Modern Physics
• What is meant by “quantized”?
– Quantity
• Specific and discrete quantity
• Packets of definite size.
• Quantized energy can be thought of as existing in
very small packets of specific size.
• Atoms absorb and emit quanta of energy.
• Let us consider light…
Electromagnetic Spectrum
Visible spectrum
Three models are used to describe light
• What model is used in geometric optics, like with
lenses and mirrors?
– Ray
• What model is used in studying diffraction and
interference?
– Wave
• What model is used to study interaction of light and
atoms?
– Particle ( photon)
But light is said to have a “dual nature”
• What is that supposed to mean?
– Wave particle duality
• Waves have both a wave and a particle component
• We describe the path of light as a ray
– Equation
• E = hf
for a single photon
• E = nhf
for multiple photons
–h = plank’s constant
»6.63x10-34 J ∙ s
(SI version)
»4.14x10-15 eV ∙ s (convenient)
–f = frequency
–n = number of photons
Conceptual checkpoint
• Which has more energy in its photons, a very bright,
powerful red laser or a small key-ring red laser?
– Neither! They both have the same energy per
photon. The big one has more power.
• Which has more energy in its photons, a red laser or
a green laser?
– The green one has shorter wavelength and higher
frequency. It has more energy per photon.
The “electron-volt” (eV)
• The electron-volt is the most useful unit on the
atomic level.
• If a moving electron is stopped by 1 V of electric
potential, we say it has 1 electron-volt (or 1 eV) of
kinetic energy.
• 1 eV = 1.602×10-19 J
• What is the frequency and wavelength of a photon
whose energy is 4.0 x 10-19J?
E = 4.0x10-19 J
h= 6.625x10-34 J ∙ s
E = hf
f = E / h= 4.0x10-19 J / 6.625x10-34 J ∙ s = 6.04x1014 Hz
λ = c/f
λ = c/f = 3x108 m/s / 6.04x1014 1 / s = 4.97x10-7 m
= 497 nm
• How many photons are emitted per second by a HeNe laser that emits 3.0 mW of power at a
wavelength of 632.8 nm?
P = 3.0 mW = 0.003 W
λ = 632.8 nm = 632.8x10-9 m
P = E /t E = P ∙ t E = 0.003 W ∙ 1 s = 0.003 J
f = c / λ = 3x108 m/s / 632.8x10-9 m = 4.74x1014 Hz
E =n(hf)
n =E / (hf)
=0.0003 J / (6.625x10-34 J ∙ s ∙ 4.74x1014 Hz) = 9.55x1014
• What are atoms composed of?
– Atoms consist of nuclei (protons and neutrons and
electrons.
• What happens when an atom encounters a photon?
– The atom usually ignores the photon, but sometimes the
atom absorbs the photon.
• If the photon is absorbed by the atom, what
happens next?
– The photon disappears and winds up giving all its energy
to the atom’s electrons.
• This is a graph of energy levels for a hypothetical atom
Atom loses an electron
0.0eV
-1.0eV
Highest energy level
-3.0eV
Higher energy
levels. (Atoms has
-5.5 eV
to absorb energy to
get from the ground
state.
Normal “unexcited”
-11.5 eV
state
Ionization level
Third excited state
Second excited state
First excited state
Ground state (lowest energy level
• What do we mean when we say the atoms energy levels
are “quantized”?
Ionization level
0.0eV
•Only certain
Third excited state
-1.0eV
energies are
allowed.
Second excited state
-3.0eV
•These are
represented by the
horizontal lines. -5.5 eV First excited state
•The atom cannot
exist at energies
not shown in this
graph!
Ground state (lowest energy level
-11.5 eV
Absorption of photon by Atom
• When a photon of light is absorbed by an atom, it
causes an increase in the energy of the atom.
• The photon disappears, and the energy of the
atom increases by exactly the amount of energy
contained in the photon.
• The photon can be absorbed ONLY if it can produce
an “allowed” energy increase in the atom.
• Absorption of photon by atom
0.0eV
Exited state
E = hf
-10.0eV
ΔE = hf
Ground state
Absorption Spectrum
• When an atom absorbs photons, it removes the
photons from the white light striking the atom,
resulting in dark bands in the spectrum.
• Therefore, a spectrum with dark bands in it is called
an absorption spectrum.
