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Chapter 10
Nuclear PhysicsChapter 10: Nuclear Physics
Homework: All questions on the “Multiple-Choice” and the odd-numbered questions on“Exercises” sections at the end of the chapter.
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Nuclear Physics
• The characteristics of the atomic nucleus areimportant to our modern society.
• Diagnosis and treatment of cancer and otherdiseases
• Geological and archeological dating
• Chemical analysis
• Nuclear energy and nuclear diposal
• Formation of new elements
• Radiation of solar energy
• Household smoke detector
Intro
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Early Thoughts about Elements
• The Greek philosophers (600 – 200B.C.) were the first people to speculateabout the basic substances of matter.
• Aristotle speculated that all matter onearth is composed of only fourelements: earth, air, fire, and water.
• He was wrong on all counts!
Section 10.1
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Symbols of the Elements
• Swedishchemist, JonsJakob Berzelius(early 1800’s)used one or twoletters of theLatin name todesignate eachelement.
Section 10.1
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Symbols of the Elements
• Since Berzelius’ time most elementshave been symbolized by the first oneor two letters of the English name.
• YOU are expected to know the namesand symbols of the 45 elements listedon Table 10.2.
Section 10.1
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Names and Symbols of Common Elements
Section 10.1
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The Atom
• All matter is composed of atoms.
• An atom is composed of three subatomicparticles: electrons (-), protons (+), andneutrons (0)
• The nucleus of the atom contains the protonsand the neutrons (also called nucleons.)
• The electrons surround (orbit) the nucleus.
• Electrons and protons have equal butopposite charges.
Section 10.2
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Major Constituents of an Atom
Section 10.2
U=unified atomic mass units
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The Atomic Nucleus
• Protons and neutrons have nearly thesame mass and are 2000 times moremassive than an electron.
• Discovery – Electron (J.J. Thomson in1897), Proton (Ernest Rutherford in1918), and Neutron (James Chadwick in1932)
Section 10.2
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Rutherford's Alpha-Scattering Experiment
• J.J. Thomson’s “plum pudding” model predicted thealpha particles would pass through the evenlydistributed positive charges in the gold atoms.
a particle = helium nucleus
Section 10.2
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Rutherford's Alpha-Scattering Experiment
• Only 1 out of 20,000 alpha particlesbounced back.
• Rutherford could only explain this byassuming that each gold atom had itspositive charge concentrated in a verysmall “nucleus.”
• Diameter of nucleus = about 10-14 m
• Electron orbit diameter = about 10-10 m
• Atomic Mass is concentrated in thenucleus (>99.97%)
Section 10.2
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Atomic Mass is Concentrated in theNucleus!
• Therefore the volume (or size) of anatom is determined by the orbitingelectrons.– The diameter of an atom is approximately
10,000 times the diameter of the nucleus.
• If only nuclear material (protons andneutrons) could be closely packed into asphere the size of a ping-pong ball itwould have the incredible mass of 2.5billion metric tons!
Section 10.2
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VisualRepresentation ofa Nucleus
• Tightly PackedProtons andNeutrons
Section 10.2
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Atomic Designations
• Atomic Number (Z) – the # of protons in thenucleus (“defines” the element – the # ofprotons is always the same for a givenelement)
• Atomic Number also designates the numberof electrons in an element.
• If an element either gains or loses electrons,the resulting particle is called and ion.
• For example, if a sodium atom (Na) loses anelectron it becomes a sodium ion (Na+.)
Section 10.2
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More Atomic Designations
• Mass Number (A) – protons + neutrons, orthe total number of nucleons
• Isotope – when the number of neutrons varyin the nucleus of a given element (alwayssame number of protons)
• Only 112 elements are known, but the totalnumber of isotopes is about 2000.
Section 10.2
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Isotopes
• Some elements have several isotopes(like carbon –12C, 13C, 14C)
• Isotopes of a single element have the‘same’ chemical properties (due to samenumber of electrons), but they havedifferent masses (due to varying numberof neutrons.)
• Due to their various masses isotopesbehave slightly different during reactions.
