(cap. 4) introdução à mecânica quântica

Introdução à Mecânica Quântica

Niels Bohr e Max Planck (pai da teoria quântica)

1Monday, March 19, 12

Produced with a Trial Version of PDF Annotator - www.PDFAnnotator.com

• Por que alguns átomos se unem para formar moléculas e outros não?

• Por que as propriedades químicas e físicas dos elementos são diferentes?

• Por que ocorre uma repetição periódicas destas propriedades conforme visto na tabela periódica?


1. Primeira Energia de Ionização (I)

I1

“A” é um átomo neutro no estado gasoso

...I3 > I2 > I1

I2


!!! " !

!!! " !

!!!!!!!!!!!!B+(1s22s2) !!! "! ___________ + e–

!E = IE2 = -E2s for B+

IE2 " second ionization energy. IE2 is always higher than the first IE.

!!!!!!!!!!!!B+2(1s22s1) !!! "! ___________ + e–

!E = IE3 = -E2s for B+2

IE3 " third ionization energy

Consider the energy required to remove electrons from the 2s orbital from B+ and B: B+(1s22s2) !!!!!!!!!!!! B+2 (1s22s1) + e–

!E = IE2 = ___________

B(1s22s22p1) !!!!!!!!!!!! B+ (1s22s12p1) + e– !E = IE2s= ___________

Are these two !E’s equal? ________ A 2s-electron in the B+ ion has less shielding.

• It therefore feels a ______________ Zeff and

• requires ___________ energy to be pulled away from the nucleus.

Periodic trends in ionization energy:

Across a row, IE _____________. Z increases, but n (the shell) stays constant. The outermost e- is bound more tightly to the nucleus and requires more E to be ejected.

Down a column, IE _____________. Although Z increases as you go down a column, so does n. Shells are well-separated in space, so electrons in larger n are farther awafrom the nucleus. A large distance from the nucleus dominates over the increased Z,making electrons less strongly bound and therefore decreasing IE.

General trends

IE (

kJ/

mol)

H

He

Li

Be

B

C

N

O

F

Ne

Na

Mg

Al Si

P

S Cl

Ar

Some “glitches” in the trend occur due to subshell structure:

3

for example, IEB < IEBe IEO < IEN

The BE gained by increased Z in B doesn't The BE gained by increased Z in Ocompensate for extra energy required to doesn't compensate for repulsionreach p state, so IE of B lower than for Be. between 2e–s in same state.

B. ELECTRON AFFINITY (EA or Eea) The ability of an atom (or ion) to gain electrons: ______ + e- # ______

Cl + e– Cl– !E = – 349 kJ/mol

energy is released --- ion is __________ stable than atom

Electron affinity, EA, is defined as EA = –!E

So, EA for Cl is EA = ___________ kJ/mol

Unlike IE (which is always positive), EA can be positive or negative.

N + e– N– !E = 7 kJ/mol

So EA = –7 kJ/mole for N --- the -1 ion is _________ stable than atom

General trends in EA:

• Noble gases (group VIII) have ___________ EA because addition of an electron would require occupying a new shell.

• Halogens (group VII) have the largest EA's

because the extra e- fills a “hole” in the p-subshell to give a complete shell.

C. ELECTRONEGATIVITY ($)

Electronegativity is the net ability of an atom to attract an electron from anotheratom. Linus Pauling first proposed this idea in 1932. His electronegativity scales arein general use today.

Mulliken's electronegativity scale developed 2 years later is conceptually easier.

4



2. Energia de Ionização e Periodicidade



Na



Chemical bonding usually involves only the outermost electrons of atoms, also calledvalence electrons. In Lewis dot representations, only the electrons in the outermost occu-pied s and p orbitals are shown as dots. Paired and unpaired electrons are also indicated.Table 7-1 shows Lewis dot formulas for the representative elements. All elements in agiven group have the same outer-shell electron configuration. It is somewhat arbitrary onwhich side of the atom symbol we write the electron dots. We do, however, represent anelectron pair as a pair of dots and an unpaired electron as a single dot.

IONIC BONDING

FORMATION OF IONIC COMPOUNDS

The first kind of chemical bonding we shall describe is ionic bonding. We recall (Section2-3) that an ion is an atom or a group of atoms that carries an electrical charge. An ionin which the atom or group of atoms has fewer electrons than protons is positively charged,and is called a cation; one that has more electrons than protons is negatively charged,and is called an anion. An ion that consists of only one atom is described as a monatomicion. Examples include the chloride ion, Cl!, and the magnesium ion, Mg2". An ion thatcontains more than one atom is called a polyatomic ion. Examples include the ammo-nium ion, NH4

"; the hydroxide ion, OH!; and the sulfate ion, SO42!. The atoms of a

polyatomic ion are held together by covalent bonds. In this section we shall discuss howions can be formed from individual atoms; polyatomic ions will be discussed along withother covalently bonded species.

