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