Section 10.2
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Carbon Isotopes - Example
SymbolProtons
(Z)Neutrons
(N)Mass #
(A)
12C 6 6 12
13C 6 7 13
14C 6 8 14
Section 10.2
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Three Isotopes of Hydrogen
In naturally occurring Hydrogen - 1 atom in 6000 isdeuterium and 1 in 10,000,000 is tritium. Heavy water = D2O
Section 10.2
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CommonIsotopes ofsome of theLighterElements
Section 10.2
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Determining the Composition of an Atom
• Atomic Number (Z) = 9
• \ protons = 9 & electrons = 9
• Mass Number (A) = 19
• A = N + Z {N = Neutron Number}
• \ N = A – Z = 19 – 9 = 10
• neutrons = 10
Section 10.2
9
• Determine the number of protons, electrons,and neutrons in the fluorine atom 19F
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Atomic Review
• Protons & Neutrons – in nucleus
• Electrons – orbit around nucleus
• Mass Number (A) = protons + neutrons
• Atomic Number (Z) = # of protons
• Neutron Number (N) = # of neutrons
• Isotope – an element with different # ofneutrons (same # of protons)
Section 10.2
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Atomic Mass
• The weighted average mass of an atom ofthe element in a naturally occurring sample
• The Atomic Mass is measured in unifiedunifiedatomic mass units (u)atomic mass units (u) – basically the weightof a proton or neutron.
• The 12C atom is used as the standard, and isassigned the Atomic Mass of exactlyexactly 12 u.
• The weighted average mass of all carbon isslightly higher than 12 (12.011) becausesome is 13C and 14C.
Section 10.2
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Schematic Drawing of a Mass Spectrometer
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The ion with the greatest mass is deflected the least, the ionwith the least mass is deflected the most.
Section 10.2
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A Mass SpectrogramShowing the Three Isotopes of Neon and their relative
abundances
Section 10.2
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Calculating an Element’s Atomic Mass
• Naturally occurring chlorine is a mixtureconsisting of 75.77% 35Cl (atomic mass =34.97 u) and 24.23% 37Cl (atomic mass =36.97 u). Calculate the atomic mass for theelement chlorine.
• Calculate the contribution of each Cl isotope.
• 0.7577 x 34.97 u = 26.50 u (35Cl)
• 0.2423 x 36.97 u = 8.96 u (37Cl)
• Total = 35.46 u = Atomic Mass for Cl
Section 10.2
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Fundamental Forces of Nature - Review
• We have previously discussed twofundamental forces of nature – gravitationaland electromagnetic.
• The electromagnetic force between a proton(+) and an electron (-) is 1039 greater thanthe gravitational forces between the twoparticles.
• Therefore the electromagnetic forces arethe only important forces on electrons andare responsible for the structure of atoms,molecules and all matter in general.
Section 10.2
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The (Strong) Nuclear Force
• Remember that the nucleus of any atom isextremely small and packed with acombination of neutrons and protons (+.)
• According to Coulomb’s Law like chargesrepel each other. Therefore the repulsiveforces in a nucleus are huge and the nucleusshould fly apart.
• There must exist a third fundamental forcethat somehow holds the nucleus together.
• For a large nucleus the forces arecomplicated.
Section 10.2
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Large Nucleus & the Nuclear Force
• An individual proton is only attracted by the 6 or 7 closestnucleons, but is repelled by all the other protons.
• When the # of protons exceeds 83, the electrical repulsionovercomes the nuclear force, and the nucleus is unstable.
• Spontaneousdisintegration ordecay occurs toadjust for the neutron-proton imbalance.
Section 10.2
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Standard Model
• Physicists have also identified a weaknuclear force within an atom.
– This is a short-range force that revealsitself principally in beta decay.
• Physicists have organized three of theknown atomic forces (electromagnetic,weak nuclear, and strong nuclear) into asingle unifying theory called thestandard model.