Ionic bonding is the attraction of oppositely charged ions (cations and anions) inlarge numbers to form a solid. Such a solid compound is called an ionic solid.

7-2

The chemical and physical propertiesof an ion are quite different from thoseof the atom from which the ion isderived. For example, an atom of Naand an Na" ion are quite different.

Lewis Dot Formulas for Representative Elements

Group

Number ofelectrons invalence shell

IA

H

Li

Na

K

Rb

Cs

Fr

1

IIA

Be

Mg

Ca

Sr

Ba

Ra

2

IIIA

B

Al

Ga

In

Tl

3

Period 7

Period 6

Period 5

Period 4

Period 3

Period 2

Period 1

IVA

4

C

Si

Ge

Sn

Pb

VA

5

N

P

As

Sb

Bi

VIA

6

O

S

Se

Te

Po

F

Cl

Br

I

At

VIIA

7

TABLE 7-1TABLE 7-1

VIIIA

8(except

He)

He

Ne

Ar

Kr

Xe

Rn

For the Group A elements, thenumber of valence electrons (dots inthe Lewis formula) for the neutralatom is equal to the group number.Exceptions: H (one valence electron)and He (two valence electrons).

For example, Al has the electronconfiguration [Ar] 3s23p1. The threedots in the Lewis dot formula for Al represent the two s electrons (thepair of dots) and the p electron (thesingle dot) beyond the noble gasconfiguration. Because of the largenumbers of dots, such formulas are notas useful for the transition and innertransition elements.


3. Espectro Eletromagnético

Sir Isaac Newton, one of the giants of science. You probably know of him from his theory ofgravitation. In addition, he made enormous contributions to the understanding of manyother aspects of physics, including the nature and behavior of light, optics, and the laws ofmotion. He is credited with the discoveries of differential calculus and of expansions intoinfinite series.

5-10 Electromagnetic Radiation 195

Figure 5-12 (a) Dispersion of visible light by a prism. Light from a source of white light is passed through a slit and then through a prism. It is spread into a continuous spectrum of all wavelengths of visible light. (b) Visible light is only a very small portion of theelectromagnetic spectrum. Some radiant energy has longer or shorter wavelengths than oureyes can detect. The upper part shows the approximate ranges of the electromagneticspectrum on a logarithmic scale. The lower part shows the visible region on an expandedscale. Note that wavelength increases as frequency decreases.

4000 5000 6000 70004.28 ! 10145.00 ! 10146.00 ! 10147.50 ! 1014

"(Å)v(Hz)

Whitelight

Prism

Increasingwavelength

Gamma rays

X-rays

Ultravioletrays

Infrared rays(heat)

Microwaves

Hertzian waves(radio, TV)

"(Å)

v(Hz)

10–3 10–1 10 103 105 107 109 1011

3 ! 1019 3 ! 1015 3 ! 1011 3 ! 107

Low energyLow frequencyLong wavelength

High energyHigh frequencyShort wavelength

Visible light

(a)

(b)

Alta energiaAlta frequênciaBaixo comprimento de onda

raios Gamma

raios - X

raios UV raios IRcalor

microondas

Rádio, TV

Baixa energiaBaixa frequênciaAlto comprimento de onda

Como converter λ (comprimento de onda) em ν (frequencia)?

Let us use a familiar kind of wave, that on the surface of water (Figure 5-11), to illustratethese terms. The significant feature of wave motion is its repetitive nature. The wave-length, !, is the distance between any two adjacent identical points of the wave, forinstance, two adjacent crests. The frequency is the number of wave crests passing a givenpoint per unit time; it is represented by the symbol " (Greek letter “nu”) and is usuallyexpressed in cycles per second or, more commonly, simply as 1/s or s#1 with “cycles”understood. For a wave that is “traveling” at some speed, the wavelength and the frequencyare related to each other by

!" $ speed of propagation of the wave or !" $ c

Thus, wavelength and frequency are inversely proportional to each other; for the samewave speed, shorter wavelengths correspond to higher frequencies.