• Absorption Spectrum
0.0eV
Ionized
Absorption
spectra always
involve atoms
going up in
energy level.
-10.0eV
Ground state
Emission of photon by atom
• When a photon of light is emitted by an atom, it
causes a decrease in the energy of the atom.
• A photon of light is created, and the energy of the
atom decreases by exactly the amount of energy
contained in the photon that is emitted.
• The photon can be emitted ONLY if it can produce
an “allowed” energy decrease in an excited atom.
• Emission of photon by atom
0.0eV
Exited state
E = hf
-10.0eV
ΔE = hf
Ground state
Emission Spectrum
• When an atom emits photons, it glows! The
photons cause bright lines of light in a spectrum.
• Therefore, a spectrum with bright bands in it is
called an emission spectrum.
• Emission of photon by atom
0.0eV
Ionized
Emission
spectra always
involve atoms
going down in
energy level.
-10.0eV
Ground state
Photoelectric effect
What is the frequency and wavelength of the light that
will cause the atom shown to transition from the ground
state to the first excited state? Draw the transition.
h = 4.14x10-15 eV ∙ s
ΔE = hf f = Δ E / h
0.0eV
-1.0eV
f = (11.5 – 3.0) / 4.14x10-15
-3.0eV
15
f = 1.45x10 Hz
λ= c / f
-5.5 eV
λ= 3x108 / 1.45x1015
λ= 2.07x10-7 m
λ= 207 nm
Ionization level
Third excited state
Second excited state
First excited state
Ground state
(lowest energy level
-11.5 eV
What is the longest wavelength of light that when
absorbed will cause the atom shown to ionize from the
ground state? Draw the transition.
h = 4.14x10-15 eV ∙ s
ΔE = hf f = Δ E / h
0.0eV
-1.0eV
f = (11.5 ) / 4.14x10-15
f=
2.78x1015
Hz
λ= c / f
-3.0eV
-5.5 eV
λ= 3x108 / 2.78x1015
λ= 1.08x10-7 m
λ= 1-8 nm
Ionization level
Third excited state
Second excited state
First excited state
Ground state
(lowest energy level
-11.5 eV
The atom shown is in the second excited state . What
frequencies of light are seen in its emission spectrum?
Draw the transition.
h = 4.14x10-15 eV ∙ s
1
ΔE = hf
0.0eV
-1.0eV
f=ΔE/h
-3.0eV
f = (11.5 – 3.0 ) / 4.14x10-15
f = 2.053x1015 Hz
-5.5 eV
Ionization level
Third excited state
Second excited state
1
2
First excited state
3
Ground state
(lowest energy level
-11.5 eV
The atom shown is in the second excited state . What
frequencies of light are seen in its emission spectrum?
Draw the transition.
h = 4.14x10-15 eV ∙ s
0.0eV
-1.0eV
2
ΔE = hf
f=ΔE/h
f = (5.5-3 ) /
4.14x10-15
f = 6.09x1014 Hz
-3.0eV
Ionization level
Third excited state
Second excited state
1
-5.5 eV
2
First excited state
3
Ground state
(lowest energy level
-11.5 eV
The atom shown is in the second excited state . What
frequencies of light are seen in its emission spectrum?
Draw the transition.
h = 4.14x10-15 eV ∙ s
3
ΔE = hf
0.0eV
-1.0eV
f=ΔE/h
f = (11.5-5.5 ) /
-3.0eV
-15
4.14x10
f = 1.45x1015 Hz
-5.5 eV
Ionization level
Third excited state
Second excited state
1
2
First excited state
3
Ground state
(lowest energy level
-11.5 eV
The atom shown is in the second excited state . What
frequencies of light are seen in its emission spectrum?
Draw the transition.
1
Ionization level
15
f = 2.053x10 Hz
0.0eV
Third excited state
-1.0eV
2
Second excited state
14
f = 6.09x10 Hz
-3.0eV
1
2
3
f =1.45x1015 Hz
-5.5 eV
First excited state
3
Ground state
(lowest energy level
-11.5 eV
Atoms absorbing photons increase in energy
•We’ve seen that if
you shine light on
atoms, they can
Ionization level
absorb photons and
0.0eV
increase in energy.