Section 10.2
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Atomic Review
• Mass Number (A) – protons + neutrons,or the total number of nucleons
• Isotope – when the number of neutronsvary in the nucleus of a given element(always same number of protons)
• Atomic Number (Z) – number of protons
Section 10.2
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Radioactivity
• Radioactivity (radioactive decay) – thespontaneous process of nuclei undergoing achange by emitting particles or rays
• Nuclide – a specific type of nucleus 238U or 14C
• Radionuclides (radioactive isotopes orradioisotopes) – nuclides whose nucleiundergo spontaneous decay (disintegration)
• Substances that give off such radiation aresaid to be radioactive
Section 10.3
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Radioactive Decay
• Parent nucleus – the original nucleusbefore decay
• Daughter nucleus (or daughter product)– the resulting nucleus after decay
• Radioactive nuclei can decay(disintegrate) in three common ways
– Alpha decay
– Beta decay
– Gamma decay
Section 10.3
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Radioactive Decay (disintegration)
• Gamma decay – occurs when a nucleusemits a gamma ray (g) and becomes aless energetic form of the same nucleus
Section 10.3
2
• Alpha decay – disintegration of a nucleusinto a nucleus of another element, w/ theemission of an alpha particle (a) - a heliumnucleus (4He)
-1
• Beta decay – a neutron is transformedinto a proton, w/ the emission of a betaparticle (b) – an electron ( 0e)
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Three Components of Radiationfrom Radionuclides
Section 10.3
Alpha(a), Beta(b), Gamma(g)
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Nuclear Decay Equations - Examples
• In a nuclear decay equation, the sums ofthe mass numbers (A) and the sums ofthe atomic numbers (Z) will be theequivalent on each side
Section 10.3
• Alpha decay = 232Th 228Ra + 4He90
88 2
• Beta decay = 14C 14N + 0e6 7 -1
• Gamma decay = 204Pb* 204Pb + g82
82
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The Products of Alpha Decay – Example
• Must determine the mass number (A), theatomic number (Z), and the chemical symbolfor the daughter product
Section 10.3
• 238 U undergoes alpha decay. Write the equationfor the process92
• 238 U ?92
• 238 U _??_ + 4He92 2
• 238 U _234_ + 4He92 290
• 238 U _234Th_ + 4He92 290
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Five Common Forms of Nuclear Radiations
Section 10.3
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Decay Series of Uranium-238 to Lead-206
Section 10.3
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Identifying Radionuclides
• Which nuclides are unstable (radioactive)and which are stable?
• An interesting pattern emerges:
– Most stable nuclides have an even number ofboth protons and neutrons (even-even nuclides)
– Most unstable nuclides have an odd number ofboth protons and neutrons (odd-odd nuclides)
• A nuclide will be radioactive if:
– Its atomic number (Z) is > than 83
Section 10.3
– n<p (except for 1H and 3He)1 2
– It is an odd-odd nuclide (except for 2H, 6Li, 10B, 14N)1 3 5 7
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Identifying Radionuclides - Example
• Identify the radionuclide in each pair,and state your reasoning.
• odd-odd
Section 10.3
a) 208Pb and 222Rn82 86 • Z above 83
• fewer n than pb) 19Ne and 20Ne10 10
c) 63Cu and 64Cu29 29
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The Pairing Effect in Stabilizing Nuclei
Section 10.3
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A Plot of Number of Neutrons (N) Versus Numberof Protons (Z) for the Nuclides
• Showing “band ofstability”
Section 10.3
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Half-Life of a Radinuclide
• Half-Life – the time it takes for half of thenuclei of a given sample to decay
• In other words – after one half-life hasexpired, only one-half of the originalamount of radionuclide remainsundecayed
• After 2 half-lives only one-quarter (½ of½) of the original amount of theradionuclide remains undecayed
Section 10.3
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Decay of Thorium-234 over Two Half-LivesThorium-234 has a Half-Life of 24 days
Section 10.3
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Decay Curvefor AnyRadionuclide
Section 10.3
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Half-Lives of Some Radionuclides
Section 10.3
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Finding the Number of Half-Livesand the Final Amount
• What fraction and mass of a 40 mg sample ofiodine-131 (half-live = 8d) will remain in 24d?
• Step 2 – Start with the given amount No =40mg, and half it 3 times (3 half-lives)– Once No/2 = 20 mg (after 8 days)
– Twice No/4 = 10 mg (after 16 days)
– Thrice No/8 = 5 mg (after 24 days)
Section 10.3
• Step 1 – find the number of half-lives that havepassed in 24 days: 24 days = 3 half-lives
8d/half-life
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Finding the Number of Half-Lives and theFinal Amount – Confidence Exercise
• What fraction of Stontium-90 produced in1963 (half-live = 29y) will remain in 2021?
• Step 2 – Start with the given amount = No
• After 1 half-life No/2 = (after 29 years)
• After 2 half-lives No /4 = (after 58 years)
\\ One fourth of the original Strontium-90 remains
Section 10.3
• Step 1 – find the number of half-lives that havepassed in 58 years: 58 years = 2 half-lives
29y/half-life
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Finding the Elapsed Time
• How long would it take a sample of 14C todecay to one-fourth its original activity?(half-live of 14C is 5730 years)
• Solution: No No/2 No/4 \ 14C wouldneed to decay for two half-lives in order tobe reduced to ¼ its original activity.
