Transcript Document

Quantum Physics
Why do the stars shine?why do the elements
exhibit the order that’s so apparent in the periodic
table? How do transistors and microelectronic devices
work?why does copper conduct electricity but glass
doesn’t?
Quantum Physics
Particle Theory
of Light
Photoelectric
effect
Compoton
De Broglie Hypothesis
Hydrogen
Uncertainty
effect
spectrum
Schrodinger equation and application
Wave
Function
Key words
Quantum
Blackbody
Radiation
Planck’s hypothesis
The photoelectric effect
The particle theory or
light
X-rays
Diffraction
Photons
Electromagnetic
Key words
The wave properties of
particles
The uncertainty principle
The scanning tunneling
microscope
Atomic Spectra
The Bohr Theory of
Hydrogen
Ultraviolet
Catastrophe
Intensity
27.1 Blackbody Radiation
and Planck’s Hypothesis
Classical
theory
Experimental
data
Wavelength
Planck’s Hypothesis
In 1900 Planck developed a formula for
blackbody radiation that was in complete
agreement with experiments at all wavelengths .
Planck hypothesized that black body
radiation was produced by submicroscopic
charged oscillators, which he called resonators.
The resonators were allowed to have only
certain discrete energies, En, given by
En  nhf
27.2 The Photoelectric Effect and
The Particle Theory of Light
1.The Photoelectric Effect
1) saturated current is proportion to intensity of light
 is fixed,I is proportion to voltage,but there’s a
saturated current which is proportion to intensity of
light
I
Is
I
A
-
K
Il higher
Il loweer
A
V
pow
-Va
o
V
inverse voltage is applied，Ic=0. Va is called cutoff
voltage or stopping voltage
KEmax  hf  
I
I
A
-
Is
Il higher
K
Il loweer
A
V
pow
-Va
o
V
2) initial kinetic energy is proportion to the
frequency of incident light,has no relation with the
intensity of light
3) there’s a cutoff frequency (threshold frequency)
to a metal,only when >o, there’s a current
4) photo current produce immediately,delay time is
not more than 10-9s。
2.Einstain’s theory
1)the hypothesis of Einstian
Light has particle nature
  h
h  6.630  10 34 J  s
Intensity of light
I  Nh
2) Einstian equation
1 2
h  mv  A
2
A work function
Notes:
a. the initial kinetic energy is linearly proportion to the
frequency of incident light
b.
1 2
mv0  0
2
0
A

h
Cutoff frequency
While ＜A/h时，no photoelectric effect
c. instantaneously effect
3.the wave-particle duality
1) Light has dual nature
2) the energy, momentum and mass
energy
  h
m
Mass:
momentum
m0
1 v2 / c2
h h
p  mc 

c 
M0=0
h
h
m 2 
c
c
  h
p
h

•
Example :the experiment result is as follows,find h
eUa
1.oq
o
q
s

2.op
3. op/oq 4. qs/os
p
solution：3
Example: the cutoff o=6500Å, the light with 
=4000Å incident on the metal
(1)the velocity of photoelectrons?
(2)stopping voltage?
Solution:
1
2
h  m  A
2
 =6.5×105(m/s)
(2)
1
2
m  eVa
2
: Va=1.19 (V)
1
hc hc
2
m 

2
 o
c =
m  9.111031
h= 6.63×10-34
Example:incident frequency is fixed；we’v experimental
curve (solid line),and then with fixed intensity of light,increase
the incident frequency，the experimental curve is as dot
line,find the correct answer in the following graphs
I
I
(A)
(B)
o
I
V
o
V
o
V
I
(D)
(C)
o
V
27.3 The Compton Effect
1.scattering
light go through medium and propagate in
different direction。


from classic view:the scatter wavelength is same
with incident wavelength。
but in graphite experiment,we found a change in
wavelength,this is called compoton scatter。
2.principle
suppose: photos collide with electrons without
loss of energy
Energy conservation
hc
ho+moc2= h +mc2
y
h 