•The transition shown
is the absorption of
Photon
-4.0eV
an 8.0 eV photon by
4.0 eV with
this atom.
largest
•You can use Planck’s
allowed
equation to calculate
energy
the frequency and
wavelength of this
Ground state
photon.
(lowest energy level
-12 eV
Question
• Now, suppose a photon with TOO MUCH ENERGY
encounters an atom?
• If the atom is “photo-active”, a very interesting and
useful phenomenon can occur…
• This phenomenon is called the
Photoelectric Effect.
Photoelectric Effect
Photon energy
•Some “photoactive” metals
can absorb photons that not
only ionize the metal, but give
Eph
the electron enough kinetic
0.0eV
energy to escape from the
atom and travel away from it.
•The electrons that escape are
often called “photoelectrons”.
•The binding energy or “work -4.0eV
function” is the energy
necessary to promote the
electron to the ionization
level.
•The kinetic energy of the
electron is the extra energy
provided by the photon.
-12 eV
e- E = W0 + KE
Kinetic energy
Ionization level
W0 = Work
function
Ground state
(lowest energy level
Photon energy
Photoelectric Effect
• Photon Energy = Work
e- E = W0 + KE
Function + Kinetic Energy
E
ph
Kinetic energy
• hf = Φ + Kmax
• Kmax = hf – Φ
Ionization level
0.0eV
• K:max Kinetic energy
of “photoelectrons”
• hf: energy of the
-4.0eV
photon
W0 = Work
• Φ: binding energy or
function
“work function” of the
metal.
-12 eV
Ground state
(lowest energy level
• Suppose the maximum wavelength a photon can
have and still eject an electron from a metal is 340
nm. What is the work function of the metal
surface?
Kmax = hf – Φ
Φ = hf
0 J = hf – Φ
f=v/λ
Φ = hv / λ
Φ = (4.14x10-15eV ∙ 3x108 m/s) / 340x10-9 m
Φ = 3.65 eV
The longest wavelength is the lowest energy, and
will provide no “extra” kinetic energy for the
electron.
Question
• Suppose you collect Kmax and frequency data for a
metal at several different frequencies. You then
graph Kmax for photoelectrons on y-axis and
frequency on x-axis. What information can you get
from the slope and intercept of your data?
Slope: Planck’s Constant
Intercept: Φ - binding energy or “work function”
The Photoelectric Effect experiment
• The Photoelectric Effect experiment is one of the
most famous experiments in modern physics.
• The experiment is based on measuring the
frequencies of light shining on a metal (which is
controlled by the scientist), and measuring the
resulting energy of the photoelectrons produced
by seeing how much voltage is needed to stop
them.
• Albert Einstein won the Nobel Prize by explaining
the results.
Photoelectric Effect experiment
Light
Collector
Metal
(+)
e-
e-
e-
e-
e-
e-
eee- e-
V
e-
e- e-
A
e-
e- eeeeee- e-
Strange results in the Photoelectric Effect experiment
• Voltage necessary to stop electrons is independent of
intensity (brightness) of light. It depends only on the
light’s frequency (or color).
• Photoelectrons are not released below a certain
frequency, regardless of intensity of light.
• The release of photoelectrons is instantaneous, even in
very feeble light, provided the frequency is above the
cutoff.
Voltage versus current for different
intensities of light.
Number of
electrons
(current)
increases with
brightness, but
energy of
electrons
doesn’t!
“Stopping
Voltage”
Vs, the voltage needed to stop the electrons,
doesn’t change with light intensity. That
means the kinetic energy of the electrons is
ndependent of how bright the light is.
Voltage versus current for different
frequencies of light.
Energy of
electrons
increases as
the energy of
the light
increases.
“Stopping
Voltage”
Vs changes with light frequency. That means the kinetic
energy of the photoelectrons is dependent on light color.
Experimental determination of the Kinetic Energy
of a photoelectron
• The kinetic energy of photoelectrons can be determined
from the voltage (stopping potential) necessary to stop
the electron.
• If it takes 6.5 Volts to stop the electron, it has 6.5 eV of
kinetic energy.
Graph of Photoelectric Equation
Kmax
KMAX = hf - Φ
y= mx + b
Slope = h
f
Cut-off Frequency
Φ (binding energy)
• Question: Does a photon have mass?
– A photon has a fixed amount of energy (E = hf).
