Transcript electric oven oven jump

Quantum Physics (or, the Physics of the very small)
Models of the Atom:
Democritus
JJ Thomson’s Plum-Pudding Model (1897):
Experimented with “cathode rays”; found that they
had a negative charge; so charges in the atom can
be separated
Plum-pudding model:
Sent cathode rays into magnetic
fields, where they were
deflected into circular paths
(more on this later)
Ernest Rutherford’s Gold-Foil Experiment (1911):
Alpha particles are
positively charged
Actual results:
< 1%
99%
<< 1%
- most of the atom is empty space
-positive charge is concentrated in
very small, very dense “nucleus”
-negative charge orbits nucleus at a
distance, separating nuclei from
each other
Nuclear radius: ~ 10-14 m
Atomic radius: ~ 10-10 m
Problem with Rutherford’s model:
If charged particles are
accelerated, they emit EM
radiation, and lose energy
Note the decreasing radius
of the particles in the
bubble chamber
Another example: the magnetron
in a microwave oven accelerates
electrons into a circular path,
emitting microwaves
As electrons orbit the nucleus,
they should radiate EM
waves, lose energy, and spiral
into the nucleus
Obviously they do not spiral in, so ????
We’ll get back to this later
JJ Thomson sent cathode rays into a region where there
was both an electric field and magnetic field
+
V
e-
+
+
+
+
+
x
x
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x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
-
-
-
-
-
-
FE = F B
qE = qvB
E = vB
v =
E
B
Open the switch:
e-
V
r
Fcent = Fmag
mv2
r
= qvB
mv2 = qvB r
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
q
m
q
m
=
v
Br
=
E
B2r
Millikan’s Oil-Drop Experiment (1911):
Ionized oil droplets,
placed them between
two charged plates,
and measured the
E field that would
make them hover
From this, could
find their charge
http://physics.wku.edu/~womble/phys260/millikan.html
Millikan Oil Drop Experiment Data
18
16
Number
14
12
10
8
6
4
2
1.0
2.0
3.0 4.0
5.0
6.0
7.0 8.0 9.0
Charge ( x 10-19 C )
Data for Black-Body Radiators:
Wien’s Law:
λ T = 2.90 x 10-3 m K
Orion
Max Planck:
Can make theory fit empirical data
if you assume energy is carried in
separate packets of energy
Energy is directly
proportional to the frequency
E = hν
E = energy
ν = frequency of IR radiation
h = Planck’s constant
= 6.63 x 10-34 J s
Photoelectric Effect: the emission of electrons from a
metallic surface when illuminated by electromagnetic
radiation
Low intensity, low frequency
metal surface
High intensity, low frequency
no emission
??????
no emission
Low intensity, moderate frequency
metal surface
Emission of
low energy
electrons
e-
High intensity, moderate frequency
eee-
More electrons emitted,
same low energy
Low intensity, high frequency
Emission of
high energy
electrons
metal surface
e-
High intensity, high frequency
eee-
More electrons emitted,
same high energy
Einstein (1905):
Extend Planck’s work to include all EM radiation
Energy of light is carried in discrete packets, called
“quanta” or “photons”; also called “wave packets”
The energy of a photon depends directly on its
frequency
E = hν
E = energy
ν = frequency of IR radiation
h = Planck’s constant
= 6.63 x 10-34 J s
ex. A photon of red light has a frequency of 7.05 x 1014 Hz.
(a) Find the energy of the photon in joules.
E = hν
= ( 6.63 x 10-34 J s )( 7.05 x 1014 Hz )
E = 4.67 x 10-19 J
(b) Find the energy of the photon in electron-volts.
