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Chapter 7

The Electronic Structure of Atoms

Outline1.From classical physics to Quantum Theory2. The photoelectric effect3. Bohr’s Theory of the Hydrogen Atom4.The Dual Nature of the electron5.Quantum Mechanism 6. Quantum number7.Atomic Orbitals8.Electron Configuration 9. The Building-up Principle

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Properties of Waves

Wavelength () is the distance between identical points on successive waves.

Amplitude is the vertical distance from the midline of a wave to the peak or trough.

Frequency () is the number of waves that pass through a particular point in 1 second (Hz = 1 cycle/s).

The speed (u) of the wave = x

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

Electromagnetic radiation is the emission and transmission of energy in the form of electromagnetic waves.

Speed of light (c) in vacuum = 3.00 x 108 m/s

All electromagnetic radiation x c

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The type of electromagnetic

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x = c = c/ = 3.00 x 108 m/s / 6.0 x 104 Hz = 5.0 x 103 m

A photon has a frequency of 6.0 x 104 Hz. Convert this frequency into wavelength (nm). Does this frequency fall in the visible region?

= 5.0 x 1012 nm

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Plank’s Quantum Theory

1. Planck's law, Blackbody radiation law ）是用於描述在任意溫度下，從一個黑體中發射的電磁輻射的輻射率與電磁輻射的頻率的關係公式。

2. 物理學家在研究輻射密度與頻率的關係 : 發現在長波長與短波長範圍與實驗觀測不同 ??

3. 古典物理認為能量是連續的 , 物質可以吸收或是發射任意大小的能量。4. Plank 提出假設，認為只吸收或發射特定能量包，稱為量子

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Mystery #1, “Heated Solids Problem”Solved by Planck in 1900

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

E = h x Planck’s constant (h)h = 6.63 x 10-34 J•s

When solids are heated, they emit electromagnetic radiation over a wide range of wavelengths.

Radiant energy emitted by an object at a certain temperature depends on its wavelength.

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Light has both:1. wave nature2. particle nature

h = KE + W

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

Photon is a “particle” of light

KE = h - W

h

KE e-

where W is the work function anddepends how strongly electrons are held in the metal

光子能量 (photon)

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Element eV Element eV Element eV Element eV Element eV Element eV

Ag 4.26 Al 4.28 As 3.75 Au 5.1 B 4.45 Ba 2.7

Be 4.98 Bi 4.22 C 5 Ca 2.87 Cd 4.22 Ce 2.9

Co 5 Cr 4.5 Cs 2.14 Cu 4.65 Eu 2.5 Fe 4.5

Ga 4.2 Gd 3.1 Hf 3.9 Hg 4.49 In 4.12 Ir 5.27

K 2.3 La 3.5 Li 2.9 Lu 3.3 Mg 3.66 Mn 4.1

Mo 4.6 Na 2.75 Nb 4.3 Nd 3.2 Ni 5.15 Os 4.83

Pb 4.25 Pt 5.65 Rb 2.16 Re 4.96 Rh 4.98 Ru 4.71

Sb 4.55 Sc 3.5 Se 5.9 Si 4.85 Sm 2.7 Sn 4.42

Sr 2.59 Ta 4.25 Tb 3 Te 4.95 Th 3.4 Ti 4.33

Tl 3.84 U 3.63 V 4.3 W 4.55 Y 3.1 Zn 4.33

Zr 4.05                    

The work function of metals

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E = h x

E = 6.63 x 10-34 (J•s) x 3.00 x 10 8 (m/s) / 0.154 x 10-9 (m)

E = 1.29 x 10 -15 J

E = h x c /

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.

EX:

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Bohr’s Model of the H Atom

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Line Emission Spectrum of Hydrogen Atoms

Line spectra(非連續光譜 , why?)

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Sodium atoms in an excited state

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showing oxygen, helium, and hydrogen ions that gush into space from regions near the Earth's poles.

Plasma: Plasma is loosely described as an electrically neutral medium of positive and negative particles (i.e. the overall charge of a plasma is roughly zero).

O2-plasma

Ar plasma

Nature Artificial

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How we using “plasma” in semiconductor fabrication?

• Etching • Deposition thin film

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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

Bohr’s Model of the Atom (1913)

En = -RH ( )1n2

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

RH (Rydberg constant) = 2.18 x 10-18J

Ground state

excited state

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E = h

E = h

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Ephoton = E = Ef - Ei

Ef = -RH ( )1n2

f

Ei = -RH ( )1n2

i

i fE = RH( )

1n2

1n2

nf = 1

ni = 2

nf = 1

ni = 3

nf = 2

ni = 3

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Ephoton = 2.18 x 10-18 J x (1/25 - 1/9)

Ephoton = E = -1.55 x 10-19 J

= 6.63 x 10-34 (J•s) x 3.00 x 108 (m/s)/1.55 x 10-19J

= 1280 nm

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 = h x c /

= h x c / Ephoton

i fE = RH( )

1n2

1n2

Ephoton =

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The Dual Nature of the Electron 電子的二元性

Physicists were intrigued by Bohr’s Theory:1.Why hydrogen electron are quantized? 2.Why electron restricted to orbiting the nucleus at certain fixed distance?

In 1924, de Broglie answer this: If the light waves can behave like a stream of particles(photons), then perhaps particles such as electrons can possess wave properties.