Momentum conservation：
hc
h o 
h h

x:
o
 cos   m cos

x
m
2h
2 
  o 
sin
mo c
2
o

h
y: 0  sin   m sin 

mo
m
c=
2
2
1 / c
2h
2
  o 
sin
mo c
2
=0， min=o ;
=180°, max=o +2c
Compton wavelength：
h
c 
 0.024 Å
mo c
y
hc
h o 
o
hc
h 



x
m
The explanation to compton effect
1)photon pass some energy to electrons
h  h 0
   0
2)compton effect is strong when the photon act on the
atom with small number atom
3) meaning：photon theory is correct show the
duality nature of light
4) micro particle obey the conservative law
5) photoelectric effect, compton effect
Example: with o =0.014Å X ray in compton
experiment, find maxmum kinetic energy in electrons？
solution:fron energy conservation
Ek 
，Emax
hc
o
Ek 
in fact

hc

hc
o
max=o +2c ，
hc
hc
Ek 

=1.1×10-13 J
o   2c
Example: o =0.1Å X ray in experiment。In the
direction of 90°,find wavelength? Kinetic energy and
momentum of electron?
solution:  =90°
2h
2
    o 
sin
mo c
2
 = o + =0.1+0.024=0.124Å
Ek 
hc
o
hc =3.8×10-15 J


From momentum conservation
h h
x:
 cos   m cos
o
y
hc
h o 
o

h
y: 0  sin   m sin 

hc
h 

 =90°


x
m
h
h
 p cos  ,
 p sin 

o
ph
1

1
-23 (SI)
=8.5×10
2

h
1

  cos ( )  38 44
o p
o
2
example:o =0.03Å X ray in experiment, the velocity
of recoil electron  =0.6c, find (1)the rate of the
scattering energy of electron to its rest energy (2)the
scattering =? And scattering  =?
solution: (1)the scattering energy
Ek  mc  mo c  mo c (
2
(2)
2
1
2
1 2 / c2
 1) =0.25moc2
hc hc
Ek  0.25mo c   ,   = 0.0434Å
o 
for   o 
2
2h
2 
sin
mo c
2
so  =63.4°
27.4 Line Spectra and Bohr Model
1.the atomic hydrogen spectrum
2.the empirical formula of Balmer
1
1
1
~
   R( 2  2 ),

2
n
n  3,4,5
R  1.0967758  10 m
7
Rydberg constant
1
H H  H  H 
The line spectrum system
1
1
1
~
Lyman series    R(  )
n  2,3,4,
2
2

1 n
ultraviolet
1
1
1
~
Balmer series    R(  ),
n  3,4,5
2
2

2
n
visible
Baschen
1
1 1
~
   R( 2  2 )

3 n
General form:
n  4,5,6,
1
1
1
~
   R( 2  2 )

k
n
infrared
Physicist: Rutherford
3.Bohr model  hypothesis
1) stationary hypothesis: electrons can be in some
certain stable orbits
2) quantum transition: h  E  E
n
3) quantization of angular momentum
PL  n
n  1,2,3
2
2
mV
e
 to hydrogen atom

r
4o r 2
m
PL  mVr  n
 0 h2
2
2
 rn 

n

r

n
, n  1,2,3,
1
2
me
r1  0.529  1010 m   first
1
1
2
E n  E np  E nk  mvn 
2
40
Bohr radus
e2
13.6ev

 E 
rn
n
n
n  1,2 ,3 ,
Notes: 1.Ground state n=1
2.excited state n>1
3.n=2,the first excited state
4.ionization: n  k  n  
5. explanation of hydrogen spectrum
En  Em
me 4
1
1


( 2 2 )
2 3
h
8 o h m
n
2
n=4
lyman
n=3
n=2
Balmer
n=1
r =a1
r =4a1
r =9a1
r =16a1
Paschen
T=R/n2

