Transcript Example 27-1

Chapter 27
Early Quantum Theory and Models of the Atom
© 2006, B.J. Lieb
Some figures electronically reproduced by permission of Pearson Education, Inc.,
Upper Saddle River, New Jersey Giancoli, PHYSICS,6/E © 2004.
Ch 27
1
Properties of the Electron
•Called “cathode ray” because appeared to come from the
cathode
•J. J. Thomson did the first experiments to discover its
properties and received Nobel Prize for work.
•First discovered in experiments where electricity is
discharged through rarefied gases.
•First experiments measured charge /mass.
•Another experiment measured e and thus mass could be
determined
Ch 27
2
Electron Charge to Mass Ratio
Magnetic Force = Centripetal Force
mv 2
ev B 
r
e
v

m
Br
The electric field is adjusted until it balances the magnetic field
and thus eE = evB . This gives E
v 
B
This is a velocity selector since all electrons that pass through
have the same velocity. The final equation is
Ch 27
e
E
 2
m
B r
3
Blackbody Radiation
•Intensity vs. wavelength is shown above
•Radiation emitted by any “hot” object
•In Ch 14, we learned Intensity  T4
•The wavelength of the peak of the spectrum P depends on the Kelvin
temperature by
 p T  2.9 103 m  K
Ch 27
4
Planck’s Quantum Hypothesis
•Attempts to explain the shape of the blackbody curve were
unsuccessful
•In 1900 Max Planck proposed that radiation was emitted in
discrete steps called quanta instead of continuously
•He did not see this as revolutionary
•Introduced a new constant-now called Planck’s constant
h =6.626 x 10-34 Js = 4.14 x 10-15 eV·s
Ch 27
5
Planck’s Quantum Hypothesis
•We can understand
quantized energy by
considering the energy of
a box on stairs vs. box on
a ramp.
E=mgy
•If the height of each step
is Δ y, can you derive an
equation for the quantized
potential energy of the
box on the steps.
E=mg(nΔy)
where n is an integer.
Ch 27
6
Photoelectric Effect
•Light shines on a metal and electrons (called photoelectrons ) are
given off
•Easy to measure kinetic energy of electrons
•There was a threshold frequency below which no electrons were
emitted
Ch 27
7
Photoelectric Effect
•Wave theory predicts that:
•Number of electrons Intensity
•Maximum electron kinetic energy intensity
•Frequency of light should not affect kinetic energy
•No threshold frequency
•This can not explain the photoelectric effect
Ch 27
8
Photon Theory of Light
•In 1905 Einstein proposed that in some experiments light
behaved like particles instead of waves
•Light consisted as stream of photons, each with energy:
E  h f
Where h is Planck’s constant
•Each photon had wavelike properties
Ch 27
9
Explanation of Photoelectric Effect
•Work function W0 is the energy necessary to free the least tightly
bound electron
•A single photon with energy hf gives all of this energy to a single
electron
Ph
•A photon with frequency below the threshold lacks sufficient energy to
free the electron, so hf0 = W0
5
0
0
2
•This electron thus escapes from the metal with kinetic energy
KEmax  hf  W0
Ch 27
10
Photons
The energy of a photon is given by
E  hf
•where h is Planck’s Constant
•The rest mass of a photon must be zero because it travels
at the speed of light
•The photon has momentum (p) because if we substitute
m0 = 0 in
E 2  p 2 c 2  m02 c 4
We get the following equation
hc
E
hf
  h
p 