• (2 half-lives)(5730 y/half-life) = 11,460years
Section 10.3
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Determining the Half-Live of aRadioactive Isotope (Radionuclide)
• In order to determine the half-life of aparticular radionuclide, we must monitor theactivity of a known amount in the laboratory
• Activity – the rate of emission of the decayparticles (usually in counts per minute, cpm)
• When (time) the initial activity rate has fallento one-half – we have reached One Half-Life
• Measured with a Geiger Counter --
Section 10.3
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Geiger Counter
When a high-energy particle from a radioactivesource enters the window it ionizes an argonatom, giving off a small pulse of current, whichis counted & amplified into the familiar “clicks”
Section 10.3
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Nuclear Reactions
• We know that radioactive nuclei canspontaneously change into nuclei of otherelements, a process called transmutation.
• Scientists wondered if the reverse waspossible.
• Could a particle (proton or neutron) be addedto a nucleus to change it into anotherelement?
• The answer is “yes,” and this process iscalled a nuclear reaction.
Section 10.4
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Nuclear Reactions
• The result was an artificial transmutationof a nitrogen isotope into an oxygenisotope.
Section 10.4
• 4 He + 14 N 17 O = 1 H2 7 8 1
• In 1919 Ernest Rutherford produced thefirst nuclear reaction by bombarding 14Nwith alpha (4He) particles.
2
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Nuclear Reaction – General Form
• Note that the conservation of mass numberand conservation of atomic number holds innuclear reactions, just like in nuclear decay.
– 18 = total mass # on each side
– 9 = total atomic # on each side
• The general form for a nuclear reaction is
• a + A B + b
– a is the particle that bombards A to form nucleus Band emitted particle b
Section 10.4
• 4 He + 14 N 17 O = 1 H2 7 8 1
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Common Particles Encounteredin Nuclear Reactions
Section 10.4
11 1
• In addition to the particles in the table below,protons (1H), deuterons (2H),and tritons (3H) arecommonly encountered in nuclear reactions.
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Completing an Equation for a NuclearReaction – an Example
• Complete the equation for the protonbombardment of lithium-7.
• Note, the sum of the mass #’s on left = 8.
• The mass # on the right must also = 8,therefore the missing particle must have amass # = 7.
• The sum of the atomic #’s on left = 4
• Therefore the sum of the atomic #’s on rightmust also equal 4.
Section 10.4
• 1H + 7Li ???? + 1n1 3 0
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Completing an Equation for a NuclearReaction – an Example (cont.)
• The missing particle must have an atomicnumber = 4
• Therefore the missing particle has a massnumber of 7 and an atomic number of 4.
• This element is 74Be (beryllium.)
• Completed equation
Section 10.4
• 1H + 7Li ???? + 1n1 3 0
• 1H + 7Li 7Be + 1n1 3 04
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Confidence Exercise
• Note the sum of the mass #’s on left = 29
• The mass # on the right must also = 29,therefore the missing particle must have amass # = 25
• The sum of the atomic #’s on left = 14
• Therefore the sum of the atomic #’s on rightmust also equal 14.
Section 10.4
• Complete the equation for the deuteronbombardment of aluminum-27
• 2H + 27Al ???? + 4He1 13 2
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Confidence Exercise (cont.)
• The missing particle must have an atomicnumber = 12
• Therefore the missing particle has a massnumber of 25 and an atomic number of 12.
Section 10.4
• 2H + 27Al 25Mg + 4He1 13 212
• 2H + 27Al ???? + 4He1 13 2
• Completed equation
• This element is 25Mg (magnesium.)12
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Nuclear Reactions
• Rutherford’s discovery of thetransmutation of 14N into 17O wasactually an accident!
– But the implications of this discovery wereenormous!
• One element could now be changed intoanother completely different element!
Section 10.4
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Nuclear Reactions
• The age-old dream of the alchemists hadcome true.– It was now possible to actually make gold
(Au) from other more common elements!
• Unfortunately, the above process to makegold is VERY expensive.– About $1,000,000 per ounce!
Section 10.4
• 1H + 200Hg 197Au + 4He1 80 79 2
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Nuclear Reactions
• Neutrons produced in nuclear reactions canbe used to induce other nuclear reactions.
– Hence, a “chain reaction” is possible.
• Since neutrons have an electrical charge,they are particularly efficient in penetratingthe nucleus and inducing a reaction.
– Without an electrical charge neutrons are notinfluenced by positive and negative atomiccharges.
Section 10.4
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Transuranium Elements
• Transuranium Elements – those withatomic number greater than 92
– All of these elements are synthetic.
– Created in the lab by bombarding a lighternucleus with alpha particles or neutrons
– For Example …
Section 10.4
• Or …
• 1n + 238U 239Np + 0e0 92 93 -1
• 58Fe + 209Bi 266Mt + 1n26 83 109 0
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Nucleosynthesis
• H, He, and Li are thought to haveformed in the Big Bang.
• Be up to Fe were likely made in thecores of stars by fusion.
• Atoms of heavier then Fe are thought tohave formed during supernovaexplosions of stars, when there was anabundance of neutrons and medium-sized atoms.
Section 10.4
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Uses of Radionuclides
• Radionuclides have many uses inmedicine, chemistry, biology, geology,agriculture, and industry.
• One medical use involves a radioactiveisotope of iodine, 123I, it is used in adiagnostic measurement of the thyroidgland.
• Americum-241, a synthetictransuranium radionuclide, is used inmost common home smoke detectors.
Section 10.4
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Smoke Detector
• A weak radioactive source ionizes the air and sets upa small current. If smoke particles enter, the currentis reduced, causing an alarm.
Section 10.4
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Uses of Radionuclides
• In both chemistry and biology,radioactive “tracers” (14C & 3H) are usedto “tag” an atom within a molecule.– In this way the reaction pathways of drugs
& hormones may be determined.
• In geology, the predictable decay rate ofradioactive elements in rocks andminerals allow age determination.
• In industry, tracer radionuclides helpmanufacturers test durability.
Section 10.4
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Nuclear Fission
• Fission – the process in which a large nucleus“splits” into two intermediate-size nuclei– With the emission of neutrons and …
– The conversion of mass into energy
• For example, 236U fissions into two smallernuclei, emits several neutrons, and releasesenergy.
Section 10.5
• Or …
• 236U 140Xe + 94Sr + 1n92 54 38 0
• 236U 132Sn + 101Mo + 3 1n92 50 42 0
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Completing the Equation for Fissionan Example
• Complete the following equation for fission.
• Atomic #’s are balanced, 92 on both sides
• Therefore the atomic # for the unknown is 0.
• Mass #’s are not balanced.– 236 on the left & 232 on the right
• Therefore 4 additional units of mass areneeded on the right.
Section 10.5
• 236U 88KR + 144Ba +????92 36 56
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Completing the Equation for Fissionan Example (cont.)
• The missing particle must have:
– An atomic number of 0
– A mass number of 4
• But no single particle exists with thoseproperties
• Therefore the missing “particle” is actually 4neutrons.
Section 10.5
• 236U 88KR + 144Ba +????92 36 56
• 236U 88KR + 144Ba +4 1n92 36 56 0
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Completing the Equation for FissionConfidence Exercise
Section 10.5
• Atomic #’s are not balanced.
– The atomic # for the unknown must be 56.
• Mass #’s are not balanced.
• The mass # for the unknown must be 146.
• Complete the following equation for fission
• 236U 88Sr + ???? + 2 1n92 36 0
• The unknown must be 146Ba.56
• 236U 88Sr + 146Ba + 2 1n92 36 56 0
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Nuclear Fission –Three Important Features
• The products of fission are alwaysradioactive.
– Some of the products have half-lives ofthousands of years.
– Nuclear waste disposal problems
• Relatively large amounts of energy areproduced.
• Neutrons are released.
Section 10.5
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Chain Reaction
• In an expanding chain reaction, one initialreaction triggers a growing number ofsubsequent reactions.
• In the case of 236U each fission emits twoneutrons.
– Each of these two neutrons can hit another 235U,resulting in the fission of two additional 235U nucleiand the release of energy and four neutrons.
• As the chain expands, the number ofneutrons emitted and the energy outputincreases.
Section 10.5
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Fission
• 235U absorbs a neutron, initially changes into 236U andthen immediately undergoes fission, releasing twoneutrons and energy.
Section 10.5
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Chain Reaction
• If all of the emittedneutrons hit another235U nucleus, thisresults in anincreasing numberof emitted neutrons,fission reactions,and energy release.
• An expanding chainreaction occurs.