6
4387cm-1 5
6855cm-1 4
En=hcR/n2
Brackett hcR/25
hcR/16
Paschen
hcR/9=-1.51eV
12186cm-13
2741cm-1
2
Balmer
hcR/4=-3.39eV
lyman
109677cm-11
Energy level diagram
hcR=13.6eV
Example:find the energy for hydrogen atom giving
longest wavelength in lyman series?
1)1.5ev
2)3.4ev
3)10.2ev
solution:3 n=2-1
4)13.6ev
Example:with 913A violent light,hydrogen
atom can be ionized,find the wavelength expression
of lyman series
n1
1)  913
n1
3)  913
solution：4
2)  913
n1
n1
n2
4)  913 2
n 1
n 1
n2  1
2
 
R
1

 R(
1
913
1
1
 2)
1
n
Example:with visible light,can we ionized the
first excited state of hydrogen atom?
Solution:   h  hc  3.1ev
紫
紫
Needed enrgy E  E  E  0  ( 3.4)  3.4ev

2
no
Example: hydrogen atom
in third excited state,find the
number of its line after
transition,name its series？
solution:
lyman:
3
Balmer: 2
Pachen： 1
-0.85
-1.51
4
3
-3.4
2
-13.6
1
Physicist: De Broglie
27.5 Wave Nature of Particles
chapter 42 Quantum Mechanics
1. De Broglie wavelength
h h
p

c

Particle nature of light:   h ,
h
  h , p 
Notes:1)

h
h
in classic situation  

p
h
h
12.2
 

A
p
2m0 eu
u
h

p
in relativity  

hc
E 2  E 20

h
p
hc
E 2 k  2 E0 E k
m0 v
Notes:1)
example：a bullet with m=0.01kg，v=300m/s
h
h 6.63  10 34
 

 2.21  10  34 m
0.01  300
p m
h is so small,the wavelength is so small
It’s difficulty to measure,behave in particle nature
On the atomic scale,however,things are quite different
Me=9.1*10-31,v=106, =0.7nm
This wavelength is of the same magnitude as interatomic
spacing in matter,and in diffraction experiment the
phenomena is evidence.
2) quantization condition of angular momentum
of Bohr is the showing of de broglie wave
h
L  rp 
n  n
2
h
p
n
2r
h

p
2r  n
Standing wave
2. experiment
diffraction by electrons
（Thomoson1927）
slit,double slit diffraction
 Experiment
Davision and Germer experiment
A

G
d
a
M
2d cos   k , k  1,2,3,
 
h

p
h
2meU
example: (1)the kinetic energy of electron
Ek=100eV；(2)momentum of bullet
p=6.63×106kg.m.s-1, find 。
solution:for the small kinetic energy,with classic
formula
1 2 p2
6
Ek  m  ,
2
2m
  5.93 10
p  m  2mEk  5.4 1024
h h

 =1.23Å
m p
h
(2)bullet   = 1.0×10-40m
p
h= 6.63×10-34
conclusion:the wave nature is only obvious in
microparticle,to macroparticle,you can’t detect the
effect
Example:with 5×104V accelerating voltage,find
the 
solution:with relativity effect
Ek  mc  mo c  mo c (
2
2
2
1
1  / c
2
2
 1)  eV
=1.24×108(m/s)
m
mo
1 2 / c2
=10×10-31 (kg)
h
  =0.0535Å
m
mo=9.11×10-31 (kg)
h
Example：suppose  c 
me c
energy equal to its rest energy，
1) 
1
2
solution：2
c
2) 

1
3
3E
?
C
hc
2
，kinetic

0
hc
3m 0 c 2
Example:the first bohr radius a,electron move along
n track，
?
solution：   h  h  hrn  hrn  2na
p mv n mv n rn n h
2
27.6 Heisenberg’s Uncertainty Relation
1.uncertainty relation
h


t  E 
2
2
It states that measured values can’t be assigned to
the position and momentum of a particle simultaneously
with unlimited precision.

x  p x 
2
Notes: 1)
represent the intrinsic uncertainties in the
measurement of the x components of and even with best
measure instruments.
2) small size of planck’s constant
guarantees uncertainty relation is
important only in atomic scale
2.the meaning of uncertainty
1)the result of dual nature
2)give the applied extent of classic mechanics，if h
can be ignored ,the uncertainty relation have no use.
xPx  0
x  0
Px  0
3）the intrinsic of uncertainty is that there’s
a uncertain action between observer and the
object, It can’t be avoided.
example:estimate vx in hydrogen atom
solution: x=10-10m(the size of atom),
From  xpx  h,
h
 x 
mx
6.63 1034
6


7
.
3

10
(m / s)
31
10
9.1110 10
so big,velocity and coordinate can’t be determined
at the same time。
example: a bullet m=0.1kg , x=10-6 m/s ，find x。
solution:  xpx  h
34
h
6.63 10
 27
x 


6
.
63

10
(
m
)

6
m x
0.110
so small,we can consider it with classic view (to
macro object)
example:=5000Å，=10-3Å，find x
solution: p  h
x

 p x  
from  xpx  h,
h
2
 x 

 2.5m
p x 
h

2

Physicist: De Broglie
27.7 Wave Function, Schrodinger Equation
1.Schrodinger equation
2 2
[
  U ( r )]  E
2m
2.the statistic explanation of wave function
wave view
particle view
Bright fringe: (x,y,z,t)2 big, possibility big;
Dark fringe: (x,y,z,t)2 small , possibility small 。
(x,y,z,t)2 is proportion to possibility density in
this point.
r1
s1
...
2a
x
r2
s2
x p
o
K=2
K=1
K=0
K=1
D
K=2
The meaning of wave function:it’s related to the
probability of finding the particle in various
regions
Character: single value, consecutive limited, be one
in whole space
Notes on Schrodinger equation
• E=Ek+Ep=p2/2m+u non relativity form
• From the Solution of equation, E  can only
take special value
• 2 express possibility density
• Quantum condition can be got in natural
way,this related with atomic state
2、the application of Schrodinger equation on
hydrogen atom
2 2
[
  U ( r )]  E
2m
U (r )  
1) quantum energy
e2
4 0 r
1 me 4
En   2 
   same with Bohr theory
2
2
n 8 0 h
n  1,2,3,     principal quantum number
2) quantum of angular momentum
h
L  l l  1
2
l  0,1,2,  n  1
Angular momentum
quantum number
3) if in magnetic field
h
m l  0,1,   l 
2
magnetic quantum number
Lz  m l 
example： L  l (l  1)  2
l=1,
L  0,
z
l=2, L  l (l  1)  6
Lz  0,,2
z
z

 2

Lz

cos  
L 0


L

L
L
0

 2
Example:n=3,the possible value of L. and Lz
Solution 1) 3
2) 5
4) possibility distribution of electron
Solution:from schudigder equation:
Ψnl m(r, , ) =Rnl(r)l m()Φ m(
)
l
l
l
Possibility density:
Ψnl m(r, , )
2
l
for example: to 1S electron，possibility density
2r

ao
2
4
p1s  3 r e
ao
,
 oh2
ao 
me2
p1s
4 2
p1s  3 r e
ao

2r
ao
ao
图20-6
r
27.8 Electron Spin Four Quantum Number
1921，(O.Stern) and (W.Gerlach) prove:except
the orbital motion,there’s a spin
S  s( s  1)
s
1
2
Spin angular momentum
3
2

z component of s
S z  ms 
Spin magnetic number
1
ms  
2
S  s( s  1)
s
1
2
3
2
1
S z  ms    
2

z
1
 
2
Sz 1
cos 

S
3
0
1
 
2


S

S
in summary:the status of atom is determined by
four quantum number。
(1)principal quantum number：n=1,2,3,…。
determine the energy of atom。
(2)angular quantum number l=0,1,2,…,(n-1)。
determine the angular momentum。
(3)magnetic
quantum
number
ml=0,±1,±2,…,±l。
determine Lz，that’s space quantum property。
1
ms  
2
determine z component of spin angular
momentum
(4)spin magnetic quantum number
Physicist: Plank