Nuclear Chemistry. Chapter 23. Radioactivity . Emission of subatomic particles or high-energy electromagnetic radiation by nuclei Such atoms/isotopes said to be radioactive. Its discovery. Discovered in 1896 by Becquerel Called strange, new emission uranic rays
Nuclear Chemistry Chapter 23
Radioactivity • Emission of subatomic particles or high-energy electromagnetic radiation by nuclei • Such atoms/isotopes said to be radioactive
Its discovery • Discovered in 1896 by Becquerel • Called strange, new emission uranic rays • Because emitted from uranium • Marie Curie discovered two new elements both of which emitted uranic rays • Po & Ra • Uranic rays became radioactivity
Types of radioactivity • Rutherford and Curie found that emissions produced by nuclei • Different types: • Alpha decay • Beta decay • Gamma ray emission • Positron emission • Electron capture
Isotopic symbolism • Remember from 139/141? • Let’s briefly go over it • Nuclide = isotope of an element • Proton = 11p • Neutron = 10n • Electron = 0-1e
Types of decay: alpha decay • Alpha () decay • Alpha () particle: helium-4 bereft of 2e- • = 42He (don’t write He+2) • Parent nuclide daughter nuclide + He-4 • 23892U 23490Th + 42He • Daughter nuclide = parent nuclide atomic # minus 2 • Sum of atomic #’s & mass #’s must be = on both sides of nuclear equation!
Alpha decay • Has largest ionizing power • Ability to ionize molecules & atoms due to largeness of -particle • But has lowest penetrating power • Ability to penetrate matter • Skin, even air, protect against -particle radiation
Beta decay • Beta () decay • Beta () particle = e- • How does nucleus emit an e-? • neutron changes into proton & emits e- • 10n 11p + 0-1e • Daughter nuclide = parent nuclide atomic number plus 1
Beta decay • Lower ionizing power than alpha particle • But higher penetration power • Requires sheet of metal or thick piece of wood to arrest penetration • more damage outside of body, but less in (alpha particle is opposite)
Gamma ray emission • Gamma () ray emission • Electromagnetic radiation • High-energy photons • 00 • No charge, no mass • Usually emitted in conjunction with other radiation types • Lowest ionizing power, highest penetrating power • Requires several inches lead shielding
Positron emission • Positron = antiparticle of e- • same mass, opposite charge • (Collision with e- causes -ray emission) • Proton converted into neutron, emitting positron • 0+1e • 11p 10n + 0+1e • 3015P 3014Si + 0+1e • Atomic # of parent nuclide decreases by 1 • Positrons have same ionizing/penetrating power as e-
Electron capture • Particle absorbed by, instead of ejected from, an unstable nucleus • Nucleus assimilates e- from an inner orbital of its e- cloud • Net result = conversion of proton into neutron • 11p+ 0-1e 10n • 9244Ru + 0-1e 9243Tc • Atomic # of parent nuclide decreases by 1
Problems • Write a nuclear equation for each of the following: • 1. beta decay in Bk-249 • 2. electron capture in I-111 • 3. positron emission in K-40 • 4. alpha decay of Ra-224
The valley of stability • Predicting radioactivity type • Tough to answer why one radioactive type as opposed to another • However, we can get a basic idea • Neutrons occupy energy levels • Too many lead to instability
Valley of stability • In determining nuclear stability, ratio of neutrons to protons (N/Z) important • Notice lower part of valley (N/Z = 1) • Bi last stable (non-radioactive) isotopes • N/Z too high • Above valley: too many n, convert n to p • Beta-decay • N/Z too low • Below valley, too many p, convert p to n • Positron emission/e—capture and, to lesser extent, alpha-decay
Predict type of radioactive decay • 1. Mg-28 • 2. Mg-22 • 3. Mo-102
Magic numbers • Actual # of n & p affects nuclear stability • Even #’s of both n & p give stability • Similar to noble gas electron configurations • 2, 10, 18, 36, etc. • Since nucleons (= n+p) occupy energy levels within nucleus • Magic numbers • N or Z = 2, 8, 20, 28, 50, 82, and N = 126
Detecting radioactivity • Particles detected through interactions w/atoms or molecules • Simplest film-badge dosimeter • Photographic film in small case, pinned to clothing • Monitors exposure • Greater exposure of film greater exposure to radioactivity
Geiger counter • Emitted particles pass through Ar-filled chamber • Create trail of ionized Ar atoms • Induced electric signal detected on meter and then clicks • Each click = particle passing through gas chamber
