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Chapter 7. Quantum Theory and Atomic Structure. 제 7 장 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. Quantum Theory and Atomic Structure.

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slide1

Chapter7

Quantum Theory and Atomic Structure

slide3
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
slide4

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

slide5

Maxwell (1873), proposed that visible light consists of electromagnetic waves.

Electric field

electromagnetic waves.

Amplitude

Wavelength(l)

Direction of wave

Magnetic field

Speed of light (c) in vacuum = 299 792 458 m/s (exact)

= 3.00* × 108 m/s

All electromagnetic radiation

l×n = c

7.1

slide6

Atom

Human

Bacteria

Ǻ

FM Radio

88-108 MHz

AM Radio

0.6-1.6 MHz

Medical X

10-0.1Ǻ

microwave oven

2.4 GHz

PET Imaging

0.1-0.01Ǻ

slide7

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

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?

7.1

mystery 1 black body problem
Mystery #1, “Black Body Problem”

Stefan–Boltzmann law

Wien’s law

Ultraviolet catastrophe!

Rayleigh-Jeans formula

7.1

mystery 1 black body problem solved by planck in 1900
Mystery #1, “Black Body Problem”Solved by Planck in 1900

Energy (light) is emitted or absorbed in discrete units (quantum).

E = h× n

Planck’s constant (h)

h = 6.626×10-34J•s

7.1

slide14

Mystery #2, “Photoelectric Effect”

7.2

from:http://www.mtmi.vu.lt/pfk/funkc_dariniai/quant_mech/wave_f.htm

slide15

Mystery #2, “Photoelectric Effect”

hn

Photoelectric Effect on Sodium Plate

KE e-

Reference: http://hyperphysics.phy-astr.gsu.edu/hbase/mod2.html

7.2

slide16

KE = hn - BE

KE

n

0

-BE

Mystery #2, “Photoelectric Effect”

Photoelectric Effect on Sodium Plate

= 1.25 eV

× 1.60217646 × 10-19J/eV

= 2.0025 × 10-19J

= 6.6 × 10-34J s

= 1.82 eV

= 1.82 × 1.60217646 × 10-19J/eV

= 2.91 × 10-19J

7.2

work function binding energy
Work function(Binding Energy)

Work Functions for Photoelectric Effect

CRC handbook on Chemistry and Physics version 2008, p. 12-114

slide18

Mystery #2, “Photoelectric Effect”Solved by Einstein in 1905

hn

  • Light has both:
  • wave nature
  • particle nature

KE e-

Photon is a “particle” of light

hn = KE + BE

KE = hn - BE

7.2

slide19

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

7.2

slide20

x-ray diffraction of aluminum foil

electron diffraction of aluminum foil

Comparing the diffraction patterns of x-rays and electrons.

from Silberberg Figure 7.14

slide22

The line spectra of several elements.

from Silberberg Figure 7.7

slide23

Line Emission Spectrum of Hydrogen Atoms

Rydberg equation

RHis the Rydberg constant = 1.096776×107m-1

Lyman

Balmer

Paschen

  • n1 = 1 and n2 = 2, 3, 4, ...

n1= 3 and n2 = 4, 5, 6, ...

n1= 2 and n2 = 3, 4, 5, ...

slide25

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).

Helium

587.49 nm

Hydrogen

slide26

Bohr’s Model of the Atom (1913)

  • e- can only have specific (quantized) energy values
  • light is emitted as e- moves from one energy level to a lower energy level

+Ze

n (principal quantum number) = 1,2,3,…

RH (Rydberg constant) = 2.17987209 ×10-18J

= 1.096776×107m-1

7.3

slide27

ni= 3

nf = 2

656.3 nm

7.3

slide28

Calculate a hydrogen atomic emission line wavelength

Ephoton =

-DE= RH

i

f

( )

1

1

n2

n2

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.