For water waves, it is the surface of the water that changes repetitively; for a vibratingviolin string, it is the displacement of any point on the string. Electromagnetic radiationis a form of energy that consists of electric and magnetic fields that vary repetitively. Theelectromagnetic radiation most obvious to us is visible light. It has wavelengths rangingfrom about 4.0 % 10#7 m (violet) to about 7.5 % 10#7 m (red). Expressed in frequencies,this range is about 7.5 % 1014 Hz (violet) to about 4.0 % 1014 Hz (red).

Isaac Newton (1642–1727) first recorded the separation of sunlight into its componentcolors by allowing it to pass through a prism. Because sunlight (white light) contains allwavelengths of visible light, it gives the continuous spectrum observed in a rainbow (Figure5-12a). Visible light represents only a tiny segment of the electromagnetic radiation spec-trum (Figure 5-12b). In addition to all wavelengths of visible light, sunlight also containsshorter wavelength (ultraviolet) radiation as well as longer wavelength (infrared) radia-tion. Neither of these can be detected by the human eye. Both may be detected andrecorded photographically or by detectors designed for that purpose. Many other familiarkinds of radiation are simply electromagnetic radiation of longer or shorter wavelengths.

In a vacuum, the speed of electromagnetic radiation, c, is the same for all wavelengths,2.99792458 % 108 m/s. The relationship between the wavelength and frequency of elec-tromagnetic radiation, with c rounded to three significant figures, is

!" $ c $ 3.00 % 108 m/s

Figure 5-11 Illustrations of thewavelength and frequency of waterwaves. The distance between anytwo identical points, such as crests, isthe wavelength, !. We couldmeasure the frequency, ", of thewave by observing how often thelevel rises and falls at a fixed point inits path—for instance, at the post—or how often crests hit the post. (a)and (b) represent two waves that aretraveling at the same speed. In (a)the wave has long wavelength andlow frequency; in (b) the wave hasshorter wavelength and higherfrequency.

One cycle per second is also called onehertz (Hz), after Heinrich Hertz(1857–1894). In 1887, Hertzdiscovered electromagnetic radiationoutside the visible range and measuredits speed and wavelengths.

The diffraction of white light by theclosely spaced grooves of a compactdisk spreads the light into itscomponent colors. Diffraction isdescribed as the constructive anddestructive interference of lightwaves.

(a)

(b)



Teoria da radiação eletromagnética (1860 - James Clerck Maxwell)







!" $ c $ 3.00 % 108 m/s




(a)

(b)







!" $ c $ 3.00 % 108 m/s




(a)

(b)







!" $ c $ 3.00 % 108 m/s




(a)

(b)

λ - comprimento de ondaν - freqüência (1/s = Hz)


4. Espectro dos átomos

Equação de Rydberg-Balmern = 3, 4, 5...



5. Fótons

Radiação do corpo negro

Se o espectro eletromagnético é contínuo como explicar as linhas espectrais?


Efeito fotoelétrico


Elétronsdifratados

Feixe deElétrons

Cristal de Ni(0)

Energia cinética = ½mv2


Energia cinética de um elétron ejetado

Energia necessária para remover um

elétron

(a) Não há ejeçãoν < ν0

(b) Há ejeçãoν ≥ν0

Funç

ãoTr

abalh

o (ϕ

)


• Frequência (ν) - energia do fóton;

• Intensidade (I) - número de fótons;


νν0(Na)ν0(Rb) ν0(K)

Rb K Na

y = mx + b

coef. angular (m) = 6.626E-34 J.s

coef. linear (b) = (6.626E-34 J.s)ν0

hν0(Rb)

hν0(K)

hν0(Na)

E.C.

Cte de Planck = h = 6.626E-34 J.s17Monday, March 19, 12


• Cada metal apresenta um valor característico de λ (comprimento de onda), abaixo do qual elétrons podem ser ejetados.

• Cada metal apresenta um valor característico de ν (freqüencia), acima da qual elétrons podem ser ejetados.


Znϕ = 6,9E-19 J

Uma lâmpada UV (λ = 254 nm) e um laser (λ = 700 nm) podem provocar a ejeção de elétrons de uma placa de Zn(0)?



1. Qual a energia de um fóton emitido pela lâmpada UV (λ = 254 nm) ?

2. Qual a energia de um fóton emitido pelo laser de um pointer (λ = 700 nm) ?

3. Qual é o número de fótons emitidos pelo laser durante 60 s se a intensidade é de 1 mW?

1 mW = 1E-3 J/s



6. Propriedades ondulatórias da matéria

• Radiação eletromagnética = comportamento ondulatório

• Fótons = comportamento de partícula

Duque Louis Victor de Broglie (1924)21Monday, March 19, 12


• Uma partícula de massa m em uma velocidade v de ter um comprimento de onda associado a ela.

p = h/λ

λ = h/p

λ = h/mv

“De Broglie destacou o primeiro parte do que estava realmente encoberto sobre dualidade onda/partícula” A. Einstein



7. Propriedades dos Elétrons

Difração (‘propriedade de ondas’) de elétrons (‘partículas’)

Davisson & Germer (1925): Prêmio Nobel

J. J. Thonsom - partículas subatômicasG. P. Thonsom - comportamento ondulatório dos elétrons


8. O átomo de Hidrogênio

Niels Bohr (1885-1962)

it emits exactly the same amount of energy it absorbed in moving from the lower to thehigher energy level. Figure 5-16 illustrates these transactions schematically. The values ofn1 and n2 in the Balmer-Rydberg equation identify the lower and higher levels, respec-tively, of these electronic transitions.

The Danish physicist Niels Bohr was one of the most influential scientists of the twentiethcentury. Like many other now-famous physicists of his time, he worked for a time inEngland with J. J. Thomson and later with Ernest Rutherford. During this period, he beganto develop the ideas that led to the publication of his explanation of atomic spectra and histheory of atomic structure, for which he received the Nobel Prize in 1922. After escapingfrom German-occupied Denmark to Sweden in 1943, he helped to arrange the escape ofhundreds of Danish Jews from the Hitler regime. He later went to the United States, where,until 1945, he worked with other scientists at Los Alamos, New Mexico, on thedevelopment of the atomic bomb. From then until his death in 1962, he worked for thedevelopment and use of atomic energy for peaceful purposes.

5-12 Atomic Spectra and the Bohr Atom 201

The Bohr Theory and the Balmer-Rydberg Equation

From mathematical equations describing the orbits for the hydrogen atom, together withthe assumption of quantization of energy, Bohr was able to determine two significant aspectsof each allowed orbit:

1. Where the electron can be with respect to the nucleus—that is, the radius, r, of thecircular orbit. This is given by

r ! n2a0

where n is a positive integer (1, 2, 3, . . .) that tells which orbit is being described anda0 is the Bohr radius. Bohr was able to calculate the value of a0 from a combinationof Planck’s constant, the charge of the electron, and the mass of the electron as

a0 ! 5.292 " 10#11 m ! 0.5292 Å

2. How stable the electron would be in that orbit—that is, its potential energy, E. Thisis given by

E ! # !$8%2hm2

a02$" ! #

where h ! Planck’s constant, m ! the mass of the electron, and the other symbolshave the same meaning as before. E is always negative when the electron is in theatom; E ! 0 when the electron is completely removed from the atom (n ! infinity).

Results of evaluating these equations for some of the possible values of n (1, 2, 3, . . .)are shown in Figure 5-17. The larger the value of n, the farther from the nucleus is theorbit being described, and the radius of this orbit increases as the square of n increases. Asn increases, n2 increases, 1/n2 decreases, and thus the electronic energy increases (becomesless negative and smaller in magnitude). For orbits farther from the nucleus, the electronicpotential energy is higher (less negative—the electron is in a higher energy level or in a less

2.180 " 10#18 J$$

n21

$n2

E n r i c h m e n t

Note: r is proportional to n2.

Note: E is proportional to #$n12$.

We define the potential energy of a setof charged particles to be zero whenthe particles are infinitely far apart.

Comportamento ondulatório proposto por de Broglie

Comportamento ondulatório proposto por Bohr

n = 1, 2, 3..


Figure 5-17 (a) The energy levels that the electron can occupy in a hydrogen atom anda few of the transitions that cause the emission spectrum of hydrogen. The numbers onthe vertical lines show the wavelengths of light emitted when the electron falls to a lowerenergy level. (Light of the same wavelength is absorbed when the electron is promoted tothe higher energy level.) The difference in energy between two given levels is exactly thesame for all hydrogen atoms, so it corresponds to a specific wavelength and to a specificline in the emission spectrum of hydrogen. In a given sample, some hydrogen atomscould have their electrons excited to the n ! 2 level. Some of these electrons could thenfall to the n ! 1 energy level, giving off the difference in energy in the form of light (the1216-Å transition). Other hydrogen atoms might have their electrons excited to the n ! 3level; subsequently some could fall to the n ! 1 level (the 1026-Å transition). Becausehigher energy levels become closer and closer in energy, differences in energy betweensuccessive transitions become smaller and smaller. The corresponding lines in theemission spectrum become closer together and eventually result in a continuum, a seriesof lines so close together that they are indistinguishable. (b) The emission spectrum ofhydrogen. The series of lines produced by the electron falling to the n ! 1 level is knownas the Lyman series; it is in the ultraviolet region. A transition in which the electron fallsto the n ! 2 level gives rise to a similar set of lines in the visible region of the spectrum,known as the Balmer series. Not shown are series involving transitions to energy levelswith higher values of n. (c) The Balmer series shown on an expanded scale. The line at6563 Å (the n ! 3 n n ! 2 transition) is much more intense than the line at 4861 Å (then ! 4 n n ! 2 transition) because the first transition occurs much more frequently thanthe second. Successive lines in the spectrum become less intense as the series limit isapproached because the transitions that correspond to these lines are less probable.

202 CHAPTER 5: The Structure of Atoms

n!543

2

1

Groundstate

6563

A°

4861

A°

4340

A°

1216

A°

1026

A°

973

A°

950

A°

°

Balmerseries

Balmer series

Exc

ited

stat

es

Lyman series

Lyman series

red Visible region violet Ultraviolet

Series limit

6563 4861 4340 4102 3646 "(A)

"(A)

°(a)

(b)

(c)

5000 2500 2000 1500 1250 1000

Incr

easi

ng e

nerg

y(Enrichment, continued)

(c)

(b)

(a)

9. Transições Eletrônicas



Teste de chama

EXAMPLE 5-6 Energy of LightA green line of wavelength 4.86 ! 10"7 m is observed in the emission spectrum of hydrogen.Calculate the energy of one photon of this green light.

PlanWe know the wavelength of the light, and we calculate its frequency so that we can then calcu-late the energy of each photon.

Solution

E # # # 4.09 ! 10"19 J/photon

To gain a better appreciation of the amount of energy involved, let’s calculate the total energy,in kilojoules, emitted by one mole of atoms. (Each atom emits one photon.)

# 4.09 ! 10"19 ! ! # 2.46 ! 102 kJ/mol

This calculation shows that when each atom in one mole of hydrogen atoms emits light ofwavelength 4.86 ! 10"7 m, the mole of atoms loses 246 kJ of energy as green light. (Thiswould be enough energy to operate a 100-watt light bulb for more than 40 minutes.)

You should now work Exercises 40 and 42.

6.02 ! 1023 atoms$$$

mol1 kJ

$$1 ! 103 J

J$atom

_?_ kJ$mol

(6.626 ! 10"34 J %s)(3.00 ! 108 m/s)$$$$

(4.86 ! 10"7 m)hc$&

Figure 5-15 Atomic spectra in the visible region for some elements. Figure 5-14a showshow such spectra are produced. (a) Emission spectra for some elements. (b) Absorptionspectrum for hydrogen. Compare the positions of these lines with those in the emissionspectrum for H in (a).

4000 5000 6000 7000!(Å)

(b)

H

4000 5000 6000 7000

Ne

Hg

H

(a)

!(Å)

EXAMPLE 5-6 Energy of LightA green line of wavelength 4.86 ! 10"7 m is observed in the emission spectrum of hydrogen.Calculate the energy of one photon of this green light.

PlanWe know the wavelength of the light, and we calculate its frequency so that we can then calcu-late the energy of each photon.

Solution

E # # # 4.09 ! 10"19 J/photon

To gain a better appreciation of the amount of energy involved, let’s calculate the total energy,in kilojoules, emitted by one mole of atoms. (Each atom emits one photon.)

# 4.09 ! 10"19 ! ! # 2.46 ! 102 kJ/mol

This calculation shows that when each atom in one mole of hydrogen atoms emits light ofwavelength 4.86 ! 10"7 m, the mole of atoms loses 246 kJ of energy as green light. (Thiswould be enough energy to operate a 100-watt light bulb for more than 40 minutes.)

You should now work Exercises 40 and 42.

6.02 ! 1023 atoms$$$

mol1 kJ

$$1 ! 103 J

J$atom

_?_ kJ$mol

(6.626 ! 10"34 J %s)(3.00 ! 108 m/s)$$$$

(4.86 ! 10"7 m)hc$&

Figure 5-15 Atomic spectra in the visible region for some elements. Figure 5-14a showshow such spectra are produced. (a) Emission spectra for some elements. (b) Absorptionspectrum for hydrogen. Compare the positions of these lines with those in the emissionspectrum for H in (a).

4000 5000 6000 7000!(Å)

(b)

H

4000 5000 6000 7000

Ne

Hg

H

(a)

!(Å)

Espectro de emissão

Espectro de absorção


(cap. 4) introdução à mecânica quântica

Documents

greek letter

comprimento

adjacent identical

wave crests

14a shows

visible region

trial version

inversely