The Structure of the Atom

Transcription

CHAPTER 5
The
Structure
of the Atom
5.4 Light and
Spectroscopy
460 – 370 BC
1808
Democritus
Atomism
Dalton
“Modern”
atomic theory
2
1870
Crookes
Cathode rays
1897 1910 1925
Thomson
Discovery
of the electron
Rutherford
Discovery
of the nucleus
Today
Pauli
Pauli exclusion
principle
5.4 Light and Spectroscopy
460 – 370 BC
1808
Democritus
Atomism
Dalton
“Modern”
atomic theory
3
1870
Crookes
Cathode rays
1897 1910 1925
Thomson Rutherford
Discovery Discovery
of the electron of the nucleus
Today
Pauli
Pauli exclusion
principle
460 – 370 BC
1808
Democritus
Atomism
Dalton
“Modern”
atomic theory
4
1870
Crookes
Cathode rays
1897 1910 1925
Thomson
Discovery
of the electron
Rutherford
Discovery
of the nucleus
Today
Pauli
Pauli
exclusion
principle
460 – 370 BC
1808
Democritus
Atomism
Dalton
“Modern”
atomic theory
1870
Crookes
Cathode rays
Today
1897 1910 1925
Thomson
Discovery
of the electron
Rutherford
Discovery
of the nucleus
Pauli
Pauli
exclusion
principle
Do we have
evidence to
support these
claims?
5
Light is a form of
electromagnetic energy that
comes from electrons in atoms
The human eye can only detect
a certain range of that energy:
the visible spectrum.
6
Light is a form of
electromagnetic energy that
comes from electrons in atoms
The human eye can only detect
a certain range of that energy:
the visible spectrum.
7
Analyzing starlight with a prism
(one of the first spectrometers)
White light from a lamp or the sun is not truly white!
8
Visible light is only a small range in the electromagnetic spectrum
9
We are surrounded by electromagnetic energy
10
Energy of a photon
Remember that light travels as bundles called photons
A very small
unit of energy
1 electron volt (eV) = 1.602 x 10–19 J.
11
Wavelength
and frequency
are related
12
The wavelength of red laser light is 652 nm.
What is its frequency?
How much energy does a photon of this light have in electron volts?
13
Asked:
Frequency and energy
λ = 652 × 10 −9 m
Relationships: c = λν , E = hν
Given:
14
Asked:
λ = 652 × 10 −9 m
Given:
Solve:
3 × 108 m / s 4.6 × 1014
c = λν therefore ν = =
=
−9
s
λ 652 × 10 m
c
(
)(
)
E = hν = 4.136 × 10 −15 eV ⋅ s 4.6 × 1014 / s = 1.9 eV
15
Asked:
λ = 652 × 10 −9 m
Given:
Solve:
3 × 108 m / s 4.6 × 1014
=
−9
s
λ 652 × 10 m
c
(
)(
)
E = hν = 4.136 × 10 −15 eV ⋅ s 4.6 × 1014 / s = 1.9 eV
16
Asked:
λ = 652 × 10 −9 m
Given:
Solve:
3 × 108 m / s 4.6 × 1014
=
−9
s
λ 652 × 10 m
c
(
)(
)
E = hν = 4.136 × 10 −15 eV ⋅ s 4.6 × 1014 / s = 1.9 eV
Answer:
17
Since 1 Hz = 1/s, the frequency is 4.6 x 1014 Hz and the
energy is 1.9 eV.
Light from an incandescent light bulb:
prism
all possible
energy levels
electron
18
Light from pure hydrogen:
prism
fixed energy
levels
electron
19
Hydrogen atoms can only absorb and emit light
of very specific energies.
20
Matter and light
Why does the atom absorb only
specific (discrete) energies?
21
Matter and light
Why does the atom absorb only
specific (discrete) energies?
Remember:
only some energy levels
are allowed.
22
Matter and light
Energy levels
Energy levels
Photon
(energy)
23
Energy of the photon matches
a gap between levels
Energy of the photon does not
match a gap between levels
Energy (light) is absorbed.
Energy (light) passes
through the atom.
Matter and light
Energy levels
Photon
(energy)
another photon is emitted
specific color
(wavelength)
Energy (light) is absorbed.
24
Each type of atom has a different electron structure.
Each element has unique energy levels like a fingerprint.
25
Spectrum cards
How to
read the
spectrum
cards
26
Spectrum cards
Combinations of elements contain spectral lines from both.
27
Energy levels
Photon
(energy)
Reemission of light has two steps:
Photon absorbed
Photon emitted
28
Range of energies
Emission spectrum
Energy levels
Photon
(energy)
29
Absorption spectrum
Visible light is only a small range of
the electromagnetic spectrum.
30
Each type of atom has a different electron structure.
Each element has unique energy levels like a fingerprint.
31

The Structure of the Atom

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