ch18 z7e nuclear

1 CHAPTER 18

Nuclear

Chemistry

CHAPTER 18

Nuclear

Chemistry18.I Nuclear 18.I Nuclear Stability & Stability &

Radioactive Decay Radioactive Decay pppp

18.I Nuclear 18.I Nuclear Stability & Stability &

Radioactive Decay Radioactive Decay pppp

I

2

Black dots are stable nuclides.

As A (atomic mass) increases, nº/p+ ratio increases.

3

Subatomic Particles

• Protons - plus charge

In the nucleus• Neutrons - neutral

• Electrons - negative charge

Outside the nucleus

4

Radiation

• Radiation comes from the nucleus of an atom.• Unstable nucleus emits a particle or energy

alpha

beta

gamma

5

He42

Types of RadiationTypes of RadiationTypes of RadiationTypes of Radiation

Alpha particle () helium nucleus paper2+

Beta particle (-) electron

e0-1 1-

leadPositron (+)

positron e01

1+

Gamma () high-energy photon 0

concrete

6

Radiation Protection

• Shielding

alpha – paper, clothing

beta – lab coat, gloves

gamma- lead, thick concrete

• Limit time exposed

• Keep distance from source

7

Radiation Protection

8 Nuclear DecayNuclear DecayNuclear DecayNuclear Decay

Alpha Emission

He Th U 42

23490

23892

parentnuclide

daughternuclide

alphaparticle

Atomic & Mass Numbers must balance!!

9 Nuclear DecayNuclear DecayNuclear DecayNuclear Decay

Beta Emission

e Xe I 0-1

13154

13153

electronPositron Emission

e Ar K 01

3818

3819

positron

10 Nuclear DecayNuclear DecayNuclear DecayNuclear Decay

Electron Capture (of inner orbital electrons)

Pd e Ag 10646

0-1

10647

electronGamma Emission

Usually follows other types of decay.

Transmutation One element becomes another.

11

12

Gamma radiation

No change in atomic or mass number

11B 11B + 0

5 5 0

boron atom in a

high-energy state

13

14 Table 18.2 Types of Nuclear Processes p. 845 Table 18.2 Types of Nuclear Processes p. 845

15

Learning Check

Write the nuclear equation for the beta emitter Cobalt-60. . .

60Co 60Ni + 0 e 27 28 -1

16

Producing Radioactive Isotopes

Bombardment of atoms produces radioisotopes = 60 = 60

59Co + 1n 56Mn + 4H e 27 0 25 2

= 27 = 27

cobalt neutron manganese alpha atom radioisotope particle

17

Learning Check NR2

What radioactive isotope is produced in the following bombardment of boron?

10B + 4He ? + 1n

5 2 0

18

Solution NR2

What radioactive isotope is produced in the following bombardment of boron?

10B + 4He 13N + 1n

5 2 7 0

nitrogen

radioisotope

19Nuclear DecayNuclear Decay ppppNuclear DecayNuclear Decay pppp

Why nuclides decay… need stable ratio of neutrons to protons

He Th U 42

23490

23892

e Xe I 0-1

13154

13153

e Ar K 01

3818

3819

Pd e Ag 10646

0-1

10647

DECAY SERIES TRANSPARENCY

2020

The decay series.The decay series.

2121

18.2 Kinetics of Radioactive Decay18.2 Kinetics of Radioactive Decay

Rate of decay is a 1st order process, which is . . .Rate of decay is a 1st order process, which is . . . ln(N/Nln(N/N00) = -kt) = -kt (memorize -- not on AP sheet) (memorize -- not on AP sheet)

NN00 = original number of nuclides at t = 0 = original number of nuclides at t = 0

N = nuclides N = nuclides remainingremaining at time t at time tHalf-life (tHalf-life (t1/21/2) = time for nuclides to reach half ) = time for nuclides to reach half

their original value.their original value. tt1/21/2 = 0.693/k = 0.693/k

22 Half-lifeHalf-lifeHalf-lifeHalf-life

Half-life (t1/2) Time required for half the atoms of a

radioactive nuclide to decay. Shorter half-life = less stable.

23

24

Examples of Half-Life

Isotope Half lifeC-15 2.4 secRa-224 3.6 daysRa-223 12 daysI-125 60 daysC-14 5700 yearsU-235 710 000 000 years

25

Learning Check NR3

The half life of Iodine-123 is 13 hr. How much of a 64 mg sample of Iodine-123 is left after 26 hours?

26

Solution NR3

t1/2 = 13 hrs

26 hours = 2 x t1/2

Amount initial = 64mg

Amount remaining = 64 mg x 1/2 x 1/2

= 16 mg

27 Half-lifeHalf-lifeHalf-lifeHalf-life

nif mm )( 2

1

mf: final massmi: initial massn: # of half-lives

28Half-lifeHalf-life ppppHalf-lifeHalf-life pppp

Fluorine-21 has a half-life of 5.0 seconds. If you start with 25 g of fluorine-21, how many grams would remain after 60.0 s?