Definition: 1 electron-volt = 1 eV = 1.6021 x 10-19 J
4.67 x 10-19 J
1 eV
1.6021 x 10-19 J
=
2.92 eV
(c) Find the wavelength of the photon.
v = fλ
c= λν
λ =
c
ν
3.00 x 108 m/s
=
λ = 6.42 x 10-7 m
= 642 x 10-9 m
= 642 nm
4.67 x 1014 Hz
Einstein (1905):
Photoelectric Effect
- Electrons are held in a metal by a work function, ϕ,
which must be overcome to eject electrons
- A photon of red light does not have enough energy
to overcome the work function
- There is a threshold frequency, at which the photon
has just enough energy to overcome ϕ
- Above the threshold frequency, photons have enough
energy to overcome ϕ and also to provide KE
h ν = ϕ + KE
Photoelectric Effect
-Intensity is related to the number of photons, not
their energy
-Each photon ejects one electron; higher intensity
results in more photoelectrons
ex. It is found that a metal will begin emitting photoelectrons
when illuminated by 6.20 x 1014 Hz – light.
(a) What is the energy of this light in joules and eV?
E = hν
= ( 6.63 x 10-34 J s )( 6.20 x 1014 Hz )
E = 4.11 x 10-19 J
1 eV
1.6021 x 10-19 J
= 2.57 eV
(b) What is the work function of this metal, in eV?
Radiation of 2.57 eV is required to eject electrons; so
electrons are held in metal by this energy
ϕ = 2.57 eV
(c) If the metal is illuminated by radiation of frequency
2.63 x 1015 Hz, what is the kinetic energy of the
photoelectrons, in J and eV?
h ν = ϕ + KE
KE = h ν - ϕ
ϕ = 4.11 x 10-19 J
ν = 2.63 x 1015 Hz
= [(6.63 x 10-34 J s)(2.63 x 1015 Hz)] - 4.11 x 10-19 J
KE = 1.33 x 10-18 J
1 eV
1.6021 x 10-19 J
= 8.32 eV
(d) What is the stopping voltage of these photoelectrons?
KE of electrons is 8.32 eV
Stopping voltage is 8.32 V
ex. When illuminated by 420 nm light, a metal ejects
photoelectrons that require a stopping voltage of 1.3 V.
(a) Find the energy of the light in joules.
E = hν
c= λν
c
ν =
=
λ
3.00 x 108 m/s
420 x 10-9 m
ν = 7.14 x 1014 Hz
E = hν
= ( 6.63 x 10-34 J s )( 7.14 x 1014 Hz )
E = 4.74 x 10-19 J
(b) Find the kinetic energy of the photoelectrons in joules.
If the photoelectrons require a
stopping voltage of 1.3 V, what is
KE = 1.3 eV
their kinetic energy in eV ?
1.3 eV
1.6021 x 10-19 J
1 eV
=
2.08 x 10-19 J
(c) Find the work function of the metal in electron-volts.
h ν = ϕ + KE
From (a), energy of light h ν = 4.74 x 10-19 J
From (b), KE of photoelectrons = 2.08 x 10-19 J
ϕ = h ν - KE
= ( 4.74 x 10-19 J ) - ( 2.08 x 10-19 J )
ϕ = 2.66 x 10-19 J
2.66 x 10-19 J
Convert to eV
1 eV
1.6021 x 10-19 J
ϕ = 1.66 eV
Emission Spectrum of Hydrogen and the Bohr Atom
Emission Spectrum of Hydrogen and the Bohr Atom
Produced lines
of definite
wavelengths
Neils Bohr used Einstein’s particle theory of light to
explain emission spectrum; also addressed
difficulties with Rutherford’s atom
The Bohr Atom
Speculated that there are certain
stable orbits for electrons
n=3
n=2
Different levels have
different energies
n=1
Electrons can jump between
levels
E1
E2
E3
h ν = E2 - E1
When they jump to a level of
lower energy, they emit a
photon of energy equal to
difference between levels
hν’
n=3
n=2
n=1
More than one transition
is possible
Will result in photons of
different energies
But energies will be of
precise values
E1
h ν = E2 - E1
n = 2 to n = 1
E2
h ν’ = E3 - E2
n = 3 to n = 2
E3
h ν’’ = E3 - E1
n = 3 to n = 1
hν’’
hν
Energy level diagram
for hydrogen
Hydrogen Energy Level Diagram
ex. To jump from the n = 2 to n = 4 energy level, a photon
of wavelength 486 nm is absorbed.
