Chapter 7
Quantum Theory and Atomic Structure
7-1
7-2
제 7장 양자론과 원자 구조
7.1 빛의 파동성, 빛의 입자성
고전물리학에서 양자론까지
광전효과
7.2 원자 스펙트럼, 수소 원자의 보어 모형
Bohr 수소
7.3 물질과 에너지의 파동-입자 이중성
하이젠베르크의 불확정성 원리
전자의 이중성
양자역학
7.4 원자의 양자역학적 모형, 원자 궤도함수
양자수
원자궤도함수
7-3
Quantum Theory and Atomic Structure
7.1 The Nature of Light
7.2 Atomic Spectra
7.3 The Wave-Particle Duality of Matter and Energy
7.4 The Quantum-Mechanical Model of the Atom
7-4
Maxwell (1873), proposed that visible light consists of electromagnetic
waves.
Electric field
electromagnetic waves.
Amplitude
Magnetic field
Direction of
wave
Speed of light (c) in vacuum = 299 792 458 m/s (exact)
= 3.00* × 108 m/s
7-5
All electromagnetic radiation
l×n=c
7.1
Human
km
m
MHz
radio
Atom
Bacteria
mm
GHz
microwave
μm
THz
THz
nm
PHz
IR
Ǻ
EHz
UV
X-ray
pm
ZHz
Gamma
FM Radio
88-108 MHz
AM Radio
0.6-1.6 MHz
microwave oven
2.4 GHz
7-6
Medical X
10-0.1Ǻ PET Imaging
0.1-0.01Ǻ
Convert frequency to wavelength
A photon has a frequency of 6.0× 104 Hz. Convert
this frequency into wavelength (nm). Does this frequency
fall in the visible region?
λ×n=c
λ = c/n
λ = 3.00 × 108 m/s / 6.0 × 104 Hz
λ = 5.0 × 103 m
AM Radio wave
λ = 5.0 × 1012 nm
7-7
7.1
Different behaviors of waves and particles.
7-8
from Silberberg Figure 7.4
The diffraction pattern caused by light passing through two
adjacent slits.
7-9
Mystery #1, “Black Body Problem”
Stefan–Boltzmann law
Wien’s law
Ultraviolet catastrophe!
Rayleigh-Jeans formula
F(l ) =
7-10
2ckT
l4
7.1
Mystery #1, “Black Body Problem”
Solved by Planck in 1900
Energy (light) is emitted or
absorbed in discrete units
(quantum).
F(l ) =
2c
l
5
2
h
e
hc
lkT
1
E=h×n
Planck’s constant (h)
h = 6.626×10-34 J•s
7-11
7.1
When (hc/lkT) << 1
(long wavelength range)
h
e
=

hc
lkT
1
h
2
 hc   hc 
1 

  ...  1
 lkT   lkT 
h
2
 hc   hc   hc  

 1 

 ...
 lkT   lkT   lkT  
=
lkT
  hc   hc  2 
c 1  

...
  lkT   lkT  


lkT
c
F(l ) =
=
7-12
=
2c
l
5
2
h
e
hc
lkT
2c lkT
= 5
l
c
1
2
2ckT
l
4
7.1
When (hc/lkT) >> 1
(short wavelength range)
h
hc
lkT
e
F(l ) =
=
7-13
2c
2c h
2
l
5
1

