The Invention of the Modern Atom
The First Atomic Theorist
(Translation)
“Everything is composed of ‘atoms’,
which are physically, but not
geometrically, indivisible. Between
atoms lies empty space; atoms are
indestructible; have always been, and
always will be, in motion. There are an
infinite number of atoms, and kinds of
atoms, which differ in shape and size.”
-Democritus, 350 BC
V1
JJ Thomson (1897)
Theorized that “cathode rays”
are actually beams of subatomic
particles.
Thomson measured the mass
of these particles by deflecting
them in an electric field.
It is now known that a cathode
ray is a beam of electrons.
V2
But Thomson noticed a large question brought up by his results.
The knowledge at the time was the following.
1) It was known that electrons are negatively charged.
2) It was also known that atoms are (usually) net neutral.
The question was:
“Where does all the mass of an atom come from if the
electrons have such small mass? How can the atom still
be net neutral?”
Thomson’s measured
mass of an electron
Known mass of lightest
element (Hydrogen)
9 × 10-31 kg
1.7 × 10-27 kg
Thomson’s “Plum Pudding” Model
An atom consists of negative
electrons surrounded by a
“pudding” of positive charge.
The positively charged
substance exactly balances out
the charge of the electrons and
is also massive enough to give
the atom its measured mass.
Ernest Rutherford (1911)
British particle physicist
who fired alpha particles at
a sheet of gold foil, and
used fluorescent paper to
detect where the alpha
particles were deflected.
V3
Rutherford’s Original Setup
Rutherford’s Surprising Setup
Some of the alpha
particles were being
deflected at extreme
angles, sometimes
straight backward!
With this one
experiment, the Plum
Pudding Model was
finished.
Predicted by Thomson’s
Plum Pudding Model
Actual Results:
Explained by Rutherford’s
Planetary Model
“The result of the experiment was
as if you had fired a cannonball at a
piece of tissue paper and had it
come back and hit you.” –E.R.
Rutherford’s results (the large angles) are in line with
the Planetary Model of the Atom, in which a dense,
positively charged nucleus is surrounded by negative,
low-mass electrons.
Big Problems with the Planetary Model
When charged particles accelerate, they radiate
electromagnetic waves – thus losing energy.
If the electron were in circular motion (with centripetal
acceleration), it would constantly radiate EM waves.
Spectrum Tube: The Fingerprint of an Element
The tube is filled with a particular gas
(Hydrogen, Helium, Neon, Mercury, etc)
A voltage is applied across the tube,
accelerating electrons through the gas.
This causes some electrons to collide with
electrons in gas atoms, transferring some
energy to the atoms.
This brings the gas atom into a higher energy
state, otherwise known as an excited state.
When the atom drops back to a lower energy,
it emits a photon (which is what you see).
S1
Spectroscope Time!
By looking at the light through a diffraction grating, we
can separate the spectrum to see which wavelengths are
actually making up the observed color.
The Emission Spectrum of Hydrogen
Hydrogen atoms, after gaining electric potential energy and and then
emitting a photon as they return to a lower energy. We detect these
frequencies of light during the process:
Niels Bohr
“The Great Dane”
Known for his dancing
ability as well as his
Nobel Prize in
Physics.
What do you think Bohr came up with?
Bohr Model
An electron can
only occupy certain
energy states
around a nucleus.
When it transitions
between energy
states, a photon
must either be
absorbed or
emitted by the
atom.
V4
Energy Level Diagram
Each “level” represents a different electric potential energy
for the electron/nucleus system.
0 eV
n = ∞ (ionization energy)
n = 2 (first excited state)
-19.36 eV
-21.5 eV
n = 1 (first excited state)
n = 0 (ground state)
Caution! Do not take these diagrams too literally!
These lines represent energy states for the atom.
Negative Potential Energies Again!
Since an atom (nucleus and electrons) is in a bound state, it has a
negative electric potential energy. This means that an input of energy is
required to separate the electron and nucleus to d = ∞ (zero electric
potential energy).
The amount of energy that is needed to completely free an electron
from the atom is called the ionization energy of the atom. (You are
making the atom into a positive ion by freeing an electron)
d≈∞
Ground State (n = 1)
Most negative EPE
First Excited State (n = 2)
Less negative EPE
Ionization Energy (n = ∞)
Zero EPE
The more negative the EPE, the closer the electron is (on average)
from the nucleus.
By putting energy into the system in the right amount, we can raise
the system to a higher “energy state”. This will bring the electron (on
average) further from the nucleus.
