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Nuclear and Particle Physics
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Nuclear Physics
Back to Rutherford and his discovery of the nucleusAlso coined the term “proton” in 1920, and described a “neutron” in 1921Neutron discovered by Chadwick in 1932
Ernest Rutherford1871-1937
me = 9.1 x 10-31 kg
mN = 1.6749 x 10-27
kg
mP = 1.6726 x 10-27
kgJames Chadwick 1891-1974
nucleons
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Nuclides and IsotopesTo specify a nuclide: XA
Z
Z is the atomic number = number of electrons or protons
A is the mass number = number of neutrons + protons
So number of neutrons = A-Z
Number of protons = ZIsotopes – same atomic number, different mass number
e.g. carbon: C126 C13
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Many isotopes do not occur naturally, also elements > U
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Sizes
We saw with the Bohr model that radius of the atom depended on atomic number
Nucleus = protons + neutrons = mass number
The volume of a nucleus is proportional to the mass number
r (1.2 x10 15m) A3
)(mA)(1.2x103
4r
3
4V 33153 π π
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MassesMass spectrometer
1 atomic mass unit (u.) = 1.6606 x 10-27 kg = 931.5 MeVFixed so that carbon = 12.00000 u
mN = 1.6749 x 10-27 kg = 1.0087 u
mP = 1.6726 x 10-27 kg = 1.0078 u
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Binding Energy
Total mass of a nucleus < sum of massesExample:Mass of helium nucleus = 6.6447 x 10-27 kg
He42
Contains 2 protons and 2 neutrons
Mass = 2 x (1.6749 x 10-27 + 1.6726 x 10-27 ) kg
= 6.6950 x 10-27 kg
Difference = (6.6950 – 6.6447) x 10-27 = 0.0503 x 10-27 kg
Energy = mc2 = 0.0503 x 10-27 x c2 = 4.53 x 10-12 J
= (4.53 x 10-12) / (1.6 x 10-19) = 2.83 x 107 eV
= 28.3 MeV
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Atomic Mass Units1 u = 931.5 MeV
mN = 1.6749 x 10-27 kg = 1.0087 u
mP = 1.6726 x 10-27 kg = 1.0078 u
Mass of helium nucleus = 4.0026 u
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Atomic Mass UnitsSame calculationMass of 2p + 2n = 2 x (1.0078 + 1.0087) =
4.0330 u
Difference = 0.0304 u
Binding energy = 0.0305 x 931.5 = 28.3 MeV
4.0330 u
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Average Binding Energy
• Graph
He – 4 nucleons, 28.3 MeV total: average = 7.075MeV
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Attractive?
How does nucleus stay together? Like charges repel!Force stronger than electric force Strong nuclear forceShort range (~10-15 m)
Stable nuclides N = Z
A > 30-40 – more neutrons
Z > 82 – no stable nuclides
Strong force can’t overcome repulsion
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Radioactivity
Becquerel, 1896Emission of radiation without external stimulus
Curies – polonium (Po) and radium (Ra) Henri Becquerel1852-1908
Marie Curie1867 - 1934
Pierre Curie1859 - 1906
1903 (Physics)1911
(Chem)
Radioactivity unaffected by heating, cooling, etc.
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ClassificationRutherford classified 3 types of radioactivity according to penetration powerAlso different charge
Important factor: Conservation of nucleon number
(neutrons + protons) = (neutrons + protons)
Video: “
People Pretending to be Alpha Particles
”
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Alpha Decay
Least penetrating – nucleus of He42
Radium 226 is an alpha emitter:
HeRnRa 42
22286
22688
Parent Daughter
transmutation
Mass of parent > mass of daughter + mass of alpha
Difference = kinetic energy
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Example
232.03714 u 228.02873 u + 4.002603 utotal = 232.03133 u
Lost mass = 232.03714 – 232.03133 = 0.00581 u0.00581u x 931.5 MeV/u = 5.4 MeV
(some recoil)
HeThU 42
22890
23292
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Beta decay One electron
eNC 01
147 14
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What is lost is NOT an orbital electron
Instead a neutron changes to a proton + electron
So (6p + 8n) => (7p + 7n) + e- - decay
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Example
Keep track of electrons!Carbon 14 has m = 14.003242 u 6 electronsNitrogen 14 has m = 14.003074 u normally 7 electronsBut in the decay, the nitrogen would have 6 electronsHowever the total on the r.h.s. of the equation has 7
So difference = 0.000168 u = 0.156 MeV = 156 keV
eNC 01
147 14
6
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Conservation of energy
• Energy of decay = 156 keV = problem!
?
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A new particle
• Proposed by Pauli (1930) - neutrino• Theory by Fermi• Discovered 1956• Zero charge, ~0 rest mass
Wolfgang Pauli
1900-1958
Enrico Fermi
1901-1954
eNC 01
147
146
antineutrino
“Zero rest mass” – speed of light
1998 – Super Kamiokande – some mass
Cosmic neutrino detection
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More on positrons
Many isotopes have more neutrons than protons® Decay by emission of electron
Other isotopes have more protons than neutrons® Decay by emission of positron Proton changes to a neutron + positron
eFNe 01
199
1910
+ decay
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Annihilation
Proton changes to a neutron + positron
eFNe 01
199
1910
+ decay
Positron annihilation
Application – positron emission tomography
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Positron Emission Tomography
PET – basis – use radio-labelled compounds, i.e. those containing a radionuclide.
Positron emitters:
I,F,C,N,O 12453
189
116
137
158
As an example, oxygen-15 can be used to look at oxygen metabolism and blood flow. Fluorine-18 is commonly used to examine cancerous tumours.
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PET - method
Annihilation produces two back-to-back 511 keV photons
Simultaneous detection
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Electron capture
• Nucleus absorbs orbiting electron
LieBe 73
01
74
Proton changes to neutron
Usually K electron
X-ray emission as outer electron jumps down to K
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Gamma decay
Most penetrating = photon. High energy*Excited nucleus lower energy state
eCB 01
126 12
5
e*CB 01
126 12
5
Energy levels far apart = keV or MeV
- (13.4 MeV)
C126
- (9.0 MeV)
(4.4 MeV)
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Homework . . .
1.p.902,#6; 2.p.908, Practice 25B; 3.p.912,Section Review4.p.928, 30-37;5.p. 930, 56,60; 6.Read through lab for next
time; answer pre-lab questions
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