GIVEN:

T1/2 = 5.0 s

mi = 25 g

mf = ?

total time = 60.0 s

n = 60.0s ÷ 5.0s =12

WORK:

mf = mi (1/2)n

mf = (25 g)(0.5)12

mf = 0.0061 g

29 Kinetics of Nuclear Decay Problems ppKinetics of Nuclear Decay Problems pp

The rate constant for 9943Tc = 1.16 x 10-1/h

What is its half life? . . . t1/2 = 0.693/k = 0.693/(1.16 x 10-1/h) = 5.98 h

It will take 5.98 hours for a given sample of technetium-99 to decrease to half the original number of nuclides.

30 Kinetics of Nuclear Decay Problems ppKinetics of Nuclear Decay Problems pp

How long for 87.5% of a sample of cobalt-60 How long for 87.5% of a sample of cobalt-60 to decay if tto decay if t1/21/2 = 5.26 years? Steps. . . = 5.26 years? Steps. . .

What % is left? . . .What % is left? . . . 12.5%12.5% How many half-lives to get to this percent?How many half-lives to get to this percent? 3. So, your answer to the problem is . . .3. So, your answer to the problem is . . . 3 x 5.26 = 15.8 years.3 x 5.26 = 15.8 years.

31 Actual AP question: 1989 MC #68 ppActual AP question: 1989 MC #68 pp

If k = 0.023 min-1 how much of X was originally present if have 40. g after 60 min.?

Your answer is . . .160. g. Solution . . . t1/2 = 0.693/k = 0.693/0.023 min-1 = 30 min.

60 minutes is 2 half-lives so going backwards 40. g to 80. g to 160. g.

32 18.3 Nuclear Transformations18.3 Nuclear Transformations

Transmutation - change of one element into another.

Particle and linear accelerators are used to synthesize new elements (currently up to element number 119).

Difficult to characterize the chemical properties because with some only a few atoms are formed with very short half-lives.

3333

A representation of a Geiger-Müller counter.A representation of a Geiger-Müller counter.

34 18.4 Detection & Uses of Radioactivity18.4 Detection & Uses of Radioactivity

Half-life measurements of radioactive elements are used to determine the age of an object

Decay rate indicates amount of radioactive material

EX: 14C - up to 40,000 years238U and 40K - over 300,000 years

35 Synthetic ElementsSynthetic ElementsSynthetic ElementsSynthetic Elements Transuranium Elements

elements with atomic #s above 92 synthetically produced in nuclear reactors and accelerators most decay very rapidly

Pu He U 24294

42

23892

36 Carbon-14 Dating Carbon-14 Dating You will have a test question like this!You will have a test question like this! pppp

Carbon-14 Dating Carbon-14 Dating You will have a test question like this!You will have a test question like this! pppp

An ancient fire in an African cave has a An ancient fire in an African cave has a 1414C decay rate of 3.1 cpm (cts per minute). C decay rate of 3.1 cpm (cts per minute). If fresh wood has 13.6 cpm how old is the If fresh wood has 13.6 cpm how old is the campfire if tcampfire if t1/21/2 = 5730 years? Steps . . . = 5730 years? Steps . . .

Decay rates are directly proportional to Decay rates are directly proportional to nuclides so their ratio = nuclides so their ratio = N/NN/N00 What is the What is the

numerical ratio? Your answer . . .numerical ratio? Your answer . . . 3.1 cpm/13.6 cpm = 3.1 cpm/13.6 cpm = 0.230.23 Use the two previous equations to solve Use the two previous equations to solve

(next slide).(next slide).

37 Carbon-14 Dating You will have a test question like this! pp

Carbon-14 Dating You will have a test question like this! pp

Ancient fire 14C decay rate 3.1 cpm, fresh wood 13.6 cpm how old if t1/2 = 5730 yrs?

3.1 cpm/13.6 cpm = 0.23 = N/N0

ln(N/N0) = -kt and t1/2 = 0.693/k

You want to solve for t (vs. t1/2) so use t1/2 to get k then plug into the 1st equation and solve for t. Your answer is . . .

The campfire is 12 000 years old. ln(N/N0) = ln(0.23) = -(0.693/5730)t

Se Ex 18.4A & 18.4B in Study Guide

38 Carbon-14 Dating You will have another test question like this! pp

Carbon-14 Dating You will have another test question like this! pp

A rock has ratio of Pb-206 to U-238 of 0.115. How old is it if t1/2 of U-238 = 4.5 x 109 yrs?

Strategy: figure out N/N0 of U-238, then use the 2 previous equations to get . . .

7.1 x 108 years. Calculations . . .Pb/U = 115/1000 so N0 U238 = 1115, N = 1000

ln(1000/1115) = -(0.693/4.5 x 109)t

39 Nuclear MedicineNuclear MedicineNuclear MedicineNuclear Medicine

Radioisotope Tracers absorbed by specific organs and used

to diagnose diseasesRadiation Treatment

larger doses are used to kill cancerous cells in targeted organs

internal or external radiation source

Radiation treatment using-rays from cobalt-60.