Section 10.5
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Self-Sustaining Chain Reaction
• A steady release of energy can be attainedwhen each fission event causes only onemore fission event - a self-sustaining chainreaction
• For a self-sustaining chain reaction toproceed, the correct amount andconcentration of fissionable material (235U)must be present.
• If there is too much fissionable materialpresent an expanding chain reaction occurs.
• If there is too little fissionable material presentthe chain reaction will stop .
Section 10.5
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Critical Mass
• Critical Mass – the minimum amount offissionable material necessary tosustain a chain reaction
– About 4kg (baseball size) of pure 235U
• Subcritical Mass – no chain reactionoccurs
• Supercritical Mass – an expandingchain reaction occurs
Section 10.5
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Subcritical and Supercritical Masses
Section 10.5
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Uranium
• Natural uranium contains 99.3% 238U and only0.7% is the fissionable 235U isotope.
• Therefore, the 235U must be concentrated or“enriched” in order to create either a self-sustaining or expanding chain reaction.
• In U.S. nuclear reactors the 235U has beenenriched to about 3%, enough for a self-sustaining chain reaction.
• In contrast, nuclear weapons require anenrichment of 90% or more 235U, enough for asudden release of energy
Section 10.5
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Atomic Bomb
• In order to create a fission bomb (“atomicbomb”), a supercritical mass of fissionablematerial must be formed and held together.
• Subcritical portions of the fissionable materialare held apart until detonation.
• A conventional explosion brings thesubcritical segments together to create asupercritical mass capable of a explosiverelease of energy.– This energy sudden energy release is due to an
expanding chain reaction of the fissionablematerial.
Section 10.5
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Nuclear Reactors
• Nuclear reactors must have a controlled andcontinual release of fission energy.
• Within the reactor core, long fuel rods andcontrol rods are placed.
• The fuel rods contain the fissionable material.
– This is the heat source.
• The control rods contain neutron-absorbingmaterial, such as B or Cd.
– These rods “control” the rate of nuclear fission andthereby the amount of heat produced.
Section 10.5
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Nuclear Reactors
• Since the fission rate in the fuel rods cannotbe directly controlled …
• The control rods are inserted or withdrawnfrom between the fuel rods to control thenumber of neutrons being absorbed
• The reactor core is basically a heat source– The heat is continually removed by the coolant
(H2O) flowing through the core
• This heat is used to generate steam in thesteam generator– The steam turns a generator – producing electricity
Section 10.5
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Nuclear Reactor
Note that thehot watercircuit iscompletely“contained,”therebyallowing noradioactivecontamination.
Hot watercircuit
Section 10.5
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Nuclear Reactors – Coolant
• The reactors coolant (H2O) performs twocritical functions:
• 1) The coolant transfers the heat from thereactor core to the steam generator.
• 2) The coolant serves as a moderator.
– The neutrons that are initially emitted from the fuelrods are moving too fast to cause 235U fissionefficiently.
– After colliding with several H2O molecules theneutrons have slowed enough to induce 235Ufission more efficiently.
Section 10.5
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Nuclear Reactors – Potential Dangers
• Potential dangers exist with a continuous-fission chain reaction.
• The amount of heat generated must becontrolled.
– The fission rate is controlled through the insertion(slow down) or withdrawal (speed up) of thecontrol rods between the fuel rods.
• The heat generated must be continuallyremoved from the core.
– The heat is removed by the circulation of thecoolant.
Section 10.5
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Nuclear Reactors – Potential Dangers
• Improper control or removal of coregenerated heat can result in the fusing or“meltdown” of the fuel rods.
• The uncontrolled fissioning mass will becomeextremely hot and will literally melt throughthe floor of the containment structure.
– At this point, the extremely hot and uncontrollableradioactive material will enter the outsideenvironment (ground, water, atmosphere.)
Section 10.5
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Nuclear Accidents
• Two major nuclear accidents have occurred:– Three Mile Island (TMI) Pennsylvania in 1979
– Chernobyl, Ukraine in 1986
• At TMI, an accidental shutdown of the coolantled to a partial meltdown.– This resulted in very little escape of radioactive
gases.
• At Chernobyl, a complete meltdown occurreddue to poor human judgment and designproblems.– This resulted in an explosion & fire in the reactor
core and significant regional contamination.
Section 10.5
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Breeder Reactor
• Besides 235U, another fissionable nuclide is239Pu.
• 239Pu is produced by the bombardment of 238U(the non-fissionable U) with “fast” neutronsduring the normal operation of a nuclearreactor.
