Ch 15: Electric force and energy
quantization determine atomic structure.
Based on experimental observations scientists propose
mental models of atom. Atom is not observed directly.
In 400 B.C. Democritus theorized that matter consisted of
tiny particles called atoms.
Dalton Model – Billiard ball
 To explain the ratio of the elements
combining to make a compound.
CO2 is 1:2 , NaCl is 1:1
 No direct or experimental evidence is provided to
support the model.
 Atom is indivisible.
15.1 The Discovery of the electron
Cathode Ray led Thomson to discover the
electron and to change Dalton’s Model to
accommodate the electron in the atom.
 Thomson discovered the electron in his cathode ray
tube, since cathode rays are deflected by electric and
magnetic fields.
1. Thomson used his cathode ray tube to measure q/m
ratio for electrons and protons. He did not know the
exact mass (m) or charge (q) for ele&proton.
Fm  Fc
mv 2
qvB 
r
q
v

m rB
 Thomson’s experiments showed that q/m for an electron
is roughly 1011 C/kg. This ratio is over a 1000 times
larger than the q/m for a hydrogen ion [explain]
Ex1: A charged object is moving at a speed of 3.5 x 104
km/s perpendicular to a magnetic field with a
magnitude of 0.60 T. If the radius of deflection is 0.33
mm, find charge-to-mass ratio and name the charged
object.
 Thomson used perpendicular crossed electric and
magnetic fields to determine the speed of the cathode
rays. For electrons passing undeflected through the fields,
forces acting on electrons are balanced:
Fe  Fm
qE  qvB
v
E
B
Ex2: A beam of charged particles moves undeflected
perpendicular to electric and magnetic fields. The
magnetic field has a magnitude of 2.0 10–3 T. The
electric field is produced by two parallel plates 3.0 cm
apart with a potential difference of 3.0 102 V. Find the
speed of the particles.
 An atom is a positive sphere of electricity in which
electrons are embedded (atom is not the smallest part of
matter and it is divisible). Thomson knew:
An atom had an equal number of protons and electrons
(neutral).
The protons had more mass than electrons but same
charge.
The size of the atom was10-10 m.
J.J. found that regardless of the type of cathode
material used, he got the same q/m, therefor C.R.
particles are contained in all types of matter.
 CR particles are building blocks of matter!
 Dalton’s billiard ball model is inadequate.
Thomson’s model was known as the plum-pudding model
or the raisin bun model.
Mass of electron and proton were found after Millikan
found the elementary charge (1.60 x 10-19C) and that is
next topic.
Tompson’s Cathode-Rays worksheet
Name: ………….
4
Ex1: A charged object is moving at a speed of 3.5 x 10 km/s perpendicular to a magnetic field with a magnitude of
0.60 T. If the radius of deflection is 0.33 mm, find charge-to-mass ratio and name the charged object.
Ex2: A beam of charged particles moves undeflected perpendicular to electric and magnetic fields. The magnetic
field has a magnitude of 2.0 10–3 T. The electric field is produced by two parallel plates 3.0 cm apart with a
potential difference of 3.0 102 V. Find the speed of the particles.
1.
What was the outcome of Thomson's experiment with cathode rays?
2.
Name the four parts of a cathode ray tube.
3.
Draw a sketch of the plates that produce an electric field.
4.
What happens to the cathode ray in the presence of an electric field?
5.
Which way is the ray deflected?
6.
What happens to the cathode ray in the presence of a magnetic field?
7.
Which way is the ray deflected?
8.
What happens to the cathode ray when both the electric field and the magnetic fields are present?
9.
Explain why the cathode ray is deflected upward in the presence of an electric field.
10.
Describe the behavior of particles in the presence of a magnetic field?
11.
How did Thomson solve the problem of deflections?
12.
What were his conclusions?
13.
What was he able to calculate?
14.
What fact did he find even though he used different sources of cathode rays?
15.
What was the mass to charge ratio?
16.
What was the name given to the new particle?
15.2 Quantization of Charge –Millikan experiment
Millikan found the charge on the electron and showed that
it was a fundamental unit of electrical charge. All charges
are multiple of 1.6 x 10-19 C.
Millikan’s Oil-drop Experiment
X-Rays to
ionize drops







Robert Millikan measured the charge on the electron.
A fine spray of oil falls through a hole into a chamber
where the drops can be observed.
The plates at the top and bottom of the chamber are
charged (the top plate is positive).
X-rays are shot onto the oil drops which cause the
drops to be negatively charged.
