chapter 21 nuclear chemistry

32
Nuclear Chemistry Chapter 21 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Upload: brucelee55

Post on 25-Jun-2015

990 views

Category:

Documents


1 download

TRANSCRIPT

Page 1: Chapter 21 Nuclear Chemistry

Nuclear ChemistryChapter 21

Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.

Page 2: Chapter 21 Nuclear Chemistry

Radioactivity

• _________________—unstable atomic nuclei spontaneously emit particles, electromagnetic radiation (EMR), or both

• ________________________—results from bombarding nuclei with neutrons, protons, or other nuclei

Page 3: Chapter 21 Nuclear Chemistry

XAZ

Mass Number

Atomic NumberElement Symbol

______________ (Z) = number of protons in nucleus

______________ (A) = number of protons + number of neutrons

= atomic number (Z) + number of neutrons

A

Z

1p11H1or

proton1n0

neutron0e-1

0-1or

electron0e+1

0+1or

positron4He2

42or

particle

21.1

Page 4: Chapter 21 Nuclear Chemistry

What is the difference between and

21.1p.673

0e+10+1

0e+1

0+1

represents an electron in or from an atomic orbital

represents an electron that is physically identical to an electron in or from an atomic orbital, but this electron comes from the decay of a neutron to a proton and an electron—it is also called a beta particle or ray

Page 5: Chapter 21 Nuclear Chemistry

21.1

Comparison of Chemical Reactions and Nuclear Reactions

Page 6: Chapter 21 Nuclear Chemistry

Balancing Nuclear Equations

1. Conserve mass number (A).

The sum of protons plus neutrons in the products must equal the sum of protons plus neutrons in the reactants.

1n0U23592 + Cs138

55 Rb9637

1n0+ + 2

235 + 1 = 138 + 96 + 2x1

2. Conserve atomic number (Z) or nuclear charge.

The sum of nuclear charges in the products must equal the sum of nuclear charges in the reactants.

1n0U23592 + Cs138

55 Rb9637

1n0+ + 2

92 + 0 = 55 + 37 + 2x021.1

Page 7: Chapter 21 Nuclear Chemistry

212Po decays by alpha emission. Write the balanced nuclear equation for the decay of 212Po.

4He242oralpha particle -

212Po 4He + AX84 2 Z

212 = 4 + A A = 208

84 = 2 + Z Z = 82

212Po 4He + 208Pb84 2 82

21.1Ex 21.1, p.673

Page 8: Chapter 21 Nuclear Chemistry

Nuclear Stability and Radioactive Decay

Beta decay

14C 14N + 0 + 6 7 -1

40K 40Ca + 0 + 19 20 -1

1n 1p + 0 + 0 -1

________ # of ________ by 1

________ # of ________ by 1

Positron decay

11C 11B + 0 + 6 5 +1

38K 38Ar + 0 + 19 18 +1

1p 1n + 0 + 1 0 +1

________ # of ________ by 1

________ # of ________ by 1

and have A = 0 and Z = 021.2

1

Page 9: Chapter 21 Nuclear Chemistry

Electron capture decay

________ # of ________ by 1

________ # of ________ by 1

Nuclear Stability and Radioactive Decay

37Ar + 0e 37Cl + 18 17-1

55Fe + 0e 55Mn + 26 25-1

1p + 0e 1n + 1 0-1

Alpha decay

________ # of ________ by 2

________ # of ________ by 2212Po 4He + 208Pb84 2 82

Spontaneous fission

252Cf 2125In + 21n98 49 021.2

Page 10: Chapter 21 Nuclear Chemistry

Nuclear Stability

• Certain numbers of neutrons and protons are extra stable

• n or p = 2, 8, 20, 50, 82 and 126

• Like extra stable numbers of electrons in noble gases (e = 2, 10, 18, 36, 54 and 86)

• Nuclei with even numbers of both protons and neutrons are more stable than those with odd numbers of neutron and protons

• All isotopes of the elements with atomic numbers higher than 83 are radioactive

• All isotopes of Tc and Pm are radioactive

21.2

Number of Stable Isotopes with Even and Odd Numbers of Protons and Neutrons

Page 11: Chapter 21 Nuclear Chemistry

n/p too large

beta decay

X

n/p too small

positron decay or electron capture

Y

21.2p.675

Page 12: Chapter 21 Nuclear Chemistry

______________________ is the energy required to break up a nucleus into its component protons and neutrons.

BE is an indication of the stability of a nucleus.

In order to compare nuclei of two different isotopes/elements, we must take into account the fact that they have different numbers of _________.

For this reason, nuclear binding energy per nucleon is more useful.

21.2p.676

Page 13: Chapter 21 Nuclear Chemistry

Mass Defect

The difference betweenthe ________________ of an atom

andthe ________________ of the masses of

protons, neutrons, and electrons

What does the mass defect tell us? How much mass was changed to energy in the formation of the atom.

p.676

Page 14: Chapter 21 Nuclear Chemistry

The Law of Charges tells us…?

So how can all those positively charged protons be crammed into the tiny space of the nucleus?

We call it the “strong nuclear force” or just the “strong force.”

Some of the mass of the nucleons is converted to energy and lost.

This is the general idea behind fusion:

Build new, larger nuclei and release great amounts of energy!