• In a breeder reactor the production of 239Pu ispromoted.
• The 239Pu is later separated and may be usedin an ordinary nuclear reactor or in weapons.
Section 10.5
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Breeder Reactor
• Therefore 239Pu is a natural by-productof a nuclear reactor.
• 20 breeder reactors running for a fullyear produce enough 239Pu to runanother reactor for a full year.
• Breeder reactors run at a highertemperature than conventional reactorsand use liquid sodium as a coolant.
Section 10.5
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Nuclear Fusion
• Fusion – the process in which smallernuclei combine to form larger nuclei
– Along with the release of energy
– Does not require a critical mass
• Fusion is the source of energy for theSun and other stars.
• In the Sun, the fusion process producesa helium nucleus from four protons(hydrogen nuclei.)
Section 10.6
– 4 1H 4He + 2 0e + energy1 2 +1
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Examples of Fusion Reactions
Section 10.6
• One deuteron and a triton form an alphaparticle and a neutron.
– Termed a D-T (deuteron-triton) reaction
• 2H + 3H 4He + 1n1 1 2 0
• Two deuterons fuse to form a triton and aproton.
– Termed a D-D (deuteron-deuteron) reaction
• 2H + 2H 3H + 1H1 1 1 1
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D-T Fusion Reaction
Section 10.6
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Nuclear Fusion – Technical Hurdles
• The repulsive force between twopositively charged nuclei is very great.
• To overcome these strong repulsiveforces and initiate fusion, the particlesmust be heated to extremetemperatures (100 million K.)
• At these extreme temperatures the Hatoms exist as a plasma .
– A plasma is gas of electrons and nucleons.
Section 10.6
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Nuclear Fusion – Technical Hurdles
• The plasma must also be confined at ahigh enough density for protons tofrequently collide.
• Even with today’s technology, it is asignificant challenge to reach thenecessary temperature andconfinement requirements.
Section 10.6
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Nuclear Fusion - Inertial Confinement
• Inertial Confinement – simultaneoushigh-energy laser pulses from all sidescause a fuel pellet of D & T to implode,resulting in compression and hightemperatures
• If the pellet can be made to stay intactfor a sufficient time, fusion is initiated.
• Research into this method is beingconducted at Los Alamos and LawrenceLivermore labs.
Section 10.6
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Nuclear Fusion – Magnetic Confinement
• Magnetic Confinement – a doughnut-shapedmagnetic field holds the plasma, while electriccurrents raise the temperature of the plasma
• Magnetic and electric fields are useful since aplasma gas is a gas of charged particles.
– Charged particles can be controlled andmanipulated with electric and magnetic fields.
• The leading labs for fusion research usingmagnetic confinement are MIT and Princeton.
Section 10.6
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Fusion Advantages over Fission
• Low cost and abundance of deuterium
– Deuterium can be extracted inexpensively fromwater.
• Dramatically reduced nuclear waste disposal
– Relatively few radioactive by-products withrelatively short half-lives
• Fusion reactors cannot get out of control
– In the event of a system failure, quick cool down
– Not dependent on a critical mass
Section 10.6
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Nuclear Reaction and Energy
• In 1905 Einstein published his specialtheory of relativity.
• This work deals with the changes thatoccur in mass, length, and time as anobject’s speed approaches the speedof light (c.)
• This theory predicts the mass (m) andenergy (E) are not separate entitiesbut rather related by his famousequation E = mc2 .
Section 10.6
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Nuclear Reaction and Energy
• Einstein’s predictions have proved correct.
• Scientists have been able to change mass intoenergy and on a very small scale, energy intomass.
• For example, using Einstein’s equation what isthe equivalent energy of 1 gram mass?
• E = mc2 = (0.001 kg)(3.00 x 108 m/s2)
• = 90 x 1012 J = 90 trillion joules
• 90 trillion joules = same amount of energyreleased by 20,000 of TNT
Section 10.6
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Nuclear Reaction and Energy
• Calculations with Einstein’s formula, E=mc2,have convinced many scientists that smallamounts of mass that are “lost” in nuclearreactions could be a tremendous source ofenergy.
• To determine the change in mass in a nuclearreaction, we simply add up the masses of allthe reactants and subtract the masses of allthe products.
• Generally during a nuclear reaction mass iseither gained or lost.