c
c
c

and thus
Ch 27
p 
h

11
Example 27-1 (20) In a photoelectric effect experiment it is observed that no
current flows unless the wavelength is less than 570 nm. What is the work function
of the material?
K Emax  h f W0
KEmax  0 (threshold )
W0  h f 0  h
c
0
(4.14 10 15 eV  s)(3.00 108 m )
s
W0 
9
570 10 m
W0  2.18 eV
What is the stopping voltage if light of wavelength 400 nm is used?
KEMAX 
(4.14 10 15 eV  s )(3.00 108 m
400 10 9 m
s
)
 2.18 eV  s  0.93 eV
So the stopping voltage is 0.93 volts.
Ch 27
12
Evidence for Photon Nature of Electromagnetic
Radiation
Compton Effect:
photon scatters off of
electron, photon looses
energy and electron
gains energy. This
effect shows that
momentum and energy
is conserved.
Ch 27
13
Further Evidence for Photon Nature of
Electromagnetic Radiation
Pair Production: a
photon passing a nucleus
is converted into an
electron-positron pair. (A
positron is a positive
particle with all the other
properties of an electron.)
Since the mass of the electron is 0.511MeV/c2 and two electrons
must be produced, the kinetic energy shared by the two electrons is
KE  E 1.022 MeV
Ch 27
14
Electron Microscopes
•Electrons accelerated through 100,000 V
have   0.004nm and can achieve a
resolution of
 0.2 nm which is a
factor of 1000x better than optical
microscopes.
•Use magnetic lenses to focus electron
beam.
•Scanning Tunneling Microscope has a
probe that moves up and down to
maintain a constant tunneling current.
Ch 27
15
Thomson Model of the Atom
•J.J. Thomson Model: had
negatively charged
electrons inside a sphere of
positive charge.
•Assumed that the electrons
would oscillate due to
electric forces.
Ch 27
•An oscillating charge
produces electromagnetic
radiation which should
match agree with atomic
16
spectra.
Rutherford Experiment
•Ernest Rutherford performed an experiment to probe
the structure of the atom.
•He aimed a beam of alpha particles at a thin gold foil and
measured how they were scattered.
•Alpha particle is the nucleus of a He atom (two protons
and two neutrons) and thus was positively charged.
•Alpha particles are emitted by a radioactive nucleus.
Ch 27
17
Rutherford Experiment Results
•Most alpha particles passed through foil without
scattering
•A few were scattered through large angles
Ch 27
18
Rutherford Experiment Results
•Concluded that foil was mainly empty space with some
small but massive concentrations of positive charge.
•An alpha particle that happened to pass near a nucleus
was repelled without ever touching the nucleus.
•Rutherford proposed a positive heavy nucleus with
radius of  10-15 m with electrons in orbit
•Problem was electrons should radiate energy away.
Ch 27
19
Hydrogen Spectra
•The visible spectra from hydrogen gas has a very distinctive pattern that
can be represented by the Balmer formula
1 
 1
 R  2  2 ,

n 
2
1
n  3,4,5....
where R = 1.0974x10-7 m-1.
Ch 27
20
Hydrogen Spectra
•The Balmer formula could also be modified to fit the
Lyman series that was discovered in the ultraviolet
1 
 1
 R 2  2 ,

n 
1
1
n  2, 3, 4....
•And the Paschen series in the infrared
1 
 1
 R 2  2 ,

n 
3
1
Ch 27
n  4,5,6....
21
Bohr Model of the Atom
•Niels Bohr was a Danish physicist who studied at the
Rutherford lab. He decided to try to add the quantum effects of
Planck and Einstein to the Rutherford planetary model of the
atom
•The discrete
wavelengths emitted
by hydrogen suggested
a quantum effect as in
the stair example
•He knew that the answer had to be the Balmer formula but
the task was to develop a set of assumptions that would lead
to it.
•Problem was that a charged particle in orbit is like an
antenna--it should emit radiation and gradually loose
energy until it fell into the nucleus
Ch 27
22
Bohr’s Assumptions
•Bohr was like a student who looked
up the answer in the back of the book
and needed to find a way to get that
answer
•He said electron could remain in
possible orbit called a stationary state
without emitting any radiation
•Each stationary state is
characterized by a definite energy En
•When electron changes from the
upper to the lower stationary state (or
orbit) it emits a photon of energy
equal to the difference in the states:
Ch 27
hf  Eu  El
23
Bohr’s Assumptions
•Radiation is only emitted when an
electron changes from one stationary
state to another
•He found he could derive the Balmer
formula if he assumed that the
electrons moved in circular orbits
with angular momentum (L) satisfied
the following quantum condition:
L  mvrn  n
h
,
2
and thus
nh
2 mr n
Ch 27
v
n  1,2,3...
24
Bohr Radius
•Z is the number of protons so Qnucleus = Ze
•The electrical force equals the centripetal force
( Ze)(e) mv2
k