In the absence of voltage, the force on the drops is
determined by their mass only, weight, Fg.
When a voltage is applied, negatively charged drops
will slow down, stop or begin moving upwards. The
behavior of the drop is determined by the applied
voltage and the charge on the oil drop and the weight
of the oil drop.
Millikan used these measurements to determine that
the charges on the drops were multiples of 1.6 x 1019
C.
[Diagrams showing: suspended, constant speed,
accelerating up &down][Copy from board]
[Note for the same mass Fg vector has same lenghth and
points downwards]
Ex- Latex sphere with a mass of 5.00 10–16 kg is
suspended in an electric field between plates directed
upwards that are 9.00 10-3 m apart. To keep the sphere
suspended, the potential difference between the plates is
55 V. How many elementary charges has the oil drop?
Ex – A positively charged particle with a mass of 4.5 x 1014
kg is accelerating upwards at 1.5 m/s2 under the
influence of an electric field between two horizontal plates
separated by 45 mm. If the potential difference across the
plates is 110 V, what is the charge on the particle? (make
FBD diagram and lines of electric fields)
The Discovery of the Nucleus
The Rutherford Model - Alpha, 2+, scattering experiment
In 1911, Ernest Rutherford
directed +2-particles at a thin
gold foil. They placed a zinc
sulphide screen near enough to
the foil to detect any alpha
particles that got through.
The zinc sulphide would give
off a flash of light whenever struck by an -particle.
They expected that the -particles would go right
through the foil with hardly any deflection because in the
Thomson model, the positive and negative electric charges
inside an atom were assumed to be uniformly distributed
through a solid atom. Consequently the positively charged
-particles would only encounter weak net electric forces
and so pass through the thin foil with only slight
deflections (less than a degree).
Thomson model
Rutherford model
The
-particles undeflected
-particles scattered
Results of the alpha-scattering experiment were quite
different. Rutherford made three observations:
1.Most alpha particles were undeflected.
2. Some were scattered (deflected)
3.Some were reflected.
Rutherford’s explanation of these observations:
1. atom must be largely empty space surrounding a tiny
nucleus,
2. atom positive charge and nearly all its mass are
concentrated in nucleus,
3. electrons revolve around the nucleus like planets revolve
around the sun. This model is known as the planetary
model of the atom. [Thomson model failed and
modified]
http://micro.magnet.fsu.edu/electromag/java/rutherford/
The fact that an occasional alpha particle can be stopped
by a gold nucleus and returned along its initial path
provides with a way of estimating an upper limit for the
size of the nucleus using conservation of energy:
Ek=Ep=Fe x d
Drawbacks
1. Emission and absorption Spectral lines cannot be
explained by Rutherford model.
Helium emission spectrum
Absorption spectrum
According to Maxwell, an accelerating charge gives off
energy. Then the orbiting
electron, with centripetal
acceleration, should continuously
radiate energy and spiral into the
nucleus, which it does not do.
The Bohr Model of the Atom
In 1913 Bohr presented a model of the atom that begins
with Rutherford's picture of an atom as a nucleus
surrounded by electrons moving in circular orbits.
Bohr's work was primarily with the hydrogen atom.
Spectroscopy
A diffraction grating can spread light out into a spectrum
with colours according to their wavelengths
continuous
spectrumcontains all
wavelengths
Hot dense, solid,
liquid, or gas,
material
emission line
spectrum bright
selected lines at
distinct
wavelengths
Hot low dense
gas
absorption line
spectrum dark
selected
wavelengths
Cold low dense
gas
 Lines are ordered from short to high wavelength.
 Each element has its unique spectrum lines. Kirchhoff
used spectra to identify unknown elements.
http://jersey.uoregon.edu/vlab/elements/Elements.html
 No one had, before Bohr, explained why this pattern
occurs. The reason why elements produce spectral lines
was still not explained.
Balmer Series
 The hydrogen atom bright line spectra were a major key
to understanding atomic structure. In this spectrum there
are four lines in the visible region. Johann Balmer, a
high school teacher, searched for a mathematical
relationship and found, by trial and error, that he could
calculate the wavelength of each of the four lines in the
hydrogen spectrum using the formula:
1

 RH (
1
1

)
2
2
nf
ni
nf = 2
RH=Rydberg's constant=1.10x107/m
ni=3,4,......(only for the Balmer series).
