Nuclear

Report
Isotopes
• Variants of a chemical element
– All isotopes of a given element have the same # of
protons
– Differ in the number of neutrons
Hydrogen
Deuterium
Tritium
Isotopes
• General Form: AXZ
– Chemical symbol for the element: X
– Atomic # (# of Protons): Z
– Protons + Neutrons: A
Hydrogen
Deuterium
1H
1
2H
1
Tritium
3H
1
Radioactive Decay
• Alpha Decay
•
238U
92
=> 234Th90 + 4He2
• Beta Decay
–
–
–
–
–
–
–
Electron Emission (β-)
3H => 3He + 0e
1
2
-1
234Th => 234Pa + 0e
90
91
-1
Electron Capture
40K + 0e => 40Ar + hv
19
-1
18
Postitron Emission (β+)
40K => 40Ar + 0e
19
18
+1
Trinity Test
• Trinity Test Video
Early History of the Bomb
• 1931—Crockroft & Walton split the atom
• 1932—Chadwick discovers the neutron (Nobel
Prize 1953
• 1934—Joliot & Curies bombard a target to
produce new elements using α particles
α
Alpha Particle
Thin Foil 7Li
11B
Early History of the Bomb
• Fermi repeats bombarding experiment with
neutrons and finds:
– Uranium produces several radioactive by
products.
α
β
γ
Neutron
N
β
Uranium
γ
Uranium-235 Fission
• 1N0 + 235U92  236U92 144Ba56 + 89Kr36 + 3N +
ENERGY
• Protons + Neutrons  1 + 235 = 236
• Protons  0 + 92 = 92
• Protons + Neutrons  144 + 89 + 3 = 236
• Protons  56 + 36 = 92
Early History of the Bomb
• 1938—Hahn & Stassmann prove that Fermi
observed fission and published 12/22/1938.
– Fermi Wins Nobel Prize.
• 1939—Frisch and Meitner describe fission and
the potential for large amounts of energy to
be released.
– The Question? Are neutrons liberated in the
process??
Early History of the Bomb
• 1939—Leo Szilard
confirms that
neutrons are
produced and an
explosive chain
reaction is possible.
Early History of the Bomb
• 1939—April 22. Letter in Nature by Joliot
confirmed that excess neutrons are produced
and a chain reaction is confirmed.
• World War II begins.
Fission of Uranium
• Mass of a Neutron = 1.008 u
Fission of Uranium
• Mass of a Neutron = 1.008 u
• Mass of 235U = 235.044 u
Fission of Uranium
• Mass of a Neutron = 1.008 u
• Mass of 235U = 235.044 u
• Mass of 144Ba56 = 143.923 u
Fission of Uranium
•
•
•
•
Mass of a Neutron = 1.008 u
Mass of 235U = 235.044 u
Mass of 144Ba56 = 143.923 u
Mass of 89Kr36 = 88.918 u
Fission of Uranium
•
•
•
•
Mass of a Neutron = 1.008 u
Mass of 235U = 235.044 u
Mass of 144Ba56 = 143.923 u
Mass of 89Kr36 = 88.918 u
• 1N0 + 235U92  236U92 144Ba56 + 89Kr36 + 3N +
ENERGY
Fission of Uranium
•
•
•
•
Mass of a Neutron = 1.008 u
Mass of 235U = 235.044 u
Mass of 144Ba56 = 143.923 u
Mass of 89Kr36 = 88.918 u
• 1N0 + 235U92  236U92 144Ba56 + 89Kr36 + 3N +
ENERGY
• 1.008 + 235.044 = 236.052
Fission of Uranium
•
•
•
•
Mass of a Neutron = 1.008 u
Mass of 235U = 235.044 u
Mass of 144Ba56 = 143.923 u
Mass of 89Kr36 = 88.918 u
• 1N0 + 235U92  236U92 144Ba56 + 89Kr36 + 3N +
ENERGY
• 1.008 + 235.044 = 236.052
• 143.923 + 88.918 + 3(1.008) = 235.865
Fission of Uranium
•
•
•
•
Mass of a Neutron = 1.008 u
Mass of 235U = 235.044 u
Mass of 144Ba56 = 143.923 u
Mass of 89Kr36 = 88.918 u
• 1N0 + 235U92  236U92 144Ba56 + 89Kr36 + 3N +
ENERGY
• 1.008 + 235.044 = 236.052
• 143.923 + 88.918 + 3(1.008) = 235.865
• Δ M = 0.187 u
Fission of Uranium
• Δ M = 0.187 u
• 1 u = 1.66 x 10-27 kg
Fission of Uranium
• Δ M = 0.187 u
• 1 u = 1.66 x 10-27 kg
• E = mc2
Fission of Uranium
•
•
•
•
Δ M = 0.187 u
1 u = 1.66 x 10-27 kg
