Transcript Document

Chapter 4: Atomic Structure
The Nuclear Atom
The Atom as the smallest division of an element
quantization of electric charge
oil drop experiments q = ne
e/m => mass of electrons
neutral atoms as “natural” state
“Plum Pudding” model
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-
-
-
BUT.....
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Rutherford scattering (alpha particles from heavy nuclei)
= test of “plum pudding” model
a alpha particles emitted in some radioactive decays
speeds ~ 2E7 m/s
q = +2e, m ~ 8000 x me (a is a He4 nucleus)
light flash
alpha source
thin foil
lead collimator
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Expected (from plum pudding): small scattering angles, no back
scattering
-
-
-
-
Results: some larger scattering angles, including some back
scattering
The Nuclear Atom:
small heavy nucleus (99.8% of atom’s mass) with positive
electric charge
~ 1/100,000 radius of atom
electron “cloud” => electrons orbit nucleus
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Rutherford Scattering (theoretical results):
N( )
ntZ2 e 4

Ni
(8 0 ) 2 r 2 KE2 sin 4 ( /2)
N( )
 fraction of incident particles scattered at
Ni

n  number of atoms per volume in foil
Z  Atomic number (number of protons in nucleus)
r  distance from foil to screen
KE  initial KE of alpha particles
t = foil thickness
45
90
135
180
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Rutherford’s ingredients:
Newtonian Mechanics (F = ma)
1 q1 q2
1 q1 q2
F
;PE 
Coulomb Interaction
2
4 0 r
4 0 r
=> Distance of closest approach
Conservation of Energy
Ei (a at large distance )  E f (a at turn around )
KE 0  0  PE
1 Ze2e
1 2Ze2
KE 
 R
4 0 R
4 0 KE
Example: The maximum KE of alpha particles from natural sources is 7.7 MeV. What is
the distance of closest approach for a gold nucleus? (ZAu = 79)
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Electron Orbits: planetary models of the atom
for the purposes of this discussion, take electron orbits to be circular
Hydrogen: single electron atom
1 e2
1 e2
FE 
; PE  2
4 0 r
4 0 r
2
mv 2
1
e
Fc 
 FE  12 mv 2  12
 KE  - 12 PE
r
4 0 r
also v 
e
4 0 mr
2
1
e
E  KE PE  - 12 PE PE  12 PE  8 0 r
Example 4.1: The ionization energy of Hydrogen is 13.6 eV (the energy required to
liberate the electron from the atom). Find the orbital radius and speed of the electron in a
hydrogen atom.
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Problems with the nuclear atom:
accelerating charges radiate
orbits cannot be stable!!
considerable problems with atomic spectra
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gas discharge tube
Atomic Spectra
emission line spectra (from thin, hot gas or vapor)
spectrum tube contains rarified gas or vapor through
which a high voltage is discharged
prism
collimating slit
screen or film
typical emission spectra
Hydrogen
Helium
emission spectra
vs.
absorption spectra
Mercury
700nm
400nm
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Hydrogen spectral series: patterns in the spectra
10
Balmer
100
 1 1
 R 2 - 2 
n
2 n 
1
1000
n  3,4,5,
10000
(visible light)
R  0.01097nm -1
Lyman
Paschen
Brackett
Pfund
1
n
1
n
1
n
1
n




1 1
R 2 - 2  n  2,3,4,
1 n 
 1 1
R 2 - 2  n  4,5,6,
3 n 
1
1

R 2 - 2  n  5,6,7,
4 n 
 1 1
R 2 - 2  n  6,7,8,
5 n 
(UV)
(IR)
(IR)
(IR)
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Bohr Atom
electron in orbit about nucleus
atomic size ~ electron orbit radius (or see example 4.1)
= 0.053 nm
compare de Broglie wavelength with radius

v
h
mv
e
4 0 m r
(from" planetaryorbit")
h 4 0 r

e
m
withr  0.053nm,  0.33nm  2 r
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h
n   n  2rn  n 
pn
...
Bohr’s original hypothesis: quantize angular momentum of
circular orbits
h
L  m vr Ln  n
 n
2
 h 
nh
  2rn  m vn rn 
n   n  n  
2
 m vn 
Bohr’s hypothesis justified by de Broglie wave theory
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Energy in the Bohr Atom
n   n  2rn
free electron
n=
...
E > 0eV
E = 0eV
...
h 4 0 rn 2rn
n 

e
m
n
n 2h 20
2
 rn 

n
a0
2
m e
h 20
a 0  r1 
 0.05292nm
2
m e
e2
m e4  1 
En  2 2  2 
8 0 rn
8 0 h  n 
E1
E n  2 , E1  -2.18  10-18 J  -13.6eV
n
n=3
E = -3.40eV
n=2
E = -13.6eV
n=1
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Origin of Line Spectra
Discrete Energy levels + conservation of energy + photons
hc
Ei  E f  h  E f 

