nuclear. band of stability number of neutrons 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10...
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Nuclear
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Band of Stability
Nu
mb
er o
f n
eutr
on
s
160
150
140
130
120
110
100
90
80
70
60
50
40
30
20
10
0
Stable nuclides
Naturally occurring radioactive nuclides
Other known nuclides
Number of protons10 20 30 40 50 60 70 80 90 100 110
n = p
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120
100
80
60
40
20
0
Neu
tro
ns
(A-Z
)
0 20 40 60 80 100 120Protons (Z)
Nuclear Decay
• 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 TRANSPARENCYