GQI-00042 & GQI-00048 Aula 02
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QUÍMICA GERAL
Escola de Engenharia Industrial Metalúrgica Universidade Federal Fluminense
Volta Redonda - RJ
Prof. Dr. Ednilsom Orestes 25/04/2016 – 06/08/2016 AULA 02
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ESTRUTURA ATÔMICA
&
PROPRIEDADES
PERIÓDICAS
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Por que entender
Teoria Quântica?
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Por que entender
Teoria Quântica?
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Por que entender
Teoria Quântica?
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Por que entender
Teoria Quântica?
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Por que entender
Teoria Quântica?
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Por que entender
Teoria Quântica?
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Por que entender
Teoria Quântica?
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Por que entender
Teoria Quântica?
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1987, Joseph John Thomson descobre o elétron em (presente em todos os materiais)
e estabelece sua relação 𝒒/𝒎.
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Tubo de Raios Catódicos
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Tubo de Raios Catódicos
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Tubo de Raios Catódicos
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Tubo de Raios Catódicos
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Tubo de Raios Catódicos
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Tubo de Raios Catódicos
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Robert Millikan estabelece a carga
do elétron em 1898
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Ernest Rutherford
Modelo Atômico Planetário -1911
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Porque o elétron não cai no núcleo?
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Joseph Jacob Balmer Espectro do Hidrogênio - 1875
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Espectro de Emissão
KNO3 CaCl2 NaCl
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Reveillon 2014
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𝑐
𝜆= 𝜈 = ℛ
1
𝑛12 −
1
𝑛22 𝑛1 = 1,2, … 𝑛2 = 𝑛1 + 1, 𝑛1 + 2, …
Calcule o comprimento de onda da radiação emitida por um átomo para n1=2 e n2=3. Dado: cte de Rydberg = 3,29x1015 Hz. Resp.: 657 nm Repita para n1=2 e n2=4 (Resp.: 486 nm) e para n1=2 e n2=5.
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1900: Catástrofe do Ultravioleta.
Max Planck: TEORIA QUÂNTICA
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Wien: Correto em altas frequências.
Rayleigh: Correto em baixas frequências.
1900: Emissão/absorção de energia dá-se em quanta.
Contraria Mec. Clássica.
𝐸 = 𝑛ℎ𝜈 E=nhν com n = 1,2,3,...
h = 6,626x10-34 J.s
ν = freqüência.
Calcule a energia (n=1) para uma radiação com ν = 6,4x1014 Hz.
Repita para n=1 mol.
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Albert Einstein
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Esperado
• Intensidade da radiação é proporcional com máxima Ke.
• Efeito fotoelétrico ocorre para qualquer freq. e compr. onda.
• Deveria haver um delay entre radiação e emissão dos elétrons.
Observado
• Intensidade luz não altera a Ke máxima dos elétrons.
• Frequência de “corte” para fotoemissão.
• Não há delay significativo entre radiação e emissão.
1905: Efeito fotoelétrico
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A. Einstein
Nasce o FÓTON!
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𝐸𝐾 = ℎ𝜈 − Φ 𝑚𝑒𝑣2
2= ℎ 𝜈 − 𝜈0
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𝐸𝐾 = ℎ𝜈 − Φ 𝑚𝑒𝑣2
2= ℎ 𝜈 − 𝜈0
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A velocidade de um elétron emitido pela superfície de uma amostra de potássio pela ação de um fóton é 668 km.s-1. (a) Qual é a energia cinética do elétron ejetado? (b) Qual é o comprimento de onda da radiação que provocou a fotoejeção do elétron no item anterior? (c) Qual é o comprimento de onda mais longo de radiação eletromagnética capaz de ejetar elétrons do potássio? A função trabalho do potássio é 2,29 eV. (1 eV = 1,602 x 10-19 J)
𝐸𝐾 =𝑚𝑒𝑣2
2= 2,03 × 10−19 J
𝑚𝑒𝑣2
2= ℎ𝜈 − Φ
ℎ𝜈 = 5,70 × 10−19 J ∴ 𝜆 = 3,49 × 10−7 m
= 349 nm
0 = ℎ𝜈 − Φ
𝜆 =𝑐ℎ
Φ
∴ 𝜆 = 5,42 × 10−7 m = 542 nm
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+
Modelo Rutherford: elegante e instável.
- -
Niels Bohr
-
Série do
Hidrgênio
H, He+, Li2+, ...
Explicava propriedades física e químicas.
Órbitas explicavam propriedades na TP.
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2
0
2
4 r
ZeFe
r
mFc
2v
-
r +
v
1. Elétron possui órbita circular em torno do núcleo.
2. Momento angular quantizado.
3. Absorve/emite energia quando muda de estado. ce FF
2
42
v
0
22
UE
r
ZemE
UTE
TE
TU
T
2
2
0
2
2
0
22
4
4v
r
ZeF
mr
rZe
c
-
+
2v
hnmrprL
r
p
r
Zem
0
22
4v
.
.
00
2
22
a42
2
n
eZE
UE
n
2
0
22
Zme
hnr
FF
n
ce
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hEE nn 12
Substituindo En tem-se:
2
2
2
100
22
00
2
1
22
00
2
2
22
11
a8
a42a42
nn
eZh
n
eZ
n
eZh
Fazendo ν = c/λ chegamos a Equação proposta por Balmer.
2
2
2
1
11
nnRH
00
22
a8hc
eZRH
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hEE nn 12
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