Quantum Physics
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Transcript of Quantum Physics
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 PhysicsQuantum Physics
Particle Theory of Light
Particle Theory of Light De Broglie HypothesisDe Broglie Hypothesis
Photo-electric
effect
Photo-electric
effect
Compoton
effect
Compoton
effect
Hydrogen
spectrum
Hydrogen
spectrumUncertaintyUncertainty Wave
Function
WaveFunction
Schrodinger equation and applicationSchrodinger equation and application
QuantumBlackbodyRadiationPlanck’s hypothesisThe photoelectric effectThe particle theory or lightX-raysDiffractionPhotonsElectromagnetic
Key words
The wave properties of particles
The uncertainty principleThe scanning tunneling
microscopeAtomic SpectraThe Bohr Theory of
Hydrogen
Key words
27.1 Blackbody Radiation and Planck’s Hypothesis27.1 Blackbody Radiation and Planck’s Hypothesis
Classical theory
Experimental data
WavelengthIn
tens
ity
Ultraviolet
Catastrophe
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
nhfEn
27.2 The Photoelectric Effect and The Particle Theory of Light27.2 The Photoelectric Effect and The Particle Theory of Light
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
AV
pow
-I KA
-Va V
I
o
IsIl higher
Il loweer
1.The Photoelectric Effect
inverse voltage is applied , Ic=0. Va is called cutoff voltage or stopping voltage
AV
pow
-I KA
-Va V
I
o
IsIl higher
Il loweer
hfKEmax
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 EinstianLight has particle nature
hNhI Intensity of light
h J s 6 630 10 34.
2) Einstian equation
Amvh 2
21 A work function
While < A/h 时, no photoelectric effect
c. instantaneously effect
Cutoff frequencyh
A00
2
1 20 mvb.
Notes: a. the initial kinetic energy is linearly proportion to the frequency of incident light
3.the wave-particle duality
h
c
h
c
hm
2Mass:
22
0
/1 cv
mm
2) the energy, momentum and mass
h
c
hmcp
h
p
h
1) Light has dual nature
energy momentum
M0=0
1.oq 2.op
3. op/oq 4. qs/os
• Example :the experiment result is as follows,find h
p
o q s
eUa
solution: 3
Example: the cutoff o=6500Å, the light with =4000Å incident on the metal (1)the velocity of photoelectrons? (2)stopping voltage? Solution:
Amh 2
2
1 o
hchcm
2
2
1
=6.5×105(m/s)(2)
aeVm 2
2
1 c =
: Va=1.19 (V)
311011.9 m
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
V
I
o
(A)
V
I
o
(B)
V
I
o
(C)
V
I
o
(D)
27.3 The Compton Effect27.3 The Compton Effect
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 。
1.scattering
2.principle
suppose: photos collide with electrons without loss of energy
x
y
oo
hch
m
hc
h Energy conservation ho+moc2= h +mc2
Momentum conservation :
coscos m
hh
o
x:
y:
sinsin0 mh
22 /1 c
mm o
c=
2sin
2 2 cm
h
oo
2sin
2 2 cm
h
oo
=0 , min=o ;
=180°, max=o +2c
Compton wavelength :
024.0cm
h
oc Å
x
y
oo
hch
m
hc
h
1)photon pass some energy to electrons
2)compton effect is strong when the photon act on the atom with small number atom
The explanation to compton effect