E = mc2
E =(1.66 x 10-27) (0.187kg) x (2.99 x 108)2
Fission of Uranium
•
•
•
•
•
Δ M = 0.187 u
1 u = 1.66 x 10-27 kg
E = mc2
E =(1.66 x 10-27) (0.187kg) x (2.99 x 108)2
E = 2.775 * 10-11 Joules
Fission of Uranium
•
•
•
•
•
Δ M = 0.187 u
1 u = 1.66 x 10-27 kg
E = mc2
E =(1.66 x 10-27) (0.187kg) x (2.99 x 108)2
E = 2.775 * 10-11 Joules
Note: A unit we like to use is the electron volt (eV). This is the energy
an electron will gain as it moves across an electric potential difference
of one volt. 1eV = 1.602 x 10-19 Joule
Fission of Uranium
•
•
•
•
•
Δ M = 0.187 u
1 u = 1.66 x 10-27 kg
E = mc2
E =(1.66 x 10-27) (0.187kg) x (2.99 x 108)2
E = 2.775 * 10-11 Joules
Note: A unit we like to use is the electron volt (eV). This is the energy
an electron will gain as it moves across an electric potential difference
of one volt. 1eV = 1.602 x 10-19 Joule
2.775 x 10-11 J x
1 eV
1.602 x 10-19 J
=
1.73 x 108 eV
=
173. x 106 eV
Fission of Uranium
•
•
•
•
•
Δ M = 0.187 u
1 u = 1.66 x 10-27 kg
E = mc2
E =(1.66 x 10-27) (0.187kg) x (2.99 x 108)2
E = 2.775 * 10-11 Joules
Note: A unit we like to use is the electron volt (eV). This is the energy
an electron will gain as it moves across an electric potential difference
of one volt. 1eV = 1.602 x 10-19 Joule
2.775 x 10-11 J x
1 eV
1.602 x 10-19 J
=
1.73 x 108 eV
=
=
173. x 106 eV
173 MeV
So What’s the Big Deal??
• If 2.775 * 10-11 Joules of energy are released
in one fission
– What if 1 Kg of Uranium 235 fissions?
– How much energy is released?
So What’s the Big Deal??
• If 2.775 * 10-11 Joules of energy are released
in one fission
– What if 1 Kg of Uranium 235 fissions?
– How much energy is released?
1 mole 235U92 = 235.044g
1 Kg = 1000 g
So What’s the Big Deal??
• If 2.775 * 10-11 Joules of energy are released
in one fission
– What if 1 Kg of Uranium 235 fissions?
– How much energy is released?
1 mole 235U92 = 235.044g
1 Kg = 1000 g
1000 g 235U92 x 1 mole 235U92
235.044 g 235U92
=
4.25 Moles 235U92
The Big Deal
4.25 Moles 235U92 x
6.022 x 1023 atoms
= 2.56*1024 Atoms 235U92
mole
The Big Deal
4.25 Moles 235U92 x
6.022 x 1023 atoms
= 2.56*1024 Atoms 235U92
mole
If 1 Kg of 235U92 fissions, we get:
2.775 X 10-11 Joule x 2.56*1024 Atoms 235U92 = 7.10* 1013 Joules
Atom
The Big Deal
4.25 Moles 235U92 x
6.022 x 1023 atoms
= 2.56*1024 Atoms 235U92
mole
If 1 Kg of 235U92 fissions, we get:
2.775 X 10-11 Joule x 2.56*1024 Atoms 235U92 = 7.10* 1013 Joules
Atom
As a comparison: 1 TON of TNT
4.184* 109 Joules
The Big Deal
4.25 Moles 235U92 x
6.022 x 1023 atoms
= 2.56*1024 Atoms 235U92
mole
If 1 Kg of 235U92 fissions, we get:
2.775 X 10-11 Joule x 2.56*1024 Atoms 235U92 = 7.10* 1013 Joules
Atom
As a comparison: 1 TON of TNT
1 Kg of
235U
92
7.10* 1013 Joules
4.184* 109 Joules
4.184* 109 Joules
=
1.7* 104 Tons of TNT
17,000 Tons of TNT
Fission Videos
•Nuclear Fission
•Chain Reaction
Fission Cross Sections
The Problem
Fission occurs in about 10-8 seconds
80 Generations pass in 0.8 microseconds
It takes less than a millionth of a second to fission a kg of 235U
The Solution
Use a gun to shoot the slug in
Little Boy Gun Type Weapon
Equivalent to 20,000 Tons of TNT
Fat Man Bomb
Fat Man Bomb
Homework
• Given the Equation:
• 1N0 + 235U92  236U92 144Ba56 + 89Kr36 + 3N + ENERGY
– What percentage of the mass is converted to energy?

similar documents