0 eV
n = ∞ (ionization energy)
n = 2 (first excited state)
-19.36 eV
-21.5 eV
n = 1 (first excited state)
n = 0 (ground state)
PhET Simulation – Hydrogen Atom
When an electron in an atom transitions from a more
excited state (less negative energy) to a less excited state
(more negative energy), it emits a photon!!
E2 = -3.4 eV
n=2
(first excited state)
The energy of the
photon will be equal to
the energy lost by the
atom during this
transition.
E1 = -13.6 eV
n=1
(ground state)
Conservation of Energy works! Again!
When an atom transitions to a lower energy level, it emits a photon.
Einitial = Efinal + Ephoton
Einitial
Einitial = E final +
hc
Ephoton
l
Efinal
Hydrogen Whiteboard
Which color light is emitted during a transition from the
n = 3 to the n = 2 energy level in hydrogen?
E3 = -1.5 eV
n=3
(second excited state)
E2 = -3.4 eV
n=2
(first excited state)
E1 = -13.6 eV
n=1
(ground state)
Einitial = E final +
hc
l
Remember to convert to
Joules! (h is given in
terms of Joules!)
E3 = -1.5 eV
n=3
(second excited state)
E2 = -3.4 eV
n=2
(first excited state)
(-1.5 eV) (1.6 x 10-19 J/eV) = (-3.4 eV) (1.6 x 10-19 J/eV) + hc/λ
λ = 650 nm
Lots of different possibilities for
energy level transitions!
For photons emitted during a drop in atomic EPE,
Ei = E f +
lemitted
hc
lemitted
hc
=
Ei - E f
More Quantum Phenomena!!
Now we have a new property that is quantized:
the energy states of an atom.
Atoms can only take on certain specific electric
potential energies, and nothing in between.
This is getting more interesting…
Predict the Spectrum of Neon!
n = ∞ (ionization energy)
-16 eV
n = 3 (first excited state)
-19.36 eV
n = 2 (first excited state)
What color will we see?
What other wavelengths could we
detect if we used IR and UV detectors?
-21.5 eV
n = 1 (ground state)
We can predict the wavelengths of the photons emitted
during the transitions 3  2, 3  1, and 2  1.
n = ∞ (ionization energy)
-16 eV
n = 3 (first excited state)
-19.36 eV
n = 2 (first excited state)
Ei = E f +
-21.5 eV
n = 1 (ground state)
hc
lemitted
PhET Simulation: Neon Lights
Lab Challenge II
By watching the simulation, determine which energy
transitions correspond with which photons (visible
range only).
Determine the changes in electric potential energy
associated with the two visible photons emitted by
excited sodium atoms.
Fill in the numbers for the energy level diagram on your
whiteboards!
Atomic Theory and the Bohr Model
Ultraviolet
Visible
Infrared
When an electron in an atom transitions from a more
excited state (less negative energy) to a less excited state
(more negative energy), it emits a photon!!
E2 = -3.4 eV
n=2
(first excited state)
The energy of the
photon will be equal to
the energy lost by the
atom during this
transition.
E1 = -13.6 eV
n=1
(ground state)
Conservation of Energy works! Again!
When an atom transitions to a lower energy level, it emits a photon.
Einitial = Efinal + Ephoton
Einitial
Einitial = E final +
hc
Ephoton
l
Efinal
Hydrogen Whiteboard
Which color light is emitted during a transition from the
n = 3 to the n = 2 energy level in hydrogen?
E3 = -1.5 eV
n=3
(second excited state)
E2 = -3.4 eV
n=2
(first excited state)
E1 = -13.6 eV
Visible Spectrum
n=1
(ground state)
Einitial = E final +
hc
l
Remember to convert to
Joules! (h is given in
terms of Joules!)
E3 = -1.5 eV
n=3
(second excited state)
E2 = -3.4 eV
n=2
(first excited state)
(-1.5 eV) (1.6 x 10-19 J/eV) = (-3.4 eV) (1.6 x 10-19 J/eV) + hc/λ
λ = 650 nm
Lots of different possibilities for
energy level transitions!
For photons emitted during a drop in atomic EPE,
Ei = E f +
lemitted
hc
lemitted
hc
=
Ei - E f
Ef
Ei
λ
More Quantum Phenomena!!
Now we have a new property that is quantized:
the energy states of an atom.
Atoms can only take on certain specific electric
potential energies, and nothing in between.