Balmer did not know why his formula worked. Balmer
predicted other lines in the UV and infra-red region.
http://www.avogadro.co.uk/light/bohr/spectra.htm
Ex - Find the wavelength and frequency of the spectral
line ni=3 in the Balmer series
1/=RH(1/nf2-1/ni2)
=1.10x107(1/4-1/9)
=6.55x10-7m
v=f
3.00x108=fx6.545454x10-7
f=4.58x1014 Hz
Bohr used Balmer idea and Planck’s quantum ideas to
establish a new atomic model.
Bohr made the following assumptions: In hydrogen atom
1. there can be only certain values
of the total energy (electron's
kinetic energy +potential energy).
Quantized energy levels.
2. These allowed energy levels
correspond to different orbits for
the electron as it moves around
the nucleus.
3. Larger orbits have larger energies. *4. When moving in an allowed orbit, called it stationary
state, the electron is exempt from the classical laws of
electromagnetism and does not radiate energy as it
moves along its orbital path.
REM: In other cases when electrons are accelerated they emit
(Radio or microwave) EMR as Maxwell stated.
5. The energy of the electron in an atom is quantized
(used Plank idea).
6. If an electron goes from a higher to a lower energy
level, energy is given off (emission spectrum). If an
electron goes from a lower level to a higher level,
energy is absorbed (absorption spectrum).
http://www.colorado.edu/physics/2000/quantumzone/bohr.html
Consequences of the Bohr Model
1. Radius of the hydrogen orbit
Bohr’s model of the hydrogen atom states that electrons
can orbit the nucleus only at specific locations given by
the formula:
rn=r1n2
rn=radius of the nth orbit (any orbit).
r1=radius of the first orbit=5.29x10-11m (see formula sheet)
‘r1 is the radius of the lowest possible energy level or
ground state of the hydrogen atom.
n=1,2,3,....
Complete the chart:
Level Radius of orbit
(n)
(m)
1
5.29x10-11
2
3
4
5
Level
(n)
1
2
3
4
5
Radius of orbit
(m)
5.29x10-11
2.12x10-10
4.76x10-10
8.46x10-10
1.32x10-9
2.Energy levels for the allowed orbits of
hydrogen
E1
En  2
n
En=Energy of nth level
E1=Energy of first level=-13.6 eV
n=1,2,3,....
Orbit Energy of Frequency
(n)
orbit (eV)
Hz
E=hf
1
-13.6 eV
2
3
4
5

Orbit
(n)
1
2
3
4
5

Energy of
orbit (eV)
-13.6
-3.40
-1.51
-0.850
-0.544
0.0
 The energy of the energy levels
is always negative (energy
levels like a well under ground
level). This is because the
energy at ∞ is 0 and the closer
to the nucleus the less the
Wavelength
(m)
c=λf
energy. At ∞ is the greatest possible energy.
 The closer the electron to the nucleus the more energy
needed to free the electron from the atom (ionization
energy).
 an electron making a transition from lower to a higher
energy level results in a dark-line spectrum
 an electron making a transition from a higher to
a lower energy level results in a bright-line spectrum
 The energy in an electron transition between energy
levels can be calculated by:
E = Ef – Ei
E = energy of transition (absorbed or emitted), J or eV
Ef = energy of the final energy level, J or eV
Ei = energy of the initial energy level, J or eV
 We can calculate the wavelength and frequency of the
absorbed or emitted photon from the transition:
Ephoton = Ef – Ei = hf =hc/λ
 The energy of the photon is always positive.
 An electron excited to higher energy levels does not
have to return to ground state in a single jump. They do
it in few intermediate steps. At each step, a photon is
emitted. Each photon emitted has a smaller energy.
Example
-0.38eV
‘n=6
-0.54eV
‘n=5
An electron in hydrogen
-0.85eV
‘n=4
atom, shown, undergoes a transition
‘n=3
-1.5 eV
th
from the 6 energy level to the 2nd
-3.4eV
energy level. What is the frequency and ‘n=2
‘n=1
-13.6eV
wavelength of light emitted? And what type?
E2 =-3.40 eV
E6 = -0.378 eV
Ephoton=Ef-Ei= E6 – E2 = -3.40-(-0.377777)=
=3.022223eV x (1.6 x 10-19)
= 4.83555 x 10-19 J
=hf
f=7.29344 x 1014Hz
v=fx
3.00x108=7.29344 x 1014
=4.11x10-7m
Type: Blue light
Example
An electron undergoes a transition
‘n=6
from the 2nd level to 4th level in a ‘n=5
‘n=4
hydrogen atom. What is the
‘n=3
frequency and wavelength of the
‘n=2
radiation absorbed?