Page 15: Chapter 21 Nuclear Chemistry

Nuclear binding energy per nucleon vs Mass number

nuclear binding energynucleon nuclear stability

21.2p.678

note

Page 16: Chapter 21 Nuclear Chemistry

Which element has the greatest net attractive forces among its nucleons? (graph)

Radioactivity: unstable nuclei spontaneously emit particles, electromagnetic radiation (EMR), or both

Main types of radioactivity: particles (He2+) particles (e-) rays (short-wavelength emr)

•positron emission

•electron capture

Often it involves a multi-step sequence, a series…. and all obey first-order kinetics.

Page 17: Chapter 21 Nuclear Chemistry

Kinetics of Radioactive Decay

N daughter

rate = -Nt

rate = N

Nt

= N-

N = N0exp(-t) lnN = lnN0 - t

N = the number of atoms at time t

N0 = the number of atoms at time t = 0

is the decay or rate constant

ln2=

21.3

The Uranium Decay Series

Page 18: Chapter 21 Nuclear Chemistry

Kinetics of Radioactive Decay

[N] = [N]0exp(-t) ln[N] = ln[N]0 - t

[N]

ln [

N]

21.3

is the first order rate constant and N is the number of radioactive nuclei present at time t

p.679f

Page 19: Chapter 21 Nuclear Chemistry

Radiometric Assumptions

The method measures the parent/daughter ratio of the elements.

1. The system must initially contain none of ____________________________________.

2. The decay rate must _______________________.

3. The amounts of the parent element and the daughter products must be affected by ___________________________________.

Page 20: Chapter 21 Nuclear Chemistry

Radiocarbon Dating

14N + 1n 14C + 1H7 160

14C 14N + 0 + 6 7 -1t½ = 5730 years

Uranium-238 Dating

238U 206Pb + 4 + 6 092 -182 2

t½ = 4.51 x 109 years

21.3

p.681f

Potassium-40 Dating

t½ = 1.2 x 109 years

40K + 0e 40Ar 19 18-1

Page 21: Chapter 21 Nuclear Chemistry

Nuclear Transmutation

Cyclotron Particle Accelerator

14N + 4 17O + 1p7 2 8 1

27Al + 4 30P + 1n13 2 15 0

14N + 1p 11C + 47 1 6 2

21.4p.683

Page 22: Chapter 21 Nuclear Chemistry

Nuclear Transmutation

21.4

The Transuranium Elements

Page 23: Chapter 21 Nuclear Chemistry

Nuclear Fission

21.5

235U + 1n 90Sr + 143Xe + 31n + Energy92 54380 0

Energy = [mass 235U + mass n – (mass 90Sr + mass 143Xe + 3 x mass n )] x c2

Energy = 3.3 x 10-11J per 235U

= 2.0 x 1013 J per mole 235U

Combustion of 1 ton of coal = 5 x 107 J

p.685f

Page 24: Chapter 21 Nuclear Chemistry

Nuclear Fission

21.5

235U + 1n 90Sr + 143Xe + 31n + Energy92 54380 0

Representative fission reaction

p.686

Page 25: Chapter 21 Nuclear Chemistry

Nuclear Fission

21.5

A ________________________ is a self-sustaining sequence of nuclear fission reactions.

The minimum mass of fissionable material required to generate a self-sustaining nuclear chain reaction is the ________________________.

Non-critical

Critical

Page 26: Chapter 21 Nuclear Chemistry

Nuclear Fission

21.5

Schematic diagram of a

nuclear fission reactor

Page 27: Chapter 21 Nuclear Chemistry

Annual Waste Production

21.5

35,000 tons SO2

4.5 x 106 tons CO2

1,000 MW coal-firedpower plant

3.5 x 106

ft3 ash

1,000 MW nuclearpower plant

70 ft3 vitrified waste

Nuclear Fission

Page 28: Chapter 21 Nuclear Chemistry

21.5

Nuclear Fission

Hazards of the radioactivities in spent

fuel compared to uranium ore

From “Science, Society and America’s Nuclear Waste,” DOE/RW-0361 TG

Page 29: Chapter 21 Nuclear Chemistry

21.6

Nuclear Fusion

2H + 2H 3H + 1H1 1 1 1

Fusion Reaction Energy Released

2H + 3H 4He + 1n1 1 2 0

6Li + 2H 2 4He3 1 2

6.3 x 10-13 J

2.8 x 10-12 J

3.6 x 10-12 J

Tokamak magnetic plasma

confinement

Page 30: Chapter 21 Nuclear Chemistry

21.6

Radioisotopes in Medicine• 1 out of every 3 hospital patients will undergo a nuclear

medicine procedure

• 24Na, t½ = 14.8 hr, emitter, _______________________

• 131I, t½ = 14.8 hr, emitter, ________________________

• 123I, t½ = 13.3 hr, ray emitter, _____________________

• 18F, t½ = 1.8 hr, emitter, ________________________

• 99mTc, t½ = 6 hr, ray emitter, _____________________

Brain images with 123I-labeled compound

Page 31: Chapter 21 Nuclear Chemistry

21.6

Biological Effects of RadiationRadiation absorbed dose (rad)

1 rad = 1 x 10-5 J/g of material

Roentgen equivalent for man (rem)

1 rem = 1 rad x Q Quality Factor-ray = 1

= 1 = 20

Average Yearly Radiation Doses for Americans

Page 32: Chapter 21 Nuclear Chemistry

Biological Effects of RadiationFormation of _______________ and/or ________________ that attack membranes, enzymes, or DNA.

Damage can be ________________ or _______________.

p.695f