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)

= 7.82 × 10 5 m-1

l = 1.28×10-6 m = 1280 nm

Ephoton = h×c/ l

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

7.3

slide29

De Broglie (1924) reasoned that e- is both particle and wave.

u= velocity of e-

m = mass of e-

7.4

slide30

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= 7.27×10-10m = 0.727 nm

7.4

slide31
Bohr 이론

전자는 원운동을 함

전자 궤도 상에 존재함

전자는 핵에 가까이 갈 수 없음

다전자 원자 적용 불가능

Bohr 수소 원자와 양자역학

양자역학

s 전자는 원운동 하지 않음

전자의 위치는 정할 수 없고 확률 분포함수

s 전자는 핵에서 발견될 확률이 최대

다전자 원자 및 분자에 적용 가능

slide32

Electron Imaging

Electron Microscopy

STM's ability to image the wave patterns of electrons on the surface of a metal

le = 0.004 nm

1993 - IBM's Quantum Corral

heisenberg uncertainty principle
Heisenberg Uncertainty Principle

위치와운동량과의 관계

위치와운동량을 동시에 정확히 알 수 없다.

slide35

Estimate uncertainty of a position

양성자의 위치를 1×10-11m의 정확도로 측정하였다. 1초 후의 양성자 위치의 불확정성은?

slide36

Estimate minimum uncertainty of momentum and speed

원자핵의 반경은 5×10-15m이다. 전자가 원자핵의 일부가 되기 위해 필요한 운동에너지는?

너무 큰 에너지로 전자가 핵 안에 있을 확률이 거의 없다.

schrodinger wave equation
Schrodinger Wave Equation
  • In 1926 Schrodinger wrote an equation that described both the particle and wave nature of the e-
  • Wave function (Y) describes:
    • . energy of e- with a given Y
    • . 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.

7.5

slide39

wave function

total quantized energy of the atomic system

mass of electron

kinetic energy

potential energy

Hamiltonian Operator

The Schrödinger Equation

HY = EY

schrodinger wave equation1
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

n=2

n=3

7.6

slide41

Where 90% of the

e- density is found

for the 1s orbital

7.6

slide42

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

l = 0 s orbital

l = 1 p orbital

l = 2 d orbital

l = 3 f orbital

n = 1, l = 0

n = 2, l = 0 or 1

n = 3, l = 0, 1, or 2

Shape of the “volume” of space that the e- occupies

7.6

slide43

l = 0 (sorbitals)

l = 1 (porbitals)

7.6

slide45

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, or1

if l = 2 (d orbital), ml = -2, -1, 0, 1, or2

orientation of the orbital in space

7.6

slide46

ml = -1

ml = 0

ml = 1

ml = -2

ml = -1

ml = 0

ml = 1

ml = 2

7.6

slide47

Schrodinger Wave Equation

Ψ= f(n, l, ml, ms)

spin quantum number ms

ms = +½ or -½

ms = +½

ms = -½

7.6

slide48

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.

Electronisa fermion. Only one fermion can occupy a quantum state at a given time;

this is the Pauli Exclusion Principle.

7.6

slide49

Schrodinger Wave Equation

Ψ = f (n, l, ml, ms)

Shell – electrons with the same value of n

Subshell – electrons with the same values of nandl

Orbital – electrons with the same values of n, l, andml

How many electrons can an orbital hold?

If n, l, and mlare fixed, then ms = ½ or - ½

Ψ = f (n, l, ml, ½)

orΨ = f (n, l, ml, -½)

An orbital can hold 2 electrons

7.6

slide50

n=2

n=3

If l = 1, then ml = -1, 0, or +1

3 orbitals

l = 1

l = 2

If l = 2, then ml = -2, -1, 0, +1, or +2

3d

5 orbitals which can hold a total of 10 e-

How many 2porbitals are there in an atom?

2p

How many electrons can be placed in the 3dsubshell?

7.6

slide51

Figure 7.16

Electron probability in the ground-state H atom.

slide52

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 mlvalues and therefore 9 orbitals with n = 3.

slide53

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

sublevel name

possible ml values

# of orbitals

(a)

2

3d

-2, -1, 0, 1, 2

3

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

slide56

The 3dorbitals.

Figure 7.20

slide57

The 3dorbitals

along z-axis

on xy-plane,

alongxy-axis

on zx-plane,

betw. zx-axis

on xy-plane,

betw. xy-axis

on yz-plane,

betw. yz-axis