spin quantum number
5. Four quantum numbers (n, l, ml, ms)
---determine the state of an electron.
(1) Principle quantum number n（n=1,2,）--determine the energy of the atom system.
(2) azimuthal  l（l=0,1,2,,n-1）---determine
the magnitude of orbital angular momentum of
an electron.
(3) magnetic  ml（ml=0,1,2,,l）--determine the orientation of orbital angular
momentum of an electron.
(4) Spin  ms（ms=1/2）---determine the
orientation of spin angular momentum of an
electron.
§20-8
Multi-electron atoms and
The Periodic table
 Multi-electron:
“Average field” theory
+Ze
+Ze protons+ (Z-1) electrons
forms a uniform field.
+Ze
moving uniform field
Hydrogenlike (类氢) motion
Conclusions
The state of each electron can be specified by
n, l , ml, ms .
Pauli exclusion principle(泡利不相容原理)：no
two electrons in any one atom can have the
same four quantum numbers n, l , ml , ms .
Corresponding same n, l , ml ,
1
ms has two values m s  
2
i.e. only 2 electrons with same n, l , ml .
Corresponding same n, l ,
ml has the values ml=0,1,2,,l
i.e. 2(2l+1) electrons with same n, l .
The electrons which have same l are said to
belong to a subshell (次壳层).
Corresponding l= 1, 2, 3, 4, 5, 6 
With code letters s, p, d, e, f, g, to specify.
So the maximum number of the electrons
with same n is
n 1
N e   22l  1  2n
2
l 0
The electrons which have same n are said to
belong to a shell(壳层).
Corresponding n= 1, 2, 3, 4, 5, 6 
With code letters K, L, M, N, O, P to specify.
 minimum energy principle：If an atom is in
ground state, the electrons must fill in such
way as to minimize the total energy of the
atom.
Minimum energy corresponds the minimum
values of n, l .
the configuration of an atom is represented
by symbol
x1
x2
( n1l1 ) ( n2 l 2 ) 
x i -- the number of the electrons with same ni , li
For example,
H-atom has one electron,
its configuration is represented by [H] 1s1
Helium-atom has 2 electrons,
its configuration is represented by [He] 1s2
Lithium-atom has 3 electrons,
its configuration is represented by [Li] 1s2 2s1
Beryllium-atom has 4 electrons,
its configuration is represented by [Be] 1s2 2s2
Boron-atom has 5 electrons,
its configuration is represented by [B] 1s2 2s2 2p1

Neon-atom has 10 electrons,
its configuration is represented by
[Ne] 1s2 2s2 2p6
Sodium-atom has 11 electrons,
its configuration is represented by
[Na] 1s2 2s2 2p6

It is a great triumph for atomic theory
to understand The periodic Table of the
Elements in terms of atomic levels (or
quantum states).