This is getting more interesting…
Predict the Spectrum of Neon!
0 eV
n = ∞ (ionization energy)
-19.36 eV
n = 3 (first excited state)
-19.6 eV
n = 2 (first excited state)
What color(s) will we see?
What other wavelengths could we
detect if we used IR and UV detectors?
-21.5 eV
n = 1 (ground state)
Visible Spectrum
We can predict the wavelengths of the photons emitted
during the transitions 3  2, 3  1, and 2  1.
n = ∞ (ionization energy)
-19.36 eV
n = 3 (first excited state)
-19.6 eV
n = 2 (first excited state)
Ei = E f +
-21.5 eV
n = 1 (ground state)
hc
lemitted
lemitted
hc
=
Ei - E f
What wavelengths can we
expect to see?
n=∞
-19.36 eV
n=3
-19.6 eV
n=2
n = 2 to n = 1
First excited state to Ground state
λ = 650 nm
n = 3 to n = 1
Second excited state to Ground state
λ = 580 nm
n = 3 to n = 2
Second excited state to Ground state
-21.5 eV
n=1
λ = 5,156 nm
(Infrared)
Neon
Careful with Joules and eV!
Use of Planck’s constant as h = 6.6 x 10-34 J*s and/or c = 3 x 108 means
that you cannot work in eV! You must then convert everything to J!
Use this as your conversion factor:
1 J = 1.6 x 10-19 eV
With quantum and atomic phenomena, you should always expect a
reasonable number of eV for an answer (0-1000 eV).
With quantum and atomic phenomena, you should always expect a
very tiny number of Joules for an answer (5 x 10-19 J).
Ultraviolet
Visible
Infrared
The larger the energy transition
made by the electron, the more
energy the emitted photon will have.
This is why the “short” transitions
of hydrogen (the Paschen series)
emit infrared light. IR waves have
long wavelengths and lowfrequencies, thus carrying less
energy.
This is why the “long” transitions of
hydrogen (the Lyman series) emit
UV light. UV waves have short
wavelengths and high-frequencies,
thus carrying more energy.
The visible portion spectrum
(Balmer series) has energies right in
the middle.
A hypothetical atom has four energy states as shown below.
1)Which of the following photon
energies could NOT be found in the
emission spectra of this atom after it
has been excited to the
n = 4 state?
(A) 1 eV (B) 2 eV (C) 3 eV
(D) 4 eV (E) 5 eV
2) Which of the following transitions
will produce the photon with the
longest wavelength?
(A) n = 2 to n = 1 (B) n = 3 to n = 1
(C) n = 3 to n = 2 (D) n = 4 to n = 1
(E) n = 4 to n = 3
Once the atom has been excited to n = 4,
all of these possible transitions can occur!
Which of the following photon
energies could NOT be found in the
emission spectra of this atom after it
has been excited to the n = 4 state?
(A) 1 eV (B) 2 eV (C) 3 eV
(D) 4 eV (E) 5 eV
None of these possible transitions
have a change in energy of 4 eV.
Emitted Photon Energies (left to right)
1 eV, 3 eV, 6 eV, 2 eV, 5 eV, 3 eV
(E)
(B)
(A)
Which of the following transitions
will produce the photon with the
longest wavelength?
(A) n = 2 to n = 1 (B) n = 3 to n = 1
(C) n = 3 to n = 2 (D) n = 4 to n = 1
(E) n = 4 to n = 3
(C)
(D)
Longest wavelength means least energy.
(Large wavelength, low frequency, less energy.)
The lowest energy photon will be emitted during the
transition that has the smallest ΔE for the electron.
(E)
(B)
(A)
Which of the following transitions
will produce the photon with the
longest wavelength?
(A) n = 2 to n = 1 (B) n = 3 to n = 1
(C) n = 3 to n = 2 (D) n = 4 to n = 1
(E) n = 4 to n = 3
(C)
(D)
Longest wavelength means least energy.
(Large wavelength, low frequency, less energy.)
The lowest energy photon will be emitted during the
transition that has the smallest ΔE for the electron.
Atoms can emit photons if they drop to a lower energy level,
or absorb energy and raise to a higher energy level.
However, they must do this is the exact perfect amount
that will bring them to an allowed energy state!
Mystery: Solved.
Final Whiteboard: Absorption and Emission
The energy level diagram below is for a hypothetical atom. A gas of
these atoms initially in the ground state is irradiated with photons
having a continuous range of energies between 7 and 10 electron
volts. One would expect photons of which of the following energies
to be emitted from the gas?