40 Other UsesOther UsesOther UsesOther Uses

Food Irradiation radiation is used to kill bacteria

Radioactive Tracers explore chemical pathways trace water flow study plant growth, photosynthesis

Consumer Products ionizing smoke detectors - 241Am

41 Radioisotopes Used As TracersRadioisotopes Used As Tracers

4218.5 Thermodynamic Stability of the Nucleus18.5 Thermodynamic Stability of the Nucleus

Mass Defect - difference from mass of an Mass Defect - difference from mass of an atomatom & the mass of its individual particles. & the mass of its individual particles.

4.00260 amu 4.03298 amu

43 Nuclear Binding Nuclear Binding EnergyEnergy

Nuclear Binding Nuclear Binding EnergyEnergy

Energy released when a nucleus is formed from nucleons.

High binding energy = stable nucleus.

E = mc2E: energy (J)m: mass defect (kg)c: speed of light

(3.00 x 108 m/s)

44 Nuclear Binding EnergyNuclear Binding EnergyNuclear Binding EnergyNuclear Binding Energy

Unstable nuclides - radioactive & undergo radioactive decay.Elements with intermediate atomic masses (e.g., Fe) have greatest binding energy, so are the most stable.

45 18.6 Nuclear Fission and Nuclear Fusion 18.6 Nuclear Fission and Nuclear Fusion

Fission - splitting

Fusion - Combining

Both produce more stable nuclides so they are exothermic processes

46 A. Nuclear FissionA. Nuclear FissionA. Nuclear FissionA. Nuclear Fission

Splitting a nucleus into two or more smaller nuclei

1 g of 235U = 3 tons of coal

U23592

47

Nuclear Fission

Fission

large nuclei break up

235U + 1n 139Ba + 94Kr + 3 1n +

92 0 56 36 0

Energy

48 Nuclear PowerNuclear PowerNuclear PowerNuclear Power

Fission Reactors Cooling Tower

4949

Schematic of the reactor core.Schematic of the reactor core.

50 Nuclear PowerNuclear PowerNuclear PowerNuclear Power

Fission Reactors

51 FissionFissionFissionFission chain reaction - self-propagating reaction critical mass -

mass required to sustain a chain reaction

52

53 Nuclear FusionNuclear FusionNuclear FusionNuclear Fusion combining of two nuclei to form one nucleus of larger mass thermonuclear reaction – requires temp of 40,000,000 K to sustain 1 g of fusion fuel =

20 tons of coal (vs. 3 in fission) occurs naturally in

stars

HH 31

21

54

Nuclear Fusion

Fusion

small nuclei combine

2H + 3H 4He + 1n +

1 1 2 0

Occurs in the sun and other stars

Energy

55 Nuclear PowerNuclear PowerNuclear PowerNuclear Power

Fusion Reactors (not yet sustainable)

56 Nuclear PowerNuclear PowerNuclear PowerNuclear Power

Fusion Reactors (not yet sustainable)

Tokomak Fusion Test Reactor

Princeton University

National Spherical Torus Experiment

57 Fission vs. FusionFission vs. Fusion ppFission vs. FusionFission vs. Fusion pp

235U is limited danger of meltdown toxic waste thermal pollution

fuel is abundant no danger of meltdown no toxic waste not yet sustainable

FISSION

FUSION

58

Learning Check NR4

Indicate if each of the following are(1) Fission (2) fusion

A. Nucleus splits B. Large amounts of energy releasedC. Small nuclei form larger nucleiD. Hydrogen nuclei react

Energy

59

Solution NR4

Indicate if each of the following are(1) Fission (2) fusion

A. 1 Nucleus splits B. 1 + 2 Large amounts of energy releasedC. 2 Small nuclei form larger nucleiD. 2 Hydrogen nuclei react

60 E. Nuclear WeaponsE. Nuclear WeaponsE. Nuclear WeaponsE. Nuclear Weapons

Atomic Bomb chemical explosion is used to form a

critical mass of 235U or 239Pu fission develops into an uncontrolled

chain reaction

Hydrogen Bomb chemical explosion fission fusion fusion increases the fission rate more powerful than the atomic bomb

61 18.7 Effects of Radiation pp18.7 Effects of Radiation pp

Somatic - damage to the organism causing sickness or death.

Genetic - damage to the genetic machinery causing birth defects.

62 Factors for Biological Effects of Radiation ppFactors for Biological Effects of Radiation pp

Energy - higher energy content (rads) causes more damage.

Penetrating Ability - > - > Ionizing Ability - > - > (eating an -particle

producer like Pu is very deadly) Chemical Properties

• Kr-85 is chemically inert, passes through quickly

• Sr-90 collects in bone and stays a long time in the body.

63

64

Radioactive Radioactive particles and particles and

rays vary rays vary greatly in greatly in penetrating penetrating

power.power.

Radioactive Radioactive particles and particles and

rays vary rays vary greatly in greatly in penetrating penetrating

power.power.

65

6666

Diagram for Diagram for the tentative the tentative plan for deep plan for deep underground underground isolation of isolation of

nuclear wastenuclear waste..