Section 10.6
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Mass Defect
• Endoergic – an increase in mass hastaken place during the reaction
– Absorbs energy by the number of atomicmass units times 931 MeV
• Exoergic – a decrease in mass hastaken place during the reaction
– Releases energy by the number of atomicmass units time 931 MeV
Section 10.6
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Calculating Mass and Energy Changes inNuclear Reactions – An Example
• Calculate the mass defect and the correspondingenergy released during this typical fission reaction(using the masses of the atoms.)
• (236.04556 u) (87.91445 u) (143.92284 u) (4 x 1.00867 u)
• Total mass on left = 236.04556 u
• Total mass on right = 235.87197 u
• Difference of 0.17359 u = mass defect
• (0.17359 u)(931 MeV/u) = 162 MeV of energy released
Section 10.6
• 236U 88Kr + 144Ba + 4 1n92 36 56 0
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Calculating Mass and Energy Changesin D-T Fusion Reaction –
Confidence Exercise
Section 10.6
• Calculate the mass defect and the correspondingenergy released during a D-T fusion reaction.
• (2.0140 u) (3.0161 u) (4.0026 u) (1.0087 u)
• Total mass on left = 5.0301 u
• Total mass on right = 5.0113 u
• Difference of 0.0188 u = mass defect
• (0.0188 u)(931 MeV/u) = 17.5 MeV of energy released
• 2H + 3H 4He + 1n1 1 2 0
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Fusion vs. Fission
• On a kilogram for kilogram comparison,more energy comes from fusion thanfrom fission.
• The fission of 1 kg of 235U providesenergy equal to the burning of 2 millionkg of coal.
• The fusion of 1 kg of deuterium releasesenergy equal to the burning of 40 millionkg of coal.
Section 10.6
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Energy Release in BothNuclear Fission and Fusion
Any reaction that leads upward on the curve releases energybecause such a reaction is accompanied by a mass defect.
Section 10.6
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Energy Release in Both Nuclear Fissionand Fusion cont.
• Fission proceeds from right to left.
• Fusion proceeds from left to right.
• Note that at the top of the curve is 56Fe. No net energy will bereleased by either splitting 56Fe or by fusing several 56Fe.
Section 10.6
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Biological Effects of Radiation
• Ionizing Radiation – radiation that is strongenough to knock electrons off atoms andform ions
• Ionizing Radiation includes; alpha particles,beta particles, gamma particles, neutrons,gamma rays, and X-rays
• These types of radiation can also harm orkill living cells, and are especially harmful ifthey affect molecules involved in cellreproduction
Section 10.7
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Effects of Radiation on Living Organisms
• Ionizing radiation cannot be seen, smelled,felt, or tasted
• Film badges worn by workers is commonlyused to measure radiation exposure
• The effects of radiation on living organismscan be classified into two categories:
• Somatic Effects – short- and long-term effectson the recipient of the radiation
• Genetic Effects – defects in the recipient’ssubsequent offspring
Section 10.7
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Radiation Units
• The rem (roentgen equivalent for man) is theunit used to discuss biological effects ofradiation
• This unit takes into consideration the relativeionizing power of each type of radiation andits affects on humans
• The average U.S. citizens receives 0.2 remper year, from a number of different sources(both natural and anthropogenic)
Section 10.7
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Sources of Exposure to Radiation
• 0.2 rem –averageannualradiationexposure forperson inU.S.
Section 10.7
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Radiation Sources
• Natural Sources – cosmic radiation (highaltitude areas), bedrock, radionuclides thatare ingested (carbon-14, potassium-40)
– Radon gas, from bedrock, varies greatly withlocation, but is thought to cause from 10,000 to130,000 lung cancers deaths per year in theU.S.
• Anthropogenic Sources – medical X-raysand treatment, TV’s, tobacco smoke,nuclear waste, certain household products
Section 10.7
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Short-Term Somatic Effectsfrom a Single Dose
Section 10.7
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Long-Term Somatic Effects
• Long-term cumulative effects ofradiation exposure are not fullyunderstood
• Without a doubt the most common long-term somatic effect is an increasedlikelihood of developing cancer
• Many early workers of radionuclidesdied of cancer – these scientists weregenerally exposed to small doses formany years
Section 10.7
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Penetration of Radiation
• Alpha and Beta particles are electrically charged andcan be easily stopped. Gamma rays, X-rays, andneutrons are more difficult to stop because they arenot charged particles.
• But – there does notappear to be any lowerlimit, below which theeffects are negligible
• \ any exposure toradiation should betaken seriously.
Section 10.7Section 10.7