2
rn
rn
v
nh
2 mr n
kZe2 k Ze 2 4 2 mrn2
rn 

2
mv
n2h2
n2
rn  r1
Z
h2
10
r1  2

0
.
529

10
m
2
4 mke
Ch 27
n2
rn  (0.529 10 10 m)
Z
25
Bohr Energy Levels
The electric potential of
the nucleus is
kQ kZe2
V

r
r
With potential energy
Ze 2
PE  eV  k
r
1 2 kZe2
En  mv 
2
rn
Substituting Bohr radius equation and values gives
Z2
En  13.6 eV 2
n
Ch 27
n  1, 2, 3
26
Summary of Bohr Model
•Electrons obit in stationary states that are characterized by a quantum
number n and a discrete energy En. Sometimes this is called a energy
level.
•En is negative indicating a bound electron
Z2
En  13.6 eV 2
n
n  1, 2, 3
•At room temperature, most H atoms have their electron in the n=1
energy level
•When electron changes to a lower n it emits a photon of energy equal
to the energy difference.
•Electron must be given energy to move to a higher n
•This formula can be used for any single electron atom or ion such as
a singly-ionized He ion in which case Z=2.
•The radius of the orbit is given by
Ch 27
n2
rn  (0.529 10 10 m)
Z
27
Energy Level Diagram
•Red arrows indicate transitions where electron emits a photon and
moves to a lower state
•Vertical scale is energy.
Ch 27
28
Example 27-2A. A hydrogen atom initially in its ground state (n=1) absorbs a
photon and ends up in the n=3 state. Calculate the energy and wavelength of the
absorbed photon.
First calculate the energy of the first three states.
 13.6 eV z 2  13.6 V
En 

2
n
n2
E1 
( z 1)
 13.6 eV
 13.6 eV
12
 13.6 eV
E2 
 3.4 eV
2
2
E3 
 13.6 eV
 1.51 eV
32
E  E31  1.51  (13.6 eV )  12.1 eV
E  h f 
Ch 27
hc

E
hc

(6.626 10 34 Js)(3.00 108 m / s)

 102 nm
19
(12.1 eV )(1.6 10 J / eV )
29
Example 27-2B. A hydrogen atom initially in its ground state (n=1)
absorbs a photon and ends up in the n=3 state. Calculate the energy and
wavelength of the absorbed photon.
When the atom returns to the ground state, what possible energy photons
could be emitted?
n 3
n2
n 1
E   1.51eV
E   3.40 eV
E  13.6 eV
 E31  1.51  (13.6 eV )  12.1 eV
E32  1.51 eV  (3.4 eV )  1.89 eV
E21  3.4 eV  (13.6eV )  10.2 eV
Ch 27
30
Example 27-3. Singly ionized 4He consists of a nucleus with two protons
and two neutrons with a single electron in orbit around this nucleus. Use
the Bohr model to calculate the energy of a photon that is emitted when
the electron goes from the first excited state to the ground state of singly
ionized 4He.
 
 13.6 eV z 2  13.6 eV  22
En 

2
n
n2
E1 
 54.4 eV
 54.4 eV
12
E2 
 54.4 eV
 13.6 eV
22
E   E  E2  E1  13.6  (54.4 eV )  40.8 eV
Calculate the ionization energy of singly ionized 4He.
Ionization energy = 0  (54.4 eV )  54.4 eV
Could you use the Bohr model for atomic 4He?
Ch 27
31
Wave Nature of Particles
Louis de Broglie proposed that if light had a wave-particle
duality, then perhaps particles, such as electrons, also had a
wave nature. He assumed that the following equation for
photons
h
p 

also applied to electrons

Ch 27
h
mv
32
Diffraction of Electrons
Later experiments showed that a beam of electrons was
diffracted just like light.

Ch 27
h
mv
33
De Broglie Hypothesis and Hydrogen
De Broglie was able to give a reason
for the Bohr quantum hypothesis by
assuming that allowed electron orbits
had to be in standing waves around
orbit.
Circumference =2 rn = n
where n = 1, 2, 3...
Combine with
h
 
mv
and we get the following equation which was Bohr’s
original quantum assumption.
mvrn 
Ch 27
nh
2
34