GHS Honors Chem

How would we write this Nuclear Reaction?
GHS Honors Chem

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Naturally occurring Uranium contains two major isotopes
Uranium-238 (99.3%)
Uranium-235 (0.7%)
As it turns out the only isotope of
Uranium that undergoes fission is
Uranium-235
Why do you think U-235 can undergo fission?
GHS Honors Chem

235
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235
92
U + 1
0 n  236
92
U * and 10 -14 seconds later...
236
92
U *  92
36
Kr + 141
56
Ba + 3 1
0 n + ENERGY
50 possible sets of fission products (sum of atomic numbers = 92)
3 neutrons released for ONE 235
92
U each neutron can split another 235
92
CHAIN REACTION POSSIBLE
U
If amount of 235
92
U is sufficient ( CRITICAL
MASS ) then the number of neutrons generated is high enough to result in a nuclear explosion )
GHS Honors Chem

235
235
92
U + 1
0 n  92
36
Kr + 141
56
Ba + 3 1
0 n + E
GHS Honors Chem

The Fission Chain
Reaction Begins …
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Representation of a fission chain process.
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The Critical Mass Needed for the Chain
Reaction
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Are there other Fuels for the Fission
Chain Reaction?
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Uranium-235 . (Keep in mind, the abundant form of
Uranium found in nature is Uranium-238 (~99%).
Plutonium-239 … but Plutonium-239 is scarce … now what?
Here’s what scientists discovered during World War II:
238
92
U + 1
0 n ---> 239
92
U + g
239
92
U
239
93
Np
---> 239
---> 239
93
94
Np + 0
Pu + 0
-1
-1 b b

• During WW2 the American set up a secret project called the “Manhattan Project” recruiting all the best physicists and nuclear chemists available
• By 1945 they developed two types of nuclear bombs
• A U-235 fission bomb
• and a Pu-239 fission bomb

Little Boy and Fat Man
10 feet

The Arrangement of Uranium in the First
Atomic Bombs

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Arranging U-235 to get a Controlled
Nuclear Reaction
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Arrangement of U-235 in a critical mass, where there’s enough U-235 in a small space so that the neutrons cause a chain reaction, would be difficult to control in a nuclear power plant.
How can this conversion of small amounts of mass into great amounts of energy be used to produce electricity?
In the 1930, Enrico Fermi, an Italian Physicist, discovered that Uranium can be arranged in what he called a Fission Pile , where the Uranium would be arranged in rows, or in a lattice, so that the fission reaction could be controlled.

Arranging U-235 to get a Controlled
Nuclear Reaction
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By arranging the Uranium fuel in a lattice structure,
CONTROL RODS could be inserted between the
Uranium fuel rods to stop the flow of neutrons flowing between the rods, and thus, slowing/stopping the reaction.
Water is used to slow the neutrons down, within the reactor, so that they are more likely to strike the
Uranium fuel rods … water is the MODERATOR , which is another discovery by Fermi.

Arranging U-235 to get a Controlled
Nuclear Reaction
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By arranging the Uranium fuel in a lattice structure, there is NO POSSIBILITY of an uncontrolled nuclear explosion , like in a nuclear warhead or bomb. Simply stated, there Uranium is not in a close-enough proximity for an uncontrolled chain reaction, or blast.
So, this is an important point: the threat of a nuclear explosion, within a nuclear power plant, is an imagined danger.

A Schematic of a Nuclear Power Plant
• Walls of reactors are made of
Graphite , which slows down the neutrons, increasing the chance that neutrons will strike U-235.
• Control Rods made of Cadmium or
Boron , which absorb large quantities of neutrons, and stop the flow of neutrons w/o dissipation of energy.
• Also use water as a moderator to slow down fast moving neutrons
• Modern reactors have a containment building make of 6 feet of concrete, used to contain radioactive material in the event of an accident.
GHS Honors Chem

A Schematic of a Nuclear Power Plant
• With a Fission
Pile, its impossible to have a Nuclear
Explosion.
• What type of
Explosion is possible?
GHS Honors Chem http://www.nrc.gov/reading-rm/basic-ref/students/reactors.html

