Nuclear Chemistry Chapter 23

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Radioactivity • Emission of subatomic particles or highenergy electromagnetic radiation by nuclei • Such atoms/isotopes said to be radioactive

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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 3

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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 6

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Isotopic symbolism • Remember from 139/141? • Let’s briefly go over it • Nuclide = isotope of an element • Proton = 11p • Neutron = 10n • Electron = 0-1e 7

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Types of decay: alpha decay • Alpha (α) decay • Alpha (α) 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! 8

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Alpha decay • Has largest ionizing power – Ability to ionize molecules & atoms due to largeness of α-particle

• But has lowest penetrating power – Ability to penetrate matter

• Skin, even air, protect against α-particle radiation 10

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Beta decay • Beta (β) decay • Beta (β) 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 11

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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)

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Gamma ray emission • Gamma (γ) ray emission • Electromagnetic radiation • High-energy photons • 00γ • No charge, no mass • Usually emitted in conjunction with other •

radiation types Lowest ionizing power, highest penetrating power – Requires several inches lead shielding 14

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Positron emission • • • • • • • • •

Positron = antiparticle of e∴ same mass, opposite charge (Collision with e- causes γ-ray emission) Proton converted into neutron, emitting positron 0 e +1 1 p⇒ 1 n + 0 e 1 0 +1 30 P ⇒ 30 Si + 0 e 15 14 +1 Atomic # of parent nuclide decreases by 1 Positrons have same ionizing/penetrating power as e-

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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 1 p+ 0 e ⇒ 1 n 1 -1 0 92 Ru + 0 e ⇒92 Tc 44 -1 43 Atomic # of parent nuclide decreases by 1 16

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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

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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

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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(non-radioactive) isotopes N/Z too high – Above valley: too many n, convert n to p

• BetaBeta-decay

• N/Z too low – Below valley, too many p, convert p to n

• Positron emission/e—capture and, to lesser extent, alphaalpha-decay

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Predict type of radioactive decay • 1. Mg-28 • 2. Mg-22 • 3. Mo-102

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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) n+p) occupy energy levels within nucleus – Magic numbers

• N or Z = 2, 8, 20, 28, 50, 82, and N = 126

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Radioactive decay series

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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 23

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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 24

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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

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Radioactive decay kinetics • All radioactive nuclei decay via 1st-order kinetics

0.693 k Ratef =kN f & Ratei =kN i t1/2 =

Ratef Nf k = Ratef = N i Ratei Ratei k Half-life = time N Ratef taken for ½ of parent Thus, f ∝ nuclides to decay to Ni Ratei daughter nuclides N R ln f = -kt & ln f = -kt Ni Ri – ∴ rate of decay ∝ to # of nuclei present • Rate = kN

•

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Decay of Rn-220

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Problem • Pu-236 is an α-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

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Solution 0.693 = 2.86 yrs 2 k k = 0.242 yrs −1 t1 =

ln

ln

Nf = -kt Ni 0.100 mg 1.35 mg

t = 10.8 yrs

L = −0.242 yrs −1 • t

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Radiometric dating: radiocarbon dating • Devised in 1949 by Libby at U of

• There is an approximately

•

• Taken up by plants via 14CO2 &

• • • •

Chicago Age of artifacts, etc., revealed by presence of CC-14 C-14 formed in upper atmosphere via: 14 N + 1 n ⇒ 14 C + 1 H 7 0 6 1 C-14 then decays back to N by βemission: 14 C ⇒ 14 N + 0 e; t 6 7 -1 1/2 = 5730 years

constant supply of CC-14

later incorporated in animals

• Living organisms have same ratio of CC-14:C14:C-12

• Once dead, no longer • • •

incorporating CC-14 ⇒ ratio decreases 5% deviation due to variance of atmospheric CC-14 Bristlecone pine used to calibrate data CarbonCarbon-dating good for 50,000 years

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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.

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Solution t 1 = 5, 730 yrs 2

t 1 = 5, 730 yrs = 2

0.693 k

k = 1.21×10−4 yrs −1 R f = 4.50 dis R i = 15.3 dis

ln

min • g carbon

Rf = -kt Ri 4.50 dis

ln

min • g carbon

15.3 dis

min • g carbon

= −1.21×10−4 yrs −1 • t

min • g carbon

t = 1.01×104 yrs

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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

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Example • A meteor contains 0.556 g Pb-206 (to every 1.00 g U-238). Determine its age.

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Solution t 1 = 4.5 × 109 yrs 2

t 1 = 4.5 × 109 yrs = 2

0.693 k

k = 1.5 × 10−10 yrs −1 U-238 has been converted to 0.556g Pb-206 1molPb-206 1mol U-238 238 g U-238 × × = 0.642g U-238 206gPb-206 1mol Pb-206 mol U-238 Thus, N i = 0.642 g + 1.00 g = 1.64g U-238

So, 0.556g Pb-206 ×

And, N f = 1.00g U-238 ln

Nf = -kt Ni

1.00 g U-238 L = −1.5 × 10−10 yrs −1 • t 1.64 g U-238 L t = 3.3 × 109 yrs ln

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Problem • A rock from Australia was found to contain 0.438 g of Pb-206 to every 1.00g of U238. Assuming that the rock did not contain any Pb-206 at the time of its formation, how old is the rock?