1


Ei 
Ei - E f
hc
E1
ni
2
(any atom )
, Ef 
E1
nf
2
-E1  1
1

 2 - 2

hc  n f
ni 
1
-E1 1


hc n 2
f
1
(hydrogen )

-E1
R
!
hc
series limit ( ni  )
n f =1 -> Lyman, n f =2 -> Balmer,
n f =3 -> Paschen, etc.
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Example 4.2: An electron collides with a hydrogen atom in its ground state(lowest
energy) and excites it to a state of n = 3. How much energy was given to the hydrogen
atom in this inelastic collision?
Example 4.3: Hydrogen atoms in state of high quantum number have been created in the
laboratory. (a) Find the quantum number of the Bohr orbit in a hydrogen atom whose
radius is 0.0100mm. (b) What is the energy of a hydrogen atom in this state?
Example 4.4: Find the longest wavelength present in the Balmer series of hydrogen
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The Correspondence Principle
A new theory should encompass an old theory where the old
theory was successful.
Quantum theory approximates the results of classical
mechanics when:
quantum numbers are large
h -> 0
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Classical treatment of radiation from “planetary” hydrogen:
frequency of emitted light = frequency of orbits (+ harmonics)
v

f 

2r
e

f 
 
e
3
2

4

m
r

v
0
4 0 m r 
n 2h 20
m e4  2  - E1  2 
withrn 
 f 

 3
2
2 3  3 
h n 
m e
8 0 h  n 
Quantum transition from n -> n- p with p << n

- E1 2np - p 2
1
1 
h  - E1 
- 2  
2
2 2
h
(
n
p
)
n
(
n
p
)
n


- E1 2 p

 p f
3
h n
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Refining the Bohr Atom
nuclear motion: electron and nucleus orbit each other (each
orbit center of mass).
Two body problem =>
center of mass motion +
relative motion (with reduced mass)
mM
m' 
mM
m' e 4
E'n  2
8 0 h 2
 1  m'  E1 
 2  2
n  m n 
m'
hydrogen:
 0.99945
m
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Example 4.6: A “positronium” atom consists of an electron and a positron. Compare the
spectrum of positronium to that of hydrogen
Example 4.7: Muons are elementary particles with mass 207me and +/-e of charge. A
muonic atom is formed by a negative muon with a proton. Find the radius of the first
Bohr orbit and the ionization energy of the atom.
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Atomic spectra
Atoms have discrete set of allowed energies
ALL changes in atom’s energy have to be to an allowed state
Absorption and emission spectra
from conservation of energy
h
DE
Franck-Hertz Experiment: inelastic
scattering of electrons by atoms
->atom only absorbs energy to DE = e DV
A
DV
DV
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The Laser: bright, monochromatic, coherent light source
Excited State: state above ground state
decays to lower states, with emission of photon (or other
mechanism for energy transfer).
Metastable State: “sort of stable” state
state with a longer life time than ordinary excited states
lifetime ~ 1E-3 s vs. 1E-8 s for ordinary states
Three kinds of transitions
h
DE
h
h
h
h
Induced
Absorption
Spontaneous
Emission
Induced
Emission
(Stimulated)
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fast emission to
metastable state
pumping
process
h
h’
h
laser transition:
stimulated emission
Energy levels for 4-level laser
h
Light
Amplification by
Stimulated
Emission of
Radiation
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Other considerations:
“recycling” inducing photons and selecting lasing transition:
the laser cavity
Fabret-Perot Interferometer = standing waves
“Tunable” Dye Lasers
Semiconductor Lasers
Chemical Lasers
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Chapter 4 exercises:3,4,5,6,7,8,11,12,13,14,15,16,18,19,21,22,29,30,31,32,33,35
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