(A) 1, 2, and 3 eV only
(B) 4, 5, and 9 eV only
(C) 1, 3, 5, and 10 eV only
(D) 1, 5, 7, and 10 eV only
(E) Since the original photons
have a range of energies, one
would expect a range of emitted
photons with no particular
energies.
How to think about this
The energy level diagram below is for a hypothetical atom. A gas of
these atoms initially in the ground state is irradiated with photons
having a continuous range of energies between 7 and 10 electron
volts.
If the photons transferred their energy to the atom,
… could jump to any
state in this range!
Electrons starting
in this state…
Since the only allowed energy
level within that range is the
second excited state (-5 eV),
the electrons in the atom will
be excited to that state!
After the electrons are excited, they fall back
down to lower energy levels
During this time, photons of energy -ΔEelectron are emitted!
9
eV
5
eV
4
eV
(A) 1, 2, and 3 eV only
(B) 4, 5, and 9 eV only
(C) 1, 3, 5, and 10 eV only
(D) 1, 5, 7, and 10 eV only
(E) Since the original photons
have a range of energies,
one would expect a range of
emitted photons with no
particular energies.
Something to Think About for Next Time
But why does an electron only have distinct
allowed atomic energies?
What prevents the atom from collapsing still?
Wouldn’t the electrons and protons
attract and there be a certain probability of
them annihilating one another?
Atomic Theory Wrap-up and Review
Announcements
Test Monday – Nuclear Physics, The Photoelectric
Effect, Photons and Matter, Atomic Physics
20 MC – All topics
2 FRQs – 1 on nuclear physics, 1 on atomic
energy levels and de Broglie wavelength
AP Test Takers! You have mandatory AP
TEST GRIDDING during 8th period Monday
in the Cafeteria.
Question 1c) from Homework
The atom is in the ground state when an electron traveling
with a speed of 1.3 x 106 m/s collides with it. Can the
electron excite the atom to the n = 2 state?
____Yes
____No
____It cannot be determined.
Justify your answer.
n=3
n=2
n=1
Does the incoming electron have
enough KE to give the atom the 3 eV it
needs to jump to the next energy
state?
(Hint: The incoming electron does not
need to transfer all of its KE!)
Question 1c) from Homework
The atom is in the ground state when an electron traveling
with a speed of 1.3 x 106 m/s collides with it. Can the
electron excite the atom to the n = 2 state?
____Yes
____No
n=3
____It cannot be determined.
The incident electron has ½ mv2
= 4.8 eV of KE.
The atom needs 3 eV to jump.
n=2
An incident electron, unlike an
incident photon, can transfer only
some of its energy to the atom.
n=1
Follow-Up Question
The atom is in the ground state when photons with energies
that range in energies from 4.5-5.5 eV are incident upon it.
To what excited state(s), if any, can the atom be excited?
n=3
n=2
n=1
An electron must absorb the full
energy of an incident photon.
No fractions or combinations!
This means that the atom will be able
to be excited to the n = 3 state.
E photon = DEelectron
Energy Level Diagram – Part I
An experiment is performed on a sample of atoms known to have a
ground state of -5.0 eV. The gas is illuminated with “white light” (400
– 700 nm). A spectrometer capable of analyzing radiation in this
range is used to measure the radiation. The sample is observed to
absorb light at only 400 nm. After the “white light” is turned off, the
sample is observed to emit visible radiation of 400 nm and 600 nm.
(a)In the space below, determine the values of the energy levels and
on the following scale sketch an energy-level diagram showing the
energy values in eV and the relative positions of:
i. The ground state
ii. The energy level to which the system was first excited.
iii. One other energy level that the experiment suggests may exist
An experiment is performed on a sample of atoms known to have a
ground state of -5.0 eV. The gas is illuminated with “white light” (400
– 700 nm). A spectrometer capable of analyzing radiation in this range
is used to measure the radiation. The sample is observed to absorb
light at only 400 nm.
E photon = hc l
400 nm light possesses
3.1 eV of energy.
State to which atom
was first excited
+3.1 eV
It can raise the atom to
a -1.9 eV level.
n = 1 (ground state)
After the “white light” is turned off, the sample is observed to emit
visible radiation of 400 nm and 600 nm.