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Currently about 103 nuclear power plants in the U.S. and about 435 worldwide.
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17% of the world’s energy comes from nuclear.
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We’ll take a closer look at Nuclear
Power Plant Mishaps in a bit …
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Nuclear Fusion
small nuclei combine
2 H + 3 H 4 He + 1 n +
1 1 2 0
Energy
Occurs in the sun and other stars
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Nuclear Fusion
Fusion
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Excessive heat can not be contained
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Attempts at “cold” fusion have FAILED.
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“Hot” fusion is difficult to contain
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Recall Mass Defect
It is experimentally observed that the mass of an atom (containing neutrons) is always slightly less than the sum of the masses of its component particles. The difference between the atomic mass and the sum of the masses of its protons, neutrons, and electrons is called the mass defect.
Can we calculate the Energy given off by a nuclear reaction based on the weight difference from reactants to products based on Einstein’s Equation,
E = mc 2 ?
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Calculation of Nuclear Energy Released
Calculate the energy released (per mole of tritium consumed) for the fusion reaction of tritium and deuterium,
3
1
H + H
1

He + n
2 0 given the following molar masses of nucleons and nuclei. (c =
3.00  10 8 m/s) mass (g/mol)
Mass Reactants = 3.01605 + 2.0140g
Mass Reactants = 5.03005g
particle proton 1.007825
Mass Products = 4.00260 + 1.008665g
Mass Products = 5.011265 g neutron 1.008665
deuterium 2.0140
tritium 3.01605
Mass Diff. = Mass
Reactants
– Mass
Products
Mass Difference = 0.018785 g
Mass Difference = 1.8785x10
-5 kg helium 4.00260
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E = mc 2 = 1.8785x10
-5 kgx(3x10 8 m/s) 2
E = 1.69 x 10 12 kg-m 2 /s 2 = 1.69 x 10 12 J

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HALF-LIFE is the time that it takes for 1/2 a sample to decompose.
The rate of a nuclear transformation depends only on the “reactant” concentration.
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Decay of 20.0 mg of 15 O. What remains after 3 half-lives?
After 5 half-lives?
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Kinetics of Radioactive Decay
For each duration (half-life), one half of the substance decomposes.
For example: Ra-234 has a half-life of 3.6 days
If you start with 50 grams of Ra-234
After 3.6 days > 25 grams
After 7.2 days > 12.5 grams
After 10.8 days > 6.25 grams
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Half–life is the time required for one-half of any given quantity of isotope to decay
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Learning Check!
The half life of I-123 is 13 hr. How much of a 64 mg sample of I-123 is left after 39 hours?
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Learning Check!
The half life of I-123 is 13 hr. Only 4 mg are present from a sample that had originally contained 128 mg sample of
I-123. How old is the sample?
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Learning Check!
The half life of I-123 is 13 hr. Only 2 mg are present after 65 days have passed. What was the original amount of sample present?
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How Old is Oetzi, the Iceman?
Oetzi the Iceman was discovered on the border between Italy and Austria in 1991, in the Otztal
Alps, partially submerged in a melting glacier.
• It’s estimated that he is from the Neolithic age, some 5300 years ago.
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How Old is Oetzi, the Iceman?
• Oetzi was discovered in an extraordinarily warm summer season, which led to abnormal melting of glacier ice.
• It’s estimated that Oetzi, along with animals and plants from his time, were covered by a sudden, deep snow, followed by a prolonged cold period.
• Examination of the body suggests that Oetzi died from an arrow wound to his back/shoulder.
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 Radioactive C-14 is formed in the upper atmosphere by nuclear reactions initiated by neutrons in cosmic radiation
14
7
N + 1 o n ---> 14
6
C + 1
1
H
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The C-14 is oxidized to CO
2
, which circulates through the biosphere.
When a plant dies, the C-14 is not replenished.
But the C-14 continues to decay with t
1/2 years.
= 5730
Activity of a sample can be used to date the sample.
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How Old is Oetzi, the Iceman?
• How much C-14 would be remaining in
Oetzi’s bones, given his estimated age of
5300 years? The T years.
1/2 of C-14 is 5730
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C-14 Dating Solves The Shroud of
Turin (Torino) Question …
GHS Honors Chem

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The Shroud bearing the image of Jesus should be approximately 2000 years old
C-14 analysis revealed the presence of 92% of the original amount of C-14
This decrease of 8% corresponds to a time period of about 700 years
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Learning Check!
The C-14 content of a fossilized bone is
12.5% of that of a live bone. Knowing that the t
1/2 of C-14 is 5730 years, how old is the artifact?
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1/2
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