(a) Find the frequency of this photon.
c= λν
ν =
c
λ
=
ν = 6.17 x 1014 Hz
3.00 x 108 m/s
486 x 10-9 m
ex. To jump from the n = 2 to n = 4 energy level, a photon
of wavelength 486 nm is absorbed.
(b) Find the energy of this photon in J and eV.
ν = 6.17 x 1014 Hz
E = hν
= ( 6.63 x 10-34 J s )( 6.17 x 1014 Hz )
E = 4.09 x 10-19 J
1 eV
1.6021 x 10-19 J
= 2.56 eV
E4 - E2 = 2.56 eV
?
E4 - E2 = 2.56 eV
E4 - E2 = ( - 0.85 eV ) - ( - 3.40 eV )
= 2.55 eV
ex. The electron then jumps down to the ground state. Find
the frequency and wavelength of the emitted photon.
E4 - E1 = ( - 0.85 eV ) - ( -13.59 eV )
= 12.73 eV
ex. The electron then jumps down to the ground state. Find
the frequency and wavelength of the emitted photon.
E4 - E1 = 12.73 eV
1.6021 x 10-19 J
1 eV
= 2.04 x 10-18 J
E= hν
ν =
E
h
=
2.04 x 10-18 J
6.63 x 10-34 J s
c= λν
c
3.00 x 108 m/s
λ =
=
ν
3.08 x 1015 Hz
=
ν = 3.08 x 1015 Hz
= λ = 9.75 x 10-8 m
= 97.5 nm
Bohr’s Assumption and Derivation of the Bohr radius
Planck and Einstein assumed that the energy of light is
quantized, or “stepped”; can have 1 photon, or 2
photons, but never 1 ½ photons
Bohr’s Assumption and Derivation of the Bohr radius
Bohr assumed the same for the angular momentum of an
electron; assumed that the angular momentum is an
integer multiple of a fundamental amount
L = I ω = ( m r2 )( v/r ) = m v r
r = radius of orbit
Assumption: m v rn = n
v =
h
2π
nh
2π m rn
Electrons are in circular orbit around nucleus; a centripetal
force is required ; provided by the electrostatic attraction
between electron and nucleus
Fcent = FE
m v2
r
=
k e Ze
charge of electron = e
charge of nucleus = Ze
r2
m v2 r2 = k e Ze r
m v2 r = k e Ze = k Ze2
r =
k Ze2
m
v2
v =
nh
2π m rn
from
earlier
r =
rn =
1 =
Bohr
orbits
k Ze2
m v2
v =
nh
2π m rn
4 π2 m2 rn2 k Ze2
m n 2 h2
4 π2 m rn k Ze2
n2 h2
rn =
n2 h2
4 π2 m k Ze2
Cancel m’s and rn’s
Solve for rn
n = integer
rn =
rn =
n2 h2
4 π2 m k Ze2
h2
4
π2
mk
n2
For hydrogen, Z = 1
Ze2
(6.63 x 10-34 J s)2
r1 =
12
4 π2 (9.11 x 10-31 kg)(9 x 109 Nm/C2)(1)(1.6021 x 10-19 C)2
r1 = 0.529 x 10-10 m
r2 = ( 0.529 x 10-10 m ) 22 = 2.12 x 10-10 m
r3 = ( 0.529 x 10-10 m ) 32 = 4.76 x 10-10 m
.
.
.