400 nm is expected! If the atom absorbed 400 nm light to jump
from -5 eV to -1.9 eV, then it should also emit 400 nm light when it
jumps back from -1.9 eV to -5 eV
600 nm light possesses
2.1 eV of energy.
In order for the atom to emit
600 nm light once it has been
excited, there must be an
energy level 2.1 eV below
the excited state to which the
atom was first raised.
n=3
-2.1 eV
-3.1 eV
n=2
n=1
Atomic Physics - Part II!
What is the wavelength of any other radiation, if any, that might
have been emitted in the experiment? Why was it not observed?
n=3
n=2
n=1
Solution - Part II!
What is the wavelength of any other radiation, if any, that might
have been emitted in the experiment? What was it not observed?
Given the energy levels that we
drew, there should be a photon
emitted during the n = 2 to n = 1
transition.
n=3
n=2
-1 eV
n=1
The energy emitted during this
transition is 1 eV
λ = 1,238 nm
Infrared
But we still have one major unanswered question:
Why are electrons in atoms only “allowed” to
occupy certain specific orbital energies?
Is this something totally unique, or can we come up
with a deeper reason for why this happens?
By combining his theory with de Broglie’s
hypothesis, Niels Bohr figured out the reason why
only certain energy level orbits are allowed!
Review: deBroglie Wavelength
On a quantum scale, all matter must be modeled by using wave
functions, which allow you to predict the probability of finding
the particle in a particular region of space.
The deBroglie wavelength, λ = h/p, defines
the “size” of the particle’s wave function.
Electron traveling at 103 m/s
Baseball traveling at 103 m/s
Orbital Standing Waves
Bohr theorized that the “allowable” orbital electron
energies were the ones in which
the circumference of the orbit is a
multiple of the deBroglie wavelength!
Bohr figured out that the electron’s orbit would only be
stable if its circumference corresponded to a standing wave
with the electron’s deBroglie wavelength!
Think of wrapping a sine wave around a circle, and only allow the ones
that have an exact whole number of waves around the circumference.
An electron’s orbit will only be stable if the circumference corresponds
to a standing wave with the electron’s deBroglie wavelength!
Think of wrapping a wave around the edge of a circle.
This would the n = 3 state (second excited state)!
Ground state is the lowest
possible energy
This is why an electron cannot get
infinitely close to the nucleus.
The closest possible orbital that the electron
can have corresponds to one full deBroglie
wavelength around the outside of the orbit.
An experimenter determines that when a beam of
monoenergetic electrons bombards a sample of pure gas,
atoms of the gas are excited if the kinetic energy of each
electron in the beam is 3.70 eV or greater.
a)Determine the deBroglie wavelength of 3.70 eV electrons.
b)Once the gas is excited by 3.70 eV electrons, it emits
monochromatic light. Determine the wavelength of this light.
Experiments reveal that two additional wavelengths are
emitted if the beam energy is raised to at least 4.90 eV.
c) In the space below construct an energy-level diagram
consistent with this information and determine the energies
of the photons associated with those two additional
wavelengths.
a)First, use K = ½ mv2 to solve for velocity: 1,140,000 m/s – then p = mv
λ = h/p gives 0.64 nm.
b) E = hc/λ gives λ = 334 nm (ultraviolet light)
c)
n=3
1.2 eV
n=2
4.9 eV
3.7 eV
n=1
Using E = hc/λ, we obtain that
the other two wavelengths
emitted are 103 nm and 253 nm.
Photoelectric Effect Review
Light of wavelength 400 nm is incident on a metal surface. Electrons
are ejected from the surface with maximum kinetic energy 1.1 x 10-19 J.
a) Calculate the frequency of the incoming light.
b) Calculate the work function of the metal surface.
c) Calculate the stopping potential for the emitted electrons.
d) Calculate the momentum of an electron with the max kinetic
energy.
Kmax = hf - F
Solution Kmax = eDVstop
a) Just use c = λf! Part (a) of these FRQs is usually a simple
plug-and-chug. Don’t overthink it! Answer =
b) Use Kmax = hf – Φ to get Φ =
c) Use Kmax = qeΔVstop!
d) K = ½ mev2 allows you to solve for velocity.
You can then plug into p = mv to solve for momentum.
Equations to Memorize
Nuclear Physics
DE = Dmc
2
Photon Model
c=lf
E = hf = hc l
Photoelectric Effect
p=h l
Kmax = hf - F
deBroglie Wavelength
Kmax = eDVstop
fthreshold
F
=
h
l=h p
Atomic Physics
E photon = DEelectron