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

Why is e- energy quantized?

u = velocity of e-

m = mass of e-

2r = n = hmu

node

Standing wave

If the electron does behave like a wave in atom,the length of wave must fit the circumference of orbital exactly!Otherwise, it will not exist~

( )hp

物質波波長

For electron, this wavelength is ~ 72 nm

光子動量

適用於所有物質與粒子

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Electrons can behave like X rays and display “wave properties”

X-ray diffraction pattern of Al-foil

electron diffraction pattern of Al-foil

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= h/mu

= 6.63 x 10-34 / (2.5 x 10-3 x 15.6)

= 1.7 x 10-32 m = 1.7 x 10-23 nm

What is the de Broglie wavelength (in nm) associated with a 2.5 g Ping-Pong ball traveling at 15.6 m/s?

m in kgh in J•s u in (m/s)

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Quantum Mechanics

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Schrodinger Wave EquationIn 1926 Schrodinger wrote an equation that described both the particle and wave nature of the e-

Wave function () describes:

1. energy of e- with a given

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.

基於德布羅伊的波動 - 粒子雙重性原理，為薛丁格方提出一種波動方程式 : 可描述粒子在空間中的波動行為。

在應用上面 , 電子在晶體的運動行為可用波函數的描述。

1. 無法準確量測能量 E 與具有此能量的時間瞬間 t2. 無法決定電子的確切位置，發展一個機率密度函數 (probability density function) ，來決定特定位置找到一個電子的機率。

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Where 90% of thee- density is foundfor the 1s orbital

Electron density

Boundary surface diagram ( 軌域輪廓 )

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Schrodinger Wave Equation is a function of four numbers called quantum numbers (n, l, ml, ms)

principal quantum number n

n = 1, 2, 3, 4, ….

n=1 n=2 n=3

distance of e- from the nucleus

主量子數量子數可以用來描述原子軌域及標示不同電子

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quantum numbers: (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 = 0n = 2, l = 0 or 1

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

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

l = 0 s orbitall = 1 p orbitall = 2 d orbitall = 3 f orbital

Schrodinger Wave Equation

角動量量子數

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l = 0 (s orbitals)

l = 1 (p orbitals)

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l = 2 (d orbitals)

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quantum numbers: (n, l, ml, ms)

magnetic quantum number ml

for a given value of lml = -l, …., 0, …. +l

orientation of the orbital in space

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

Schrodinger Wave Equation

磁量子數

軌域在空間中的方向

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ml = -1, 0, or 1 3 orientations is space

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ml = -2, -1, 0, 1, or 2 5 orientations is space

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(n, l, ml, ms)

spin quantum number ms

ms = +½ or -½

Schrodinger Wave Equation

ms = -½ms = +½

自旋量子數

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Existence (and energy) of electron in atom is described by its unique wave function .

Pauli exclusion principle - no two electrons in an atomcan have the same four quantum numbers.

Schrodinger Wave Equation

quantum numbers: (n, l, ml, ms)

Each seat is uniquely identified (E, R12, S8)Each seat can hold only one individual at a time

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Schrodinger Wave Equationquantum numbers: (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

= (n, l, ml, ½)or= (n, l, ml, -½)

An orbital can hold 2 electrons

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How many 2p orbitals are there in an atom?

2p

n=2

l = 1

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

3 orbitals

How many electrons can be placed in the 3d subshell?

3d

n=3

l = 2

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

5 orbitals which can hold a total of 10 e-

EX:

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Energy of orbitals in a single electron atom

Energy only depends on principal quantum number n

En = -RH ( )1n2

n=1

n=2

n=3

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Energy of orbitals in a multi-electron atom

Energy depends on n and l

n=1 l = 0

n=2 l = 0n=2 l = 1

n=3 l = 0n=3 l = 1

n=3 l = 2

需要考慮到多電子間的能量排斥

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2s 軌域的 density 比 2p 高 ,軌域的穿透力高，因此被 1s電子屏蔽的效應小。

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Order of orbitals (filling) in multi-electron atom

1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s

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Electron configuration is how the electrons are distributed among the various atomic orbitals in an atom.

1s1

principal quantumnumber n

angular momentumquantum number l

number of electronsin the orbital or subshell

Orbital diagram

H

1s1

電子在原子的地址

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Paramagnetic

unpaired electrons

2p

Diamagnetic

all electrons paired

2p

順磁 逆磁

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What is the electron configuration of Mg?

Mg 12 electrons

1s < 2s < 2p < 3s < 3p < 4s

1s22s22p63s2 2 + 2 + 6 + 2 = 12 electrons

Abbreviated as [Ne]3s2 [Ne] 1s22s22p6

What are the possible quantum numbers for the last (outermost) electron in Cl?

Cl 17 electrons 1s < 2s < 2p < 3s < 3p < 4s

1s22s22p63s23p5 2 + 2 + 6 + 2 + 5 = 17 electrons

Last electron added to 3p orbital

n = 3 l = 1 ml = -1, 0, or +1 ms = ½ or -½

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“Fill up” electrons in lowest energy orbitals (Aufbau principle)

H 1 electron

H 1s1

He 2 electrons

He 1s2

Li 3 electronsLi 1s22s1

Be 4 electronsBe 1s22s2

B 5 electronsB 1s22s22p1

? ?

構築原理 : 由低至高能階，每個軌域填入兩個電子

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The most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins (Hund’s rule).

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Outermost subshell being filled with electrons

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HW 77.7 (a) What is the wavelength (in nanometers) of light having a frequency of 8.6 × 1013 Hz? (b) What is the frequency (in Hz) of light having a wavelength of 566 nm?

7.25 Explain why elements produce their own characteristic colors when they emit photons.

7.66 Why do the 3s, 3p, and 3d orbitals have the same energy in a hydrogen atom but different energies in a many-electron atom?

7.82 Indicate the number of unpaired electrons present in each of the following atoms: B, P, Mn, Se, Kr, Fe, I.

5. 基於放射光譜的概念，敘述雷射 (laser)的產生機制與實際應用 ?