Chapters
3 & 22
Kishan Alluri
Christina Costeas
Sandy Jiang
Patrick Morgan
Ben Wong

http://www.youtube.com/watch?v=XU8nMKk
zbT8
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

Born in Manchester, England in December
1856
Studied physics and mathematics
Won a nobel prize in 1906
1.
2.
3.
4.
An object placed between the cathode and
the opposite end of the tube cast a
shadow on the glass.
A paddle wheel placed on rails between
the electrodes rolled along the rails from
the cathode toward the anode.
Cathode rays were deflected by a magnetic
field in the same manner as a wire
carrying electric current, which was known
to have a negative charge.
The rays were deflected away from a
negatively charged object.




Experiments supported hypothesis that
the particles that compose cathode rays
are negatively charged
Measured the ratio of cathode-ray
particles to their mass—found it was
always the same
Concluded that all cathode rays are
composed of identical negatively charged
particles called electrons
Experiments revealed the electron has a
very large charge for its tiny mass
By: Katrina Leung,
Jesse Stathis, Jenna Taormina,
Danielle Zhang, Matt Odea




Electrons are really small
Mass of electron 1/2000 the mass of a
hydrogen atom
Electron has a mass of 9.109 * 10^-31kg or
1/1837 the mass of a hydrogen atom
Electrons carry a negative charge
http://www.youtube.com/watch?v=XMfYHag7Liw
1.
2.
Because atoms are electrically neutral, they
must contain a positive charge to balance
the negative electrons.
Because electrons have so much less mass
than atoms, atoms must contain other
particles that account for most of their
mass.
As flawlessly explained by: Adam, Jane, Justin, Jeremy,
rob
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Born 1871 in New
Zealand
In early work,
discovered radioactive
half-life
Had an element named
after himrutherfordium
Became known as the
father of nuclear
physics



Assistants Geiger and Marsden
bombarded a thin piece of gold foil
with a narrow beam of alpha
particles.
Some of the particles were
redirected by the gold foil back
towards their source.
Rutherford thus concluded that the
force must be caused by a very
densely packed bundle of matter
with a positive charge, which he
called the nucleus.

Rutherford had discovered that the volume of the nucleus was
very small compared to the total volume of the atom,
suggesting that there was a lot of empty space.
By Monica, Santosh, Jennie, Antonella, and
Drew

In 1932, Chadwick discovered a previously
unknown particle in the atomic nucleus.3
This particle became known as the neutron
because of its lack of electric charge.
Chadwick's discovery was crucial for the
fission of uranium 235.

mass # = protons + neutrons
 always a whole
number
 NOT on the
Periodic Table!
© Addison-Wesley Publishing Company, Inc.

Atoms of the same element with different
mass numbers.
 Nuclear symbol:
Mass #
Atomic #
12
6
 Hyphen notation: carbon-12
C
© Addison-Wesley Publishing Company, Inc.

Chlorine-37
◦ atomic #: 17
◦ mass #: 37
◦ # of protons:17
◦ # of electrons:17
◦ # of neutrons:20
37
17
Cl

Chlorine-35
◦ atomic #: 17
◦ mass #: 35
◦ # of protons:17
◦ # of electrons:17
◦ # of neutrons:18
35
17
Cl
 12C
atom = 1.992 × 10-23 g
 atomic mass unit (amu)
 1 amu = 1/12 the mass of a
12C
atom
 1 p = 1.007276 amu
1 n = 1.008665 amu
1 e- = 0.0005486 amu
C. Johannesson
© Addison-Wesley Publishing Company, Inc.



weighted average of all isotopes
on the Periodic Table
round to 2 decimal places
Avg.
Atomic
Mass
(mass)(% )  (mass)(% )

100

EX: Calculate the avg. atomic mass of oxygen if its
abundance in nature is 99.76% 16O, 0.04% 17O, and
0.20% 18O.
Avg.
(16)(99.76 )  (17)(0.04)  (18)(0.20)
 16.00
Atomic 
100
amu
Mass

EX: Find chlorine’s average atomic mass if
approximately 8 of every 10 atoms are chlorine-35
and 2 are chlorine-37.
Avg.
Atomic
Mass
(35)(8)  (37)(2)

 35.40 amu
100


Atom = nuclide
Nucleus = nucleon

Difference between the mass of an atom
and the mass of its individual particles.
4.00260 amu
4.03298 amu



Protons - 1.007276 u
Electrons – 0.0005486 u
Neutrons – 1.008665 u
Ex. Helium-4 has an atomic mass of 4.002602
Calculate the mass defect:
2 Protons = (2x1.007276)= 2.014552u
2 Electrons = (2x0.0005486) = 0.001097u
2 Neutrons = (2x1.008665)= 2.017330u
Total =
4.032979 u
Mass defect = 4.032979 - 4.002602= 0.030377


