Isotopes Variants of a chemical element – All isotopes of a given element have the same # of...

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Isotopes

• Variants of a chemical element– All isotopes of a given element have the same # of

protons– Differ in the number of neutrons

TritiumHydrogen Deuterium

Isotopes

• General Form: AXZ

– Chemical symbol for the element: X– Atomic # (# of Protons): Z – Protons + Neutrons: A

TritiumHydrogen Deuterium

1H13H1

2H1

Radioactive Decay• Alpha Decay • 238U92 => 234Th90 + 4He2

• Beta Decay– Electron Emission (β-)– 3H1 => 3He2 + 0e-1

– 234Th90 => 234Pa91 + 0e-1

– Electron Capture– 40K19 + 0e-1 => 40Ar18 + hv– Postitron Emission (β+)– 40K19 => 40Ar18 + 0e+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

Thin Foil 7Li 11BAlpha Particle

Early History of the Bomb

• Fermi repeats bombarding experiment with neutrons and finds:– Uranium produces several radioactive by products.

Neutron

Uranium

β

γ

N

β

γ

α

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 JoulesNote: 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 JoulesNote: 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

1.602 x 10-19 J

1 eV = 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 JoulesNote: 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 = 1.73 x 108 eV =

=

173. x 106 eV

173 MeV1.602 x 10-19 J

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 g1000 g 235U92 x 1 mole 235U92

235.044 g 235U92

= 4.25 Moles 235U92

The Big Deal4.25 Moles 235U92 6.022 x 1023 atoms

molex

= 2.56*1024 Atoms 235U92

The Big Deal4.25 Moles 235U92 6.022 x 1023 atoms

molex

= 2.56*1024 Atoms 235U92

If 1 Kg of 235U92 fissions, we get:

2.775 X 10-11 Joule Atom

x 2.56*1024 Atoms 235U92 = 7.10* 1013 Joules

The Big Deal4.25 Moles 235U92 6.022 x 1023 atoms

molex

= 2.56*1024 Atoms 235U92

If 1 Kg of 235U92 fissions, we get:

2.775 X 10-11 Joule Atom

x 2.56*1024 Atoms 235U92 = 7.10* 1013 Joules

As a comparison: 1 TON of TNT 4.184* 109 Joules

The Big Deal4.25 Moles 235U92 6.022 x 1023 atoms

molex

= 2.56*1024 Atoms 235U92

If 1 Kg of 235U92 fissions, we get:

2.775 X 10-11 Joule Atom

x 2.56*1024 Atoms 235U92 = 7.10* 1013 Joules

As a comparison: 1 TON of TNT 4.184* 109 Joules

1 Kg of 235U92 7.10* 1013 Joules

4.184* 109 Joules1.7* 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 seconds80 Generations pass in 0.8 microsecondsIt takes less than a millionth of a second to fission a kg of 235U

The Solution

Equivalent to 20,000 Tons of TNT

Use a gun to shoot the slug in

Little Boy Gun Type Weapon

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?

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