– We can calculate how much mass would have to
be destroyed to create a photon (E=mc2).
• Calculate the mass that must be destroyed to
create a photon of 340 nm light.
λ = 340x10-9 m
h = 6.63x10-34 J∙s
h = 6.63x10-34 (kg ∙m2 / s2)∙s
E = hf
E = mc2
hf = mc2
f=c/λ
hc / λ = mc2
h / λ = mc
m = h / (λc)
m = 6.63x10-34 / (340x10-9 ∙3x108 ) =
Momentum of a Photon
• Does a photon have momentum?
Yes
• A photon’s momentum is calculated by
– p = E / c = hf / c = h / λ
We have experimental proof of
the momentum of photons
• Compton scattering
– Proof that photons have momentum.
– High-energy photons collided with electrons
exhibit conservation of momentum.
– Work Compton problems just like other
conservation of momentum problems – except
the momentum of a photon uses a different
equation.
• What is the frequency of a photon that has the
same momentum as an electron with speed 1200
m/s?
melectron = 9.11x10-31 Kg h = 6.63x10-34 J∙s
p = mv
h = 6.63x10-34 (kg ∙m2 / s2)∙s
p = 9.11x10-31 Kg ∙ 12 m / s =
p = hf / c
f=Pc/h
f = ( kg ∙ m/s) ∙ (3x108 m/s) / (6.63x10-34 (kg ∙m2 / s2)∙s)
f =
Wave-Particle Duality
• Waves act like particles sometimes and particles
act like waves sometimes.
• This is most easily observed for very energetic
photons (gamma or x-Ray) or very tiny particles
(elections or nucleons)
Particles and Photons both have Energy
• A moving particle has kinetic energy
– E = K = ½ mv2
• A particle has most of its energy locked up in its
mass.
– E = mc2
• A photon’s energy is calculated using its frequency
– E = hf
Particles and Photons both have Momentum
• For a particle that is moving
– p = mv
– kg ∙ m/s
• For a photon
– p = h/λ
– (kg ∙m2 / s2)∙s / m = kg ∙ m / s
– Check out the units! They are those of momentum.
Particles and Photons both have a Wavelength
• For a photon
– l = c/f
• For a particle, which has an actual mass, this
equation still works
– λ= h/p where p = mv
– This is referred to as the deBroglie wavelength
We have experimental proof that
particles have a wavelength
• Davisson-Germer Experiment
– Verified that electrons have wave properties by
proving that they diffract.
– Electrons were “shone” on a nickel surface and
acted like light by diffraction and interference.
– We’ll study diffraction in the next unit, and
return to this experiment then…
• What is the momentum of photons that have a
wavelength of 620 nm?
λ = 620x10-9 m
h = 6.63x10-34 J∙s
p = h / λ = 6.63x10-34 J∙s / 620x10-9 m =
• What is the wavelength of a 2,200 kg elephant
running at 1.2 m/s?
m = 2200 kg v = 1.2 m / s
p = mv = 2200 kg ∙1.2 m/s =
p=h/λ
λ = h / p = 6.63x10-34 J∙s / 12 m / s =
Naming a Nucleus
Mass number
12
6
Atomic number
C
• What are isotopes?
• What characteristics do isotopes of the same
element share?
What characteristics do
isotopes of the same
element share?
Atomic number
What characteristics do
isotopes of the same
element not share?
Atomic mass
Radioactivity
•
•
•
•
Naming
Isotopes
a Nucleus
Isotopes have the same atomic number and
different atomic mass.
Isotopes have similar or identical chemistry.
Isotopes have different nuclear behavior.
Examples:
12
6
C C C
13
14
6
6
Elementary Particles
mass 1
Proton
charge 1
mass 1
Neutron
charge 0
mass 0
Electron
charge -1
p
n
e
Negative chare
0
+1
e
Positive chare
Nuclear reactions
• Nuclear Decay: a spontaneous process in which an
unstable nucleus ejects a particle and changes to
another nucleus.
– Alpha decay
– Beta decay
• Beta Minus
• Positron
• Fission: a nucleus splits into two fragments of
roughly equal size.
• Fusion: Two nuclei combine to form another
nucleus.
Decay Reactions
• Alpha decay
– A nucleus ejects an alpha particle, which is just a
helium nucleus.
• Beta decay
– A nucleus ejects a negative electron.