It is nuclear binding: (Energy released when a
nucleus is formed from nucleons)
According to E=mc2:
◦ Mass can be converted into energy and energy to
mass

The mass defect is caused by the conversion
of mass to energy upon formation of the
nucleus.
E=mc2

◦
◦
◦
E-energy (J, Joules)
m- mass (kg)
c- speed of light (3.00x108 m/s)
J= kg x m2
s2
Steps:
1. Calculate the mass defect in (u).
2. Convert to kg using 1u=1.6605X10-27kg
3. Plug in to energy equation
◦
1. Calculate the binding energy of Sulfur-32.
The measured atomic mass is 31.972070 u.
(4.36 x 10-11 J)
2. Calculate the nuclear binding energy per
mole of Oxygen-16. The measured atomic
mass of oxygen is 15.994915u.
(1.23 x 1013 J)
3. Calculate the binding energy per nucleon
of a Manganese-55 atom. It’s measured
atomic mass is 54.938047 u.
(1.41 x 10-12J)


Energy released when a nucleus is
formed from nucleons.
High binding energy = stable nucleus.
E=
2
mc
E: energy (J)
m: mass defect (kg)
c: speed of light
(3.00×108 m/s)

Calculate the binding energies of the
following two nuclei, and indicate which
nucleus releases more energy when formed.
A. Potassium–35, Atomic mass 34.988011
B. Sodium – 23, Atomic Mass 22.989767
A.
Potassium–35, Atomic mass 34.988011
4.47 X 10-11 J
B. Sodium – 23, Atomic Mass 22.989767
2.99 x 10 -11J
Potassium releases more energy
Unstable nuclides are radioactive and
undergo radioactive decay.
- Binding energy per nucleon, is the binding energy of the
nucleus divided by its number of nucleons.
- The higher the binding energy per nucleon, the more
tightly the nucleons are held together.
- Elements with intermediate atomic masses have the
greatest binding energies per nucleon and are therefore
the most stable.
What
observations
can you make?

Atoms having low atomic numbers, are the
most stable.
◦ N:P = 1:1

As atomic number increases the N:P increases
(1.5:1)

Why does this happen?
- Explained by the relationship btwn nuclear force
and electrostatic forces btwn protons.



Protons in the nucleus repel all other protons
through electrostatic repulsion.
As #p increase, the repulsive electrostatic
force between protons increase faster than
the nuclear force
More neutrons are required to increase the
nuclear force to stabilize the nucleus.


Beyond Bismuth (#83), the repulsive forces of
protons are so great that no stable nuclide
exists.
Stable nuclei tend to have even number of
nucleons

Alpha particle ()
◦ helium nucleus
◦ 2 protons and 2 neutrons bound together and
emitted from the nucleus
◦ Restricted nearly to heavy atomic nuclei
◦ Mostly because protons numbers need to be reduced
to stabilize the nuclei

Beta (-)
◦ Is an electron emitted from the nucleus during some
kinds of radioactive decay.
◦ Usually for the nuclides above the band of stability,
to decrease the number of neutrons.
◦ Electron is emitted from the nucleus as a beta
particle.

Positron particle (+)
◦ Particle has the same mass as an electron but has a
positive charge and emitted from the nucleus
◦ Proton is converted into a neutron

Alpha particle ()
◦ helium nucleus
 Beta particle (-)
 electron
 Positron (+)
 positron
4
2
He
0
-1
e
0
1
e
 Gamma ()
 high-energy photon
2+
11+
0

Alpha Emission
238
92
parent
nuclide
U
Th  He
234
90
daughter
nuclide
4
2
alpha
particle
Numbers must balance!!

Beta Emission
131
53
I
Xe  e
131
54
0
-1
electron
 Positron Emission
38
19
K  Ar  e
38
18
0
1
positron

Electron Capture
106
47
Ag  e 
0
-1
106
46
Pd
electron
 Gamma Emission
 Usually follows other types of
decay.
 Transmutation
 One element becomes another.
238
92
U
234
90
I
131
54
131
53
38
19
106
47
Th  He
4
2
Xe  e
0
-1
K  Ar  e
38
18
Ag  e 
0
-1
0
1
106
46
Pd
238
92
U
234
90
I
131
54
131
53
38
19
106
47
Th  He
4
2
Xe  e
0
-1
K  Ar  e
38
18
Ag  e 
0
-1
0
1
106
46
Pd
238
92
U
234
90
I
131
54
131
53
38
19
106
47
Th  He
4
2
Xe  e
0
-1
K  Ar  e
38
18
Ag  e 
0
-1
0
1
106
46
Pd
238
92
U
234
90
I
131
54
131
53
38
19
106
47
Th  He
4
2
Xe  e
0
-1
K  Ar  e
38
18
Ag  e 
0
-1
0
1
106
46
Pd
238
92
U
234
90
I
131
54
131
53
38
19
106
47
Th  He
4
2
Xe  e
0
-1
K  Ar  e
38
18
Ag  e 
0
-1
0
1
106
46
Pd

