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1
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2
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- 1919: Ernest Rutherford
experimented with bombarding nitrogen gas molecules with alpha particles
emitted from bismuth-214
- Discovery: faster moving
particles were produced, and these could travel farther than the alpha
particles!
- “New” particles also deflected in a magnetic field like a positive
particle
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3
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- Conclusion: The faster moving
particles were protons
- Artificial Transmutation:
- The change of one element to another through the bombardment of a
nucleus
- More experiments to determine exact nature of the particles and how they
were “created” done with a cloud chamber…
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4
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- Invented ~1911 by a Scottish Atmospheric Physicist (C.T.R. Wilson) to
experiment with rain clouds.
- Enclosed environment made to be supersaturated (originally with water
vapor, now commonly ethanol)
- Ions introduced to this environment would attract water molecules
(which are polar), forming clouds…
- Earned a share in the
- 1927 Nobel Prize in
- Physics for the invention…
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5
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- Why would this be useful for Rutherford?
- Water vapor condenses around ions
- An alpha particle is ionizing radiation, thus leave a LOT of ions in
its path
- Water vapor would condense around these ions, leaving a vapor trail
showing where an alpha particle had been…
- Video
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6
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- If proton was simply “chipped off” the Nitrogen nucleus by the alpha
particle, there should be 4 visible tracks in the cloud chamber:
- The original alpha particle BEFORE collision
- The alpha particle AFTER the collision
- The “chipped off” proton
- The Nitrogen nucleus, now charged, as it recoiled after the collision
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7
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- If alpha particle was absorbed, and that caused the proton to be pushed
out, then there should be 3 visible tracks:
- The alpha particle before collision
- The proton emitted after the collision
- The path of the recoiling Nitrogen nucleus
- This theory was proven true in 1925
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8
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- Note: Deuteron = Hydrogen-2
atom, a.k.a Deuterium
- Example problem:
- A sample of Oxygen-16 is bombarded with neutrons. If one of the resulting products is a
deuteron, what is the resulting nucleus?
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9
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- A unit adopted by scientists that is
more appropriate for masses along the order of magnitude of
atomic masses
- 1 u = 1.66 x 10-27 kg
- Mass of an electron (me) = 0.000549 u
- Mass of a proton (mp) = 1.007277 u
- Mass of a neutron (mn) = 1.008665 u
- Mass of 1 H atom (mH) = 1.007825 u
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10
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- Einstein hypothesized a relationship between mass and energy in 1905
- Many years later, data from nuclear reactions showed that his hypothesis
was indeed true
- c = 3.00 x 108 m·s-1
- m = mass (kg)
- E = Energy (J)
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11
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- Used to calculate the Rest Energy of a mass
- Used to calculate the amount of energy released in nuclear reactions
- For Example:
- Calculate the amount of energy released when 1.00 kg of fuel is used up
in a nuclear reactor…
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- All atomic nuclei have a total mass that is lower than the sum of the
masses of each individual particle
- For example: The EXPECTED mass
of an atom of Helium would be the sum of the mass of 2 neutrons, 2
protons, and 2 electrons:
- 2(0.000549 u) + 2(1.007277 u) + 2(1.008665 u) = 4.032982 u
- The MEASURED mass of an atom of helium has been found to be 4.002602 u
- àa difference of 0.03038
u
- This difference is known as the Mass Defect of the atom
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13
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- …a measure of the energy needed to keep a nucleus together
- Binding Energy is the energy equivalent of the mass defect
- E = mc2
- E = (1.66 x 10-27 kg)(3.00 x 108 m·s-1)2
- E = 1.49 x 10-10 J = 931 MeV
- (Since 1 eV = 1.6 x 10-19 J)
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14
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- Calculate the binding energy of Oxygen-16. The measured mass of Oxygen-16 is
15.994915 u
- 8 protons+8 electrons+8 neutrons
- 8mH + 8mn = mexpected
- = 8(1.007825 u) + 8(1.008665 u)
- = 8.062600 u + 8.069320 u
- = 16.131920 u
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15
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- Calculate the binding energy of Oxygen-16. The measured mass of Oxygen-16 is
15.994915 u
- mdefect = mexpected – mmeasured
- = 16.131920 u – 15.994915 u
- = 0.137005 u
- Eb = mdefect · (931 MeV·u-1)
- Eb = (0.137005)(931)
- = 127 MeV
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16
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- Fission: A reaction that involves
the splitting of a large, unstable nucleus into 2 or more smaller, more
stable nuclei
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17
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- Fusion: A reaction that joins two very light nuclei to form a heavier
nucleus
- Picture source: www.atomicarchive.com
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- Nuclei with higher amounts of binding energy per nucleon are more stable
than those with lower amounts of binding energy per nucleon.
- Fission and fusion processes each release large amounts of energy as the
nuclei join or split to form more stable products.
- To predict how much energy can result from a nuclear reaction, we use a
binding energy curve…
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19
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- Example: Use the binding energy
curve to predict the amount of energy released when Uranium-235
undergoes fission to produce two Palladium-117 fragments.
- Eb for 235U = 7.6 MeV/nucleon
- Eb for 117Pd = 8.4 MeV/nucleon
- The difference between these values, multiplied by the total number of
nucleons, is equal to the amount of energy released in the reaction:
- (0.8 MeV/nucleon) x (235 Nucleons) = 188 MeV
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- Only takes place in certain very heavy elements, such as Uranium-235
- Fissile Uranium-235 is used in nuclear reactions:
- Nucleus bombarded with a neutron to begin a chain reaction…
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- Self-sustaining (chain) reactions:
when enough neutrons are produced to naturally enable the
reaction to continue until all fissile material is gone
- Examples: Nuclear Reactors in
Power Plants; Bombs dropped on Hiroshima and Nagasaki in WWII
- Critical Mass: The amount of
fissile material required to sustain a fission reaction
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23
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24
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- Conditions required for fusion reactions:
- Very high temperatures (because nuclei need very high kinetic
energies)
- Very densely packed (to ensure that enough collisions will occur), therefore:
- Very high pressures
- Problems with creating fusion on Earth:
- Containment is a huge problem
- At temps required, atoms would ionize and technically would become a
plasma
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25
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- Proton-Proton Cycle = the fusion reaction that is the source of energy
in young/cool stars such as the sun:
- The first two reactions in the cycle must occur twice
- Total energy released = 24.7 MeV
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26
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- Calculate the energy released when a proton and a deuteron undergo
fusion to produce helium-3.
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Notes