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Solution t 1 = 4.5 × 109 yrs 2

t 1 = 4.5 × 109 yrs = 2

0.693 k

k = 1.5 × 10 −10 yrs −1 U-238 has been converted to 0.556g Pb-206 1molPb-206 1mol U-238 238 g U-238 × × = 0.506g U-238 So, 0.438g Pb-206 × 206gPb-206 1mol Pb-206 mol U-238 Thus, N i = 0.506 g + 1.00 g = 1.51g U-238 And, N f = 1.00g U-238 ln

Nf = -kt Ni

1.00 g U-238 L = −1.5 × 10−10 yrs −1 • t 1.51 g U-238 L t = 2.7 × 109 yrs ln

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Fission • Meitner, Strassmann, and Hahn discovered fission

– Splitting of uraniumuranium-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 selfsustaining rxn 38

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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) ! 39

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Nuclear power plant • Controlled fission through • • •

U fuel rods (3.5% UU-235) Rods absorb neutrons Retractable Heat boils water, making steam, turning turbine on generator to make electricity

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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 42

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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

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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 UU-235 fission per 1 mole of UU-235 is 1.7 x 1013 J! – More than a million times more energy per mole than a chemical rxn! rxn! 44

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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) 45

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Some more stuff • Nuclear physicists use eV or MeV (mega eV) 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 46

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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 47

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Solution Mass defect = 4.03298 amu - 4.00260 amu = 0.03038 amu 931.5MeV = 28.30 MeV 1 amu Since He-4 has 4 nucleons (2 protons + 2 neutrons): nuclear binding energy = 0.03038 amu ×

binding energy per nucleon =

28.30 MeV = 7.075 MeV per nucleon 4 nucleons

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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 1 n 0 • C-16 mass = 16.014701 amu

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Solution mass defect = 6 × 11 H + 10 × 01 n - 166 C = 6 × 1.00783 amu + 10 × 1.00866amu - 16.014701 amu = 0.11888 amu 931.5 MeV = 110.74 MeV 1 amu 110.74 MeV = 6.9213 MeV/nucleon nuclea binding energy per nucleon = 16 nucleons

0.11888 amu ×

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Curve of binding energy • Measure of stability of nucleus (binding energy/nucleon) – Reaches max at FeFe-56

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Fusion • 21H + 31H ⇒ 42He + 10n – Ten times more energy per gram than fission

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Transmutation • Transforming one element into another • In 1919, Rutherford bombarded N-17 to make • •

O-17 The Joliot-Curie’s bombarded Al-27 to form P-30 In 1930’s, devices needed that could accelerate particles to high velocities: – Linear accelerator – Cyclotron

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Linear accelerator • ChargedCharged-particle • • •

accelerated in evacuated tube Alternating current causes particle to be pulled into next tube Continues, allowing velocity = 90% speed of light! 2 miles long ⇒⇒⇒ 54

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Cyclotron • Similar alternating • •

voltage used But applied between two semicircular halves of cyclotron Particle spirals due to magnets – Hits target

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Radiation on life • 1. 2. 3.

3 divisions Acute radiation Increased cancer risk Genetic effects

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The first • Quickly dividing cells at greatest risk: – Intestinal lining – Immune response cells

• Likelihood of death depends on dose & duration

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The second • Cancer = uncontrolled cell growth leading to tumors – Dose? • Unknown

• Cancer is a murky illness

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The third • Causes genetic defects ⇒ teratogenic

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How to measure radiation exposure? • Decay events exposure

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– Curie (Ci (Ci)) • 3.7 x 1010 decay events/second – Based on one gram of RaRa-226 decay events/second Or, amount of energy absorbed by body tissue – Gray (Gy (Gy)) • 1 joule of energy absorbed/kg body tissue – Rad (radiation absorbed dose) = 0.01 Gy • 1 rad = 0.01 J/kg body tissue

• SI unit – Becquerel (Bq (Bq)) • 1 Bq = 1 decay/second – 1 Ci = 3.7 × 1010 Bq

• All measure radiation but none account for amount of damage to life life

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Ah ha! • Biological effectiveness factor (RBE = relative biological effectiveness) – Dose in rads x RBE = dose in rems

• Rems = (roentgen equivalent man) – Roentgen = amt of radiation producing 2.58 x 10-4 C of charge/kg air

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Average American 360 mrem/yr

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Good site: let’s take a look • http://www.deq.idaho.gov/inl_oversight/r adiation/radiation_guide.cfm

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More facts • 20 rem – Decreased white blood cell count after instantaneous exposure

• 100100-400 rem – Vomiting, diarrhea, lesions, cancercancer-risk increase

• 500500-1000 – Death w/in 2 months

• 10001000-2000 – Death w/in 2 weeks

• Above 2000 – Death w/in hours 64

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Diagnostic and therapeutic radiation • Radiotracer – Radioactive nuclide in brew to track movement of brew in body

• Tc-99 ⇒ bones • I-131 ⇒ thyroid • Tl-201 ⇒ heart • F-18 ⇒ heart, brain • P-31 ⇒ tumors 65

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PET • Positron emission tomography • Shows both rate of glucose metabolism and structural features of imaged organ

• F-18 emits positrons – Positron and e- produce two gamma rays • Rays detected – Imaged

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PET

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Radiotherapy • Using radiation to treat cancer • Depending on duration/dose can develop symptoms of radiation sickness – Vomiting, diarrhea, skin burns, hair loss

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Other applications • Irradiating foods • Nuking bugs like fruit flies and screwworm flies

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