Why nuclides decay…
◦ need stable ratio of neutrons to protons
DECAY SERIES TRANSPARENCY



When number of protons is plotted against
number of neutrons, a belt-like graph is
obtained.
Atoms of low atomic numbers have stable
nuclei with a ratio of 1:1, p:n
As atomic number increases, the p:n
increases

Explained by nuclear and electrostatic forces
between protons.
◦ Protons in the nucleus repel each other bc of
electrostatic repulsion
◦ However Nuclear forces allow protons to attract to
other protons in close proximity

As #p increase the repulsive electrostatic
force between protons increase faster than
the nuclear force. Therefore creating an
unstable nucleus.
◦ More neutrons are required to stabilize this force.


The spontaneous disintegration of a nucleus
into a slightly lighter nucleus, accompanied
by the emission of particles, electromagnetic
radiation or both.
Leads to more stable nucleons.

Half-life (t½)
◦ Time required for half the atoms of a
radioactive nuclide to decay.
◦ Shorter half-life = less stable.
mf  m ( )
1 n
i 2
mf:
mi:
n:
n:
final mass
initial mass
# of half-lives
t/T t = time elapsed
T = Length of half life
 Fluorine-21 has a half-life of 5.0 seconds. If you
start with 25 g of fluorine-21, how many grams
would remain after 60.0 s?
GIVEN:
WORK:
t½ = 5.0 s
mf = mi (½)n
mi = 25 g
mf = (25 g)(0.5)12
mf = ?
mf = 0.0061 g
total time = 60.0 s
n = 60.0s ÷ 5.0s

The half-life of carbon-14 is 5715 years. How
long will it be until only half of the carbon-14
in a sample remains?
(5715y)

The EPA is concerned about the levels of
radon gas in homes. The half-life of radon222 isotope is 3.8days. If a sample of gas
taken from a basement contained 4.38ug of
radon-222, how much will remain after 15.2
days?
(0.274ug)

Uranium-238 decays through alpha decay
with a half life of 4.46x109years. How long
would it take for 7/8 of a sample of uranium238 to decay?
(three half-lives or 1.34 x 1010 years)

Parent Nuclide:
The heaviest nuclide of each decay series

Daughter Nuclide:
The nuclide produced by decay of the parent
nuclide


splitting a nucleus into two or more smaller
nuclei
1 g of 235U =
3 tons of coal
235
92
U


chain reaction - self-propagating
reaction
critical mass mass required
to sustain a
chain reaction




2
1
combining of two nuclei to form one nucleus of larger
mass
thermonuclear reaction – requires temp of 40,000,000
K to sustain
1 g of fusion fuel =
20 tons of coal
occurs naturally in
stars
H H
3
1
FISSION
 235U





is limited
danger of
meltdown
toxic waste
thermal pollution
FUSION


fuel is abundant
no danger of
meltdown
no toxic waste
not yet sustainable

Fission Reactors
Cooling
Tower

Fission Reactors

Fusion Reactors (not yet sustainable)

Fusion Reactors (not yet sustainable)
National Spherical
Torus Experiment
Tokamak Fusion Test Reactor
Princeton University

Transuranium Elements
◦ elements with atomic #s above 92
◦ synthetically produced in nuclear reactors and
accelerators
◦ most decay very rapidly
238
92
U  He  Pu
4
2
242
94



half-life measurements of radioactive
elements are used to determine the age of an
object
decay rate indicates amount of radioactive
material
EX: 14C - up to 40,000 years
238U and 40K - over 300,000 years

Radioisotope Tracers
◦ absorbed by specific organs and used to diagnose
diseases

Radiation Treatment
◦ larger doses are used
to kill cancerous cells
in targeted organs
◦ internal or external
radiation source
Radiation treatment
using
-rays from cobalt60.

Atomic Bomb
◦ chemical explosion is used to form a critical mass of
235U or 239Pu
◦ fission develops into an uncontrolled chain reaction

Hydrogen Bomb
◦ chemical explosion  fission  fusion
◦ fusion increases the fission rate
◦ more powerful than the atomic bomb

Food Irradiation
◦  radiation is used to kill bacteria

Radioactive Tracers
◦ explore chemical pathways
◦ trace water flow
◦ study plant growth, photosynthesis

Consumer Products
◦ ionizing smoke detectors -
241Am