00 hh
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
hchc
Eo
k
in fact max=o +2c ,
, Emax
ok
hcE
cok
hchcE
2 =1.1×10-13 J
Example: o =0.1Å X ray in experiment 。 In the direction of 90°,find wavelength? Kinetic energy and momentum of electron? solution: =90°
2sin
2 2 cm
h
oo
= o + =0.1+0.024=0.124Å
hchc
Eo
k =3.8×10-15 J
coscos mhh
o
x:
y:
sinsin0 mh
=90°
,cos
ph
o
sinph
22
11
o
hp =8.5×10-23 (SI)
4438)(cos 1
p
h
o
x
y
oo
hch
m
hc
h
From momentum conservation
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
)1/1
1(
22
222
c
cmcmmcE ook
=0.25moc2
(2) ,25.0 2
hchc
cmEo
ok = 0.0434Å
2sin
2 2 cm
h
oo for so =63.4°
1.the atomic hydrogen spectrum
H H H H
5,4,3),1
2
1(
1~22
nn
R
17100967758.1 mR
Rydberg constant
2.the empirical formula of Balmer
27.4 Line Spectra and Bohr Model27.4 Line Spectra and Bohr Model
,6,5,4)1
3
1(
1~22
nn
R
Baschen
infrared
,4,3,2)1
1
1(
1~22
nn
R
Lyman series
ultraviolet
5,4,3),1
2
1(
1~22
nn
R
Balmer series
visible
The line spectrum system
)11
(1~
22 nkR
General form:
Physicist: Rutherford
3.Bohr model
1) stationary hypothesis: electrons can be in some certain stable orbits
3) quantization of angular momentum
mnEEh 2) quantum transition:
3,2,1 nnPL
to hydrogen atom2
22
4 r
e
r
mV
o
hypothesis
,3,2,1nn
nnknpn r
emvEEE
2
0
2
4
1
2
1
2
6.13n
evE
n
4.ionization:
Notes: 1.Ground state n=1 2.excited state n>1 3.n=2,the first excited state
nkn
5. explanation of hydrogen spectrum
)n
1
m
1(
h8
me
h
EE2232
o
4mn
nmVrPL
,3,2,1,21
22
20
nnrnme
hrn
radusBohrfirstmr 10529.0 10
1
n=4
n=3
n=2
n=1
r =a1
r =4a1
r =9a1
r =16a1
lyman
Balmer
Paschen
En=hcR/n2
hcR/25hcR/16
hcR/9=-1.51eV
hcR/4=-3.39eV
hcR=13.6eV1
2
65
3
4
lyman
Balmer
Paschen
Brackett
T=R/n2
109677cm-1
2741cm-1
12186cm-1
6855cm-1
4387cm-1
Energy level diagram
Example:find the energy for hydrogen atom giving longest wavelength in lyman series?
1)1.5ev 2)3.4ev 3)10.2ev 4)13.6ev
solution:3 n=2-1
Example:with 913A violent light,hydrogen atom can be ionized,find the wavelength expression of lyman series
1
1913)1
n
n 1
1913)2
n
n
1
1913)3
2
2
n
n1
913)42
2
n
n
solution : 4
913
1
)1
1
1(
12
R
nR
Example:with visible light,can we ionized the first excited state of hydrogen atom?
Solution: evhc
h 1.3紫
紫
Needed enrgy evEEE 4.3)4.3(02
no
Example: hydrogen atom in third excited state,find the number of its line after transition,name its series ? solution:
-13.6
-3.4
-1.51-0.85
1
2
34
lyman: 3 Balmer: 2 Pachen : 1
Physicist: De Broglie
Particle nature of light: h
c
hph ,
chapter 42 Quantum Mechanics
h
ph ,
27.5 Wave Nature of Particles27.5 Wave Nature of Particles
vm
h
p
hsituationclassicin
0
p
h A
ueum
h
p
h 2.12
2 0
k0k2
022 EE2E
hc
EE
hc
p
hrelativityin
Notes:1)
1. De Broglie wavelength
Notes:1)
example : a bullet with m=0.01kg , v=300m/s
mmh
ph 341021.2
30001.0
341063.6
h is so small,the wavelength is so smallIt’s difficulty to measure,behave in particle natureOn 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.
Standing wave
nr 2
p
hn
r
hp
2
nnh
rpL 2
2) quantization condition of angular momentum of Bohr is the showing of de broglie wave
2. experiment
diffraction by electrons
slit,double slit diffraction
( Thomoson1927 )
M
Experiment
Davision and Germer experiment
G
A
a
,3,2,1,cos2 kkd
d
meU
h
p
h
2
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
,22
1 22
m
pmEk
24104.52 kmEmp
61093.5
p
h
m
h
=1.23Å
(2)bulletp
h
h= 6.63×10-34
= 1.0×10-40m
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
eVc
cmcmmcE ook
)1/1
1(
22
222
=1.24×108(m/s)
22 /1 c
mm o
=10×10-31 (kg)
m
h =0.0535Å mo=9.11×10-31 (kg)
Example : suppose , kinetic energy equal to its rest energy , cm
h
ec
c2
1)1 C
3
1)2
Example:the first bohr radius a,electron move along n track ,
?
solution : nahn
hr
rmv
hr
mv
h
p
h n
nn
n
n
2
2
solution : 2 200
2 33 cm
hc
E
hc
?
27.6 Heisenberg’s Uncertainty Relation27.6 Heisenberg’s Uncertainty Relation
1.uncertainty relation
2
xpx
2h
2
Et
It states that measured values can’t be assigned to the position and momentum of a particle simultaneously with unlimited precision.
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
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.
0
0
0
x
x
P
x
Px
3 ) the intrinsic of uncertainty is that there’s a uncertain action between observer and the object, It can’t be avoided.
2.the meaning of uncertainty
example:estimate vx in hydrogen atom solution: x=10-10m(the size of atom),
From xpx h,
xm
hx
)/(103.7101011.9
1063.6 61031
34
sm
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: xpx h
xm
hx
)(1063.6
101.0
1063.6 276
34
m
so small,we can consider it with classic view (to macro object)
example:=5000Å , =10-3Å , find x
solution:
from xpx h,
h
px
2
hpx
mp
hx
x
5.22
Physicist: De Broglie
1.Schrodinger equation
ErUm
)](2
[ 22
27.7 Wave Function, Schrodinger Equation27.7 Wave Function, Schrodinger Equation
2.the statistic explanation of wave function
wave view particle viewBright 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.
x
x
s2
s1
p
o
D
2a r2
r1
. .. K=0
K=1
K=1
K=2
K=2
The meaning of wave function:it’s related to the probability of finding the particle in various regionsCharacter: single value, consecutive limited, be one in whole space
• 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,t
his related with atomic state
Notes on Schrodinger equation
2 、 the application of Schrodinger equation on hydrogen atom
re
rU0
2
4)(
ErU
m)](
2[ 2
2
1) quantum energy
numberquantumprincipaln
theoryBohrwithsameh
men
En
,3,2,1
81
22
0
4
2
2) quantum of angular momentum
1,2,1,02
1
nlh
llL
numberquantummagnetic
lmh
mLllz
,1,02
3) if in magnetic field
Angular momentum quantum number
example :l=1, 2,,0 zL
6)1( llLl=2, 2)1( llL
,0zL
z
L0
2
2
z
0
L
Lzcos L
L
Example:n=3,the possible value of L. and Lz
Solution 1) 3 2) 5
4) possibility distribution of electron
Solution:from schudigder equation:
lm lm lmΨnl (r, , ) =Rnl(r)l ()Φ ()
Possibility density:
Ψnl (r, , ) lm
2
for example: to 1S electron , possibility density
,4
2
231
oa
r
o
s era
p
2
2
me
ha o
o
oa
r
o
s era
p2
231
4
ao r
p1s
图 20-6
1921 , (O.Stern) and (W.Gerlach) prove:except the orbital motion,there’s a spin
)1( ssS 2
1s
2
3
z component of s
sz mS
Spin magnetic number2
1sm
Spin angular momentum
27.8 Electron Spin Four Quantum Number27.8 Electron Spin Four Quantum Number
3
1cos
S
S z
2
1 sz mS)1( ssS
2
32
1s
z
0
2
1
2
1
S
S
(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 。
determine z component of spin angular momentum
(4)spin magnetic quantum number 2
1sm
in summary:the status of atom is determined by four quantum number 。
Physicist: Plank