e
2
l
5
e
h
 lhc
kT
hc
lkT
= he
h
e
hc
lkT
1

2c
l
5
2
 lhc
kT
he
 lhc
kT
7.1
Mystery #2, “Photoelectric Effect”
7-14
from:http://www.mtmi.vu.lt/pfk/funkc_dariniai/quant_mech/wave_f.htm
7.2
Mystery #2, “Photoelectric Effect”
Photoelectric Effect on Sodium Plate
hn
KE e-
Reference: http://hyperphysics.phy-astr.gsu.edu/hbase/mod2.html
7-15
7.2
Mystery #2, “Photoelectric Effect”
KE = hn - BE
KE
Photoelectric Effect on Sodium Plate
= 1.25 eV
= 6.6 ×
10-34
Js
× 1.60217646 × 10-19 J/eV
= 2.0025 × 10-19 J
n
0
= 1.82 eV
-BE
7-16
= 1.82 × 1.60217646 × 10-19 J/eV
= 2.91 × 10-19 J
7.2
Work Functions for Photoelectric Effect
Work function
(Binding
Energy)
CRC handbook on Chemistry and Physics version 2008, p. 12-114
7-17
Element
Aluminum
Beryllium
Cadmium
Calcium
Carbon
Cesium
Cobalt
Copper
Gold
Iron
Lead
Magnesium
Mercury
Nickel
Niobium
Potassium
Platinum
Selenium
Silver
Sodium
Uranium
Zinc
Work Function(eV)
4.06-4.26
4.98
4.08
2.87
~5
2.14
5.0
4.53-4.95
5.1 -5.47
4.67-4.81
4.25
3.66
4.475
5.014-5.35
3.95-4.87
2.29
5.12-5.93
5.9
4.52-4.74
2.36
3.63-3.90
3.63-4.9
Mystery #2, “Photoelectric Effect”
Solved by Einstein in 1905
Light has both:
1. wave nature
2. particle nature
hn
KE e-
Photon is a “particle” of light
hn = KE + BE
KE = hn - BE
7-18
7.2
Convert wavelength to energy
When copper is bombarded with high-energy electrons, X
rays are emitted. Calculate the energy (in joules) associated
with the photons if the wavelength of the X rays is 0.154 nm.
E=h×n
E=h×c/l
E = 6.63×10-34 (J•s) × 3.00×10 8 (m/s) / 0.154×10-9 (m)
E = 1.29×10 -15 J
7-19
7.2
Comparing the diffraction patterns of x-rays and electrons.
x-ray diffraction of aluminum foil
electron diffraction of aluminum foil
from Silberberg Figure 7.14
7-20
The diffraction pattern caused by electron passing through two
adjacent slits.
7-21
The line
spectra of
several
elements.
7-22
from Silberberg Figure 7.7
Line Emission Spectrum of Hydrogen Atoms
Rydberg equation
 1
1 
= RH  2  2 
l
n2 
 n1
1
RH is the Rydberg constant = 1.096776×107 m-1
Lyman
Balmer
n1 = 1 and n2 = 2, 3, 4, ...
n1 = 2 and n2 = 3, 4, 5, ...
7-23
Paschen
n1 = 3 and n2 = 4, 5, 6, ...
7-24
7.3
Chemistry Mystery: Discovery of Helium
In 1868, Pierre Janssen detected a new dark line in the solar emission spectrum
that did not match known emission lines
Mystery element was named Helium (from helios)
In 1895, William Ramsey discovered helium in a mineral of uranium (from alpha
decay).
Hydrogen
Waveleng
th(nm)
Color
435
purple
486
blue
657
red
7-25
587.49 nm
Helium
Wavelength
(nm)
Color
447
purple
469
blue
472
blue
493
blue-green
501
blue-green
505
blue-green
587
yellow
669
red
Bohr’s Model of the Atom (1913)
1. e- can only have specific
(quantized) energy values
2. light is emitted as e- moves
from one energy level to a
lower energy level
+Ze
 Z2 
En =  RH  2 
n 
2
n
rn = a0
Z
n (principal quantum number) = 1,2,3,…
RH (Rydberg constant) = 2.17987209 ×10-18J
= 1.096776×107 m-1
7-26
7.3
 1 
En =  RH  2 
n 
1
𝐸𝑓 = −𝑅H ( 2 )
𝑛𝑓
1
𝐸𝑖 = −𝑅H ( 2 )
𝑛𝑖
ni = 3
nf = 2
7-27
656.3 nm
∆𝐸 = 𝐸𝑓 − 𝐸𝑖 = 𝑅H (
1
𝑛𝑖
2
−
1
𝑛𝑓
2)
7.3
Calculate a hydrogen atomic emission line wavelength
Calculate the wavelength (in nm) of a photon
emitted by a hydrogen atom when its electron
drops from the n = 5 state to the n = 3 state.
1
1
Ephoton =DE = RH ( 2
)
2
ni
nf
Ephoton = - 2.18×10-18 J × (1/25 - 1/9)
= 1.55 ×10-19 J
1/ l = -1.10×107 m-1 × (1/25 - 1/9)
Ephoton = h×c / l
= 7.82 × 10 5 m-1
l = 1.28×10-6 m = 1280 nm
l = h×c/ Ephoton
l = 6.63×10-34 (J•s) × 3.00×108 (m/s)/1.55×10-19J = 1.28×10-6 m
l = 1280 nm
7-28
7.3
De Broglie (1924) reasoned
that e- is both particle and
wave.
2 r = nl , n = 1, 2,3,...
h
l=
mu
u = velocity of e-
m = mass of e-
7-29
7.4
Calculate the de Broglie wavelength
What is the de Broglie wavelength (in nm) associated
with an electron traveling at 1.00× 106m/s?
λ = h/mu
h in J•s m in kg u in (m/s)
λ = 6.63×10-34 / (9.11×10-31 × 1.00×106)
λ = 7.27×10-10 m = 0.727 nm
7-30
7.4
Bohr 수소 원자와 양자역학
Bohr 이론
양자역학
전자는 원운동을 함
전자 궤도 상에 존재함
s 전자는 원운동 하지 않음
전자의 위치는 정할 수 없고 확률
분포함수
s 전자는 핵에서 발견될 확률이 최대
전자는 핵에 가까이 갈 수
없음
다전자 원자 적용 불가능
7-31
다전자 원자 및 분자에 적용 가능
Electron Imaging
Electron Microscopy
le = 0.004 nm
STM's ability to image the wave patterns of
electrons on the surface of a metal
1993 - IBM's Quantum Corral
7-32
Classical Theory to QUANTUM THEORY
Observation
Explanation
by
when
blackbody radiation
Energy is quantized; only certain
values allowed
Planck
1900
photoelectric effect
Light has particulate behavior
(photons)
Einstein
1905
atomic line spectra
Energy of atoms is quantized;
photon emitted when electron
changes orbit.
Bohr
1913
Davisson/Germer:
electron diffraction
by metal crystal
All matter travels in waves; energy de Broglie
of atom is quantized due to wave
motion of electrons
Compton scattering(1923)
photon wavelength
increases (momentum
decreases) after colliding
with electron
Mass and energy are equivalent;
particles have wavelength and
photons have momentum
7-33
Einstein/
de Broglie
1924
Heisenberg Uncertainty Principle
위치와 운동량과의 관계
Dx  Dp 