Energy : electron has both electric potential energy and
kinetic energy as it orbits the nucleus
V =
W
q
=
PE
-e
PE = - e V = - e
kq
r
k Z e2
= r
q = Ze
KE = ½ m v2
Energy = KE + PE
For nth orbit, Energy = ½ m vn2 +
(
k Z e2
rn
)
Energy = ½ m vn2
k Z e2
rn
vn =
rn =
Energy = -
= -
2 π2 m k2 Z2 e4
n2 h2
2 π2 m k2 Z2 e4
1
h2
n2
nh
2π m rn
n2 h2
4 π2 m k Ze2
Energy = -
E1 = -
2 π2 m k2 Z2 e4
1
h2
n2
2π2(9.11 x 10-31 kg)(9 x 109 Nm/C2)2(1)2(1.6021 x 10-19 C)4
(6.63 x 10-34 J s)2
E1 = - 2.18 x 10-18 J
1 eV
1.6021 x 10-19 J
E1 = - 13.6 eV
1
12
Energy = - ( 13.6 eV )
E2 = - ( 13.6 eV )
E3 = - ( 13.6 eV )
1
22
1
32
1
n2
=
E2 = - 3.40 eV
=
E3 = - 1.51 eV
If a wave (such as light) can act like a particle, can a
particle (such as an electron) act like a wave?
Einstein: Although photons have no
mass, they do have
momentum
p =
E
c
Photons act like particles
For particles, p = m v
Louis de Broglie
(1892-1987)
p =
so
mv =
mv =
mv =
λ =
E
c
E
c
hν
c
h
λ
h
mv
p = mv
For photons, E = h ν
c = λν
λmv = h
de Broglie
Wavelength
1
λ
=
ν
c
ex. Find the de Broglie wavelength of a baseball,
mass 0.150 kg, moving at 45 m/s.
λ =
=
h
mv
( 6.63 x 10-34 J s )
( 0.150 kg )( 45.0 m/s )
λ = 9.82 x 10-35 m
ex. Find the de Broglie wavelength of an electron moving
at a speed of 1.86 x 106 m/s.
λ =
=
h
me = 9.11 x 10-31 kg
mv
( 6.63 x 10-34 J s )
( 9.11 x 10-31 kg )( 1.86 x 106 m/s )
λ = 3.91 x 10-10 m
Electron Diffraction Experiment
Send electrons, one-by-one, through an atomic
crystal lattice whose atoms are separated by
about 10-10 m
An interference pattern emerges, as if the electrons
went through more than one space at the same time,
and interfered with itself
λ =
h
λ =
mv
h
2π m rn
m
nh
Bohr assumption: Angular momentum is quantized
L = I ω = ( m r2 )( v/r ) = m v r
m v rn = n
h
2π
v =
nh
2π m rn
λ =
λ =
h
mv
λ =
h
2π m rn
m
nh
2π rn
n
2π rn = n λ
2π rn : circumference of
nth Bohr orbit
Gives explanation for stability of Bohr orbits
Electrons orbit at a location such that their de Broglie
wavelengths form standing waves around the nucleus
stable
not stable
Heisenberg Uncertainty Principle
Enter a room that is completely dark
In the room is a ping pong ball suspended from
the ceiling; your task is to find the ball
To find the ball, feel around the room with your
hands until you touch the ball
When you touch the ball (and locate it) you
cause it to move
The act of determining its position changed its
state (from stationary to moving, or from
moving slowly to moving quickly, or…)
To locate an electron, bounce a photon from it (the
photon will enter your eye, and you will “see” the
electron)
But by bouncing the photon from the electron, you
change the state of the electron; you are no longer
certain of its original state of motion
Heisenberg: you cannot know the exact position and
motion of an object at the same time the act of
determining one alters the other to a certain degree
Δx Δp ≤
h
4π
Δx : uncertainty in position
Δp : uncertainty in momentum
Electron Tunneling
Quantum mechanics tells us
that electrons have both wave and
particle like properties. Tunneling
is an effect of the wavelike nature.
The top image shows us that when
an electron (the wave) hits a
barrier, the wave doesn't abruptly
end, but tapers off very quickly exponentially. For a thick barrier,
the wave doesn't get past.
The bottom image shows the
scenario if the barrier is quite thin
(about a nanometer). Part of the
wave does get through, and
therefore some electrons may
appear on the other side of the
barrier.