‘n=1
E=Ef – Ei =E2-E4
=-0.850-(-3.40)
f=6.15x1014 Hz
c=f
3.00x108=6.15x1014
=4.88x10-7 m
-0.38eV
-0.54eV
-0.85eV
-1.5 eV
-3.4eV
-13.6eV
States of the atom
1. When an electron has the smallest allowable amount of
energy, it is in the lowest energy level called the ground
state(stationary). The atom is stable.
2. If an electron absorbs energy, it can make transition to a
higher energy level called an excited state. Atomic
electrons remain in excited levels only a fraction of a
second before returning to the ground state and emitting
energy (neon light is an exited neon gas).
3. The ionization state level is the state at which the
electron is totally removed from the atom to E=0. This
is equivalent to work function in photoelectric effect
http://www.avogadro.co.uk/light/bohr/spectra.htm
Exact match – which photon will be absorbed?
http://www.stmary.ws/physics/home/animations3/modernPhysics/bohr_transitions.html
The Northern Lights and the Emission Line
Spectrum of Oxygen
Alberta skies often display aurora
borealis, or northern lights. At
high altitudes above the surface
of Earth, high-energy electrons,
trapped by Earth’s magnetic field,
interact with oxygen and nitrogen
atoms. During these interactions,
the electrons in these atoms are
excited and move into higher
energy levels. Eventually, the
excited electrons return to their
ground states. In doing so, they
emit light that forms the aurora
borealis.
Physics 30 Practice Examples:
1. Find the energy needed to ionize a hydrogen atom
whose electron is in n=4.
(E=Ef-Ei=0-(-0.850)=
0.850 eV
2. A hydrogen atom absorbs a photon of wavelength
434.1 nm.
(a)How much energy did the atom absorb?
(b)What were the initial and final states of the hydrogen
electron?
(a)E=hc/(=6.63x10-343.00x108/434.1x10-9
=2.86eV
(b) Using the energy level
diagram n=2 to n=5
3. An electron dropped from n=4 to n=1. How many
spectral lines could be produced ?
6 lines
4. If the wavelength of emitted spectrum is 486 nm,
what are the corresponding stationary states?
E=hc/λ(=6.63x10-34 x3.00x108/486x10-9
=4.09259x10-19 J
=2.55 eV. The two levels that have a difference of 2.55 eV
are the second and the fourth.
5. Find the longest wavelength photons that will be
absorbed by a hydrogen atom in its ground state.
Longest wavelength=smallest energy absorbed=n=1to
n=2.
(=1.22x10-7m
n=1to n=3 (=1.02x10-7 m
n=1to n=4 (=9.75x10-8m
Physics 30 Practice Examples:
Name………………..
1. An electron in a hydrogen atom drops from the 4th to 2nd energy level. Is this an absorption or
emission photon? Find the frequency and wavelength of the photon. (Make a diagram to show the
transition)
2. Find the energy needed to ionize a hydrogen atom whose electron is in n=4.
3. A hydrogen atom absorbs a photon of wavelength 434.1 nm.
a) How much energy did the atom absorb?
b) What were the initial and final states of the hydrogen electron?
4. An electron dropped from n=4 to n=1. How many possible spectral lines could be produced? (make
diagram)
5. Find the longest wavelength photons that will be absorbed by a hydrogen atom in its ground state.
‘n=6
5. An electron undergoes a transition from the 2nd level to 4th level in a
hydrogen atom. What is the frequency and wavelength of the radiation ‘n=5
‘n=4
absorbed?
‘n=3
-1.5 eV
‘n=2
-3.4eV
‘n=1
-13.6eV
-0.38eV
-0.54eV
-0.85eV
Successes of the Bohr model
1. It explained the properties of elements.
2. It explained the hydrogen spectra.
Drawbacks of the Bohr model
1. It could not explain the spectra of elements containing
many electrons.
2. It could not explain why only certain orbits were
allowed (see 5).
3. It does not explain why each spectral line split into
several lines when put in a magnetic field or electric
field.
4. It does not explain the relative intensities of spectral
lines.
5. The wave nature of the electron is not used in this
model.
The quantum mechanical model
In about 1920 Heisenberg and
Schroedinger developed a quantum
mechanical model to describe the nature
of the atom.
 It is a mathematical model.
 The atom cannot be visualized.
 There are no exact levels. The position of the
electron is given in terms of probabilities. These
probabilities are called orbital (clouds).
END OF CHAPTER 15