h
, where  =
2
2
위치와 운동량을 동시에 정확히 알 수 없다.
7-34
Estimate uncertainty of a position
양성자의 위치를 1×10-11m의 정확도로
측정하였다. 1초 후의 양성자 위치의 불확정성은?
Dx  Dp =
2
Dx = D v t 
7-35
,
Dp 
2mDx0
t
2Dx0
,
Dv =
 3.15 103 m
Dp

m
2mDx0
Estimate minimum uncertainty of momentum and speed
원자핵의 반경은 5×10-15m이다. 전자가 원자핵의
일부가 되기 위해 필요한 운동에너지는?

Dx  Dp =
2

Dp 
 1.110  20 kg  m / s
2Dx
Dp = mDv = 1.11020 kg  m / s
Dv = 1.11020 kg  m / s /(9.11031 kg) = 1.2 1010 m / s  c
KE = p 2c 2  m 2c 4  m c2  pc
KE = pc  1.11020 kgm / s  3 108 m / s = 3.31012 J
 20 MeV
너무 큰 에너지로 전자가 핵 안에 있을 확률이 거의 없다.
7-36
Estimate minimum uncertainty of momentum and speed
수소원자의 반경이 5.3×10-11m이다. 전자의 최소
운동에너지는?
Dx = 5.31011 m