Scintillation counter • Particles pass through material (NaI or CsI) that emits UV or visible light due to excitation • Atoms excited to higher E state • E releases as light, measured on meter
Radioactive decay kinetics • All radioactive nuclei decay via 1st-order kinetics • rate of decay to # of nuclei present • Rate = kN • Half-life = time taken for ½ of parent nuclides to decay to daughter nuclides
Problem • Pu-236 is an -emitter w/half-life = 2.86 years. If sample initially contains 1.35 mg, what mass remains after 5.00 years? • How long would it take for 1.35 mg sample of Pu-236 above to decay to 0.100 mg? • Assume 1.35 mg/1 L air
Devised in 1949 by Libby at U of Chicago Age of artifacts, etc., revealed by presence of C-14 C-14 formed in upper atmosphere via: 147N + 10n 146C + 11H C-14 then decays back to N by -emission: 146C 147N + 0-1e; t1/2 = 5730 years There is an approximately constant supply of C-14 Taken up by plants via 14CO2 & later incorporated in animals Living organisms have same ratio of C-14:C-12 Once dead, no longer incorporating C-14 ratio decreases 5% deviation due to variance of atmospheric C-14 Bristlecone pine used to calibrate data Carbon-dating good for 50,000 years Radiometric dating: radiocarbon dating
Problem • Artifact is found to have C-14 decay rate of 4.50 disintegration/min g of carbon. • If living organisms have a decay rate of 15.3, how old is the artifact? • Given decay rate is to amount of C-14 present.
Radiometric dating: uranium/lead dating • Relies on ratio of U-238:Pb-206 w/in igneous rocks (rocks of volcanic origin) • Measures time that has passed since rock solidified • t1/2 = 4.5 x 109 years
Example • A meteor contains 0.556 g Pb-206 (to every 1.00 g U-238). Determine its age.
Problem • A rock from Australia was found to contain 0.438 g of Pb-206 to every 1.00g of U-238. Assuming that the rock did not contain any Pb-206 at the time of its formation, how old is the rock?
Fission • Meitner, Strassmann, and Hahn discovered fission • Splitting of uranium-235 • Instead of making heavier elements, created a Ba and Kr isotope plus 3 neutrons and a lot of energy • Sample rich in U-235 could create a chain rxn • To make a bomb, however, need critical mass = enough mass of U-235 to produce a self-sustaining rxn
Nuclear power • In America, about 20% electricity generated by nuclear fission • Imagine: • Nuclear-powered car • Fuel = pencil-sized U-cylinder • Energy = 1000 20-gallon tanks of gasoline • Refuel every 1000 weeks (about 20 years) • !
Nuclear power plant • Controlled fission through U fuel rods (3.5% U-235) • Rods absorb neutrons • Retractable • Heat boils water, making steam, turning turbine on generator to make electricity
Comparing • Typical nuclear power plant makes enough energy for city of 1,000,000 people and uses about 50 kg of fuel/day • No air pollution/greenhouses gases • But, nuclear meltdown (overheating of nuclear core) is a potential threat • No problem! • Also, waste disposal • Location, containment problems
Comparing • Coal-burning power plant uses about 2,000,000 kg of fuel to make same amount of energy • But, releases huge amounts of SO2, NO2, CO2
Mass to energy • E = mc2 • Explains relationship between energy formation and matter loss • Amount of energy released in U-235 fission per atom of U-235 is 2.8 x 10-11 J • BUT, amount of energy released in U-235 fission per 1 mole of U-235 is 1.7 x 1013 J! • More than a million times more energy per mole than a chemical rxn!
Mass defect • Mass products < mass reactants • Difference in mass due to conversion of mass into energy • Called mass defect • Nuclear binding energy is energy corresponding to mass defect • Amount of energy required to break apart nucleus into nucleons (n + p)
Some more stuff • Nuclear physicists use eV or MeV (mega eV) instead of joules • 1 MeV = 1.602 x 10-13 J • 1 amu = 931.5 MeV • Energy per nucleus and not per mole • To compare energy of 1 nucleus to another, calculate binding energy per nucleon • = nuclear binding energy of nuclide per #of nucleons (n + p) in nuclide • As binding energy per nucleon increases so does stability of species
Example • Calculate the mass defect and nuclear binding energy per nucleon (in MeV and in J) for: • 42He • Made from: 211H + 210n • 11H = 2 x 1.00783 amu • 210n = 2 x 1.00866 amu • Net mass = 4.03298 amu
Problem • Calculate the mass defect and nuclear binding energy per nucleon (in MeV) for C-16 • Consider C-16 being made from 6 11H & 10 10n • C-16 mass = 16.014701 amu