• Positron decay
– A nucleus ejects a positive electron.
• Simulations
– http://library.thinkquest.org/17940/texts/radioactivity
/radioactivity.html
Alpha (α) Decay
Alpha particle
(helium nucleus) is
released. Alpha
decay only occurs
with very heavy
elements.
239
94
Pu
235
92
U
+
4
2
He
Beta (β-)Decay
A beta particle
(negative electron) is
released. Beta decay
occurs when a nucleus
has too many neutrons
for the protons present.
A neutron converts to a
proton. An antineutrino
is also released.
14
6
C
14
7
N
+
0
-1
e
+γ
Beta (β+)Decay
Positron (positive
electron) is released.
Positron decay occurs
when a nucleus has too
many protons for the
neutrons present. A
proton converts to a
neutron. A neutrino is
also released.
2
2
He
2
1
H
+
0
1
e
+γ
Neutrino and Anti-Neutrino
• Proposed to make beta and positron decay obey
conservation of energy.
• These particles possess energy and spin, but do
not possess mass or charge.
• They do not react easily with matter, and are
extremely hard to detect.
Gamma Radiation, γ
•Gamma radiation is
electromagnetic in nature.
•Gamma photons are
released by atoms which
have just undergone a
nuclear reaction when the
excited new nucleus drops
to its ground state.
•The high energy in a
gamma photon is calculated
by E = hf.
• Complete the reaction, identify the type of decay.
234
90
Th
234
Pa
91
+
0
e
-1
+
γ
• Complete the reaction for the alpha decay of
Thorium-232
232
90
Th
228
88
Ra
+
4
2
He
Nuclear Fission and Fusion
Fission
•Fission occurs when an unstable heavy nucleus splits apart
into two lighter nuclei, forming two new elements.
•Fission can be induced by free neutrons.
•Mass is destroyed and energy produced according to E =
mc2.
• http://library.thinkquest.org/17940/texts/fission/fission.html
• http://www.atomicarchive.com/Movies/index.shtml
Neutron-induced fission
• Neutron-induced fission produces a “chain
reaction.” What does that mean?
• Nuclear power plants operate by harnessing the
energy released in fission in by controlling the
chain reaction.
• Nuclear weapons depend upon the initiation of an
uncontrolled fission reaction.
Critical Mass
• The neutrons released from an atom that has
undergone fission cannot immediately be
absorbed by other nearby fissionable nuclei until
they slow down to “thermal” levels.
• How can this concept be used to explain why a
chain reaction in nuclear fission will not occur
unless a “critical mass” of the fissionable element
is present at the same location?
Nuclear Reactors
•Nuclear reactors produce electrical energy through fission.
•Advantages are that a large amount of energy is produced
without burning fossil fuels or creating greenhouse gases.
•A disadvantage is the production of highly radioactive
waste.
•Another simulation appears at
http://www.howstuffworks.com/nuclear-power.htm
Nuclear Weapons
•Nuclear weapons have been used
only twice, although they have
been tested thousands of times.
•Weapons based on nuclear
fission involve slamming together
enough material to produce an
uncontrolled fission chain
reaction.
Little Boy was dropped on
Hiroshima and contained U-235
produced in Oak Ridge, TN.
Fission
• Fission occurs only with very heavy elements, since
fissionable nuclei are too large to be stable.
• A charge/mass calculation is performed to balance
the nuclear equation.
• Mass is destroyed and energy produced according
to E = mc2.
•
•
•
•
Fusion
Fusion occurs when two light nuclei come
together to form a new nucleus of a new element.
Fusion is the most energetic of all nuclear
reactions.
Energy is produced by fusion in the sun.
Fusion of light elements can result in nonradioactive waste.
Fusion
•Fusion is the reaction that powers the sun, but it
has not been reliably sustained on earth in a
controlled reaction.
•Advantages to developing controlled fusion would
be the tremendous energy output and the lack of
radioactive waste products.
•Disadvantages are – we don’t know if we’ll be
technically able to do it on earth!
Mass defect
• This strange term is used to indicate how much
mass is destroyed when a nucleus is created from
its component parts.
• The mass defect is generally much, much less than
the mass of a proton or neutron, but is significant
nonetheless.
• The loss of mass results in creation of energy,
according to E = mc2.
89