 9.9 10 25 kgm / s
2Dx
p2
KE =
= 5.4 1019 J = 3.4 eV
2m
Dp =
7-37
Schrodinger Wave Equation
In 1926 Schrodinger wrote an equation that
described both the particle and wave nature of the eWave function (Y) describes:
1. energy of e- with a given Y
2. probability of finding e- in a volume of space
Schrodinger’s equation can only be solved exactly
for the hydrogen atom. Must approximate its
solution for multi-electron systems.
7-38
7.5
The Schrödinger Equation
wave function
Hamiltonian
Operator
kinetic energy
HY = EY
potential energy
total quantized energy of
the atomic system
H = T V
 2 2
2
2
=
( 2  2  2 )  V ( x, y, z )
2me x
y
z
mass of electron
 2 2Y 2Y 2Y
( 2  2  2 )  (V ( x, y, z)  E)Y = 0
2me x
y
z
7-39
Schrodinger Wave Equation
Ψ = f(n, l, ml, ms)
principal quantum number n
n = 1, 2, 3, 4, ….
distance of e- from the nucleus
n=1
7-40
n=2
n=3
7.6
Where 90% of the
e- density is found
for the 1s orbital
7-41
7.6
Schrodinger Wave Equation
Ψ = f(n, l, ml, ms)
angular momentum quantum number l
for a given value of n, l = 0, 1, 2, 3, … n-1
n = 1, l = 0
n = 2, l = 0 or 1
n = 3, l = 0, 1, or 2
l=0
l=1
l=2
l=3
s orbital
p orbital
d orbital
f orbital
Shape of the “volume” of space that the e- occupies
7-42
7.6
l = 0 (s orbitals)
l = 1 (p orbitals)
7-43
7.6
l = 2 (d orbitals)
7-44
7.6
Schrodinger Wave Equation
Ψ = f(n, l, ml, ms)
magnetic quantum number ml
for a given value of l
ml = -l, …., 0, …. +l
if l = 1 (p orbital), ml = -1, 0, or 1
if l = 2 (d orbital), ml = -2, -1, 0, 1, or 2
orientation of the orbital in space
7-45
7.6
ml = -1
ml = -2
7-46
ml = 0
ml = -1
ml = 0
ml = 1
ml = 1
ml = 2
7.6
Schrodinger Wave Equation
Ψ = f(n, l, ml, ms)
spin quantum number ms
ms = +½ or -½
ms = +½
7-47
ms = -½
7.6
Schrodinger Wave Equation
Ψ = f(n, l, ml, ms)
Existence (and energy) of electron in atom is described
by its unique wave function Ψ.
Pauli exclusion principle - no two electrons in an atom
can have the same four quantum numbers.
Electron is a fermion. Only one fermion can
occupy a quantum state at a given time;
this is the Pauli Exclusion Principle.
7-48
7.6
Schrodinger Wave Equation
Ψ = f (n, l, ml, ms)
Shell – electrons with the same value of n
Subshell – electrons with the same values of n and l
Orbital – electrons with the same values of n, l, and ml
How many electrons can an orbital hold?
If n, l, and ml are fixed, then ms = ½ or - ½
Ψ = f (n, l, ml, ½)
7-49
or Ψ = f (n, l, ml, -½)
An orbital can hold 2 electrons
7.6
How many 2p orbitals are there in an atom?
n=2
If l = 1, then ml = -1, 0, or +1
2p
3 orbitals
l=1
How many electrons can be placed in the 3d subshell?
n=3
3d
l=2
7-50
If l = 2, then ml = -2, -1, 0, +1, or +2
5 orbitals which can hold a total of 10 e-
7.6
Figure 7.16
Electron probability in the
ground-state H atom.
7-51
Determining Quantum Numbers for an Energy Level
What values of the angular momentum (l) and magnetic (ml) quantum
numbers are allowed for a principal quantum number (n) of 3? How
many orbitals are allowed for n = 3?
For n = 3, l = 0, 1, 2
For l = 0 ml = 0
For l = 1 ml = -1, 0, or +1
For l = 2 ml = -2, -1, 0, +1, or +2
There are 9 ml values and therefore 9 orbitals with n = 3.
7-52
Determining Sublevel Names and Orbital Quantum Numbers
Give the name, magnetic quantum numbers, and number of
orbitals for each sublevel with the following quantum numbers:
(a) n = 3, l = 2 (b) n = 2, l = 0 (c) n = 5, l = 1 (d) n = 4, l = 3
n
l
(a)
3
2
3d
-2, -1, 0, 1, 2
5
(b)
2
0
2s
0
1
(c)
5
1
5p
-1, 0, 1
3
(d)
4
3
4f
-3, -2, -1, 0, 1, 2, 3
7
7-53
sublevel name possible ml values # of orbitals
Figure 7.18
7-54
The 2p orbitals.
7-55
Figure 7.20
7-56
The 3d orbitals.
The 3d orbitals
along z-axis
on xy-plane,
betw. xy-axis
7-57
on xy-plane,
along xy-axis
on yz-plane,
betw. yz-axis
on zx-plane,
betw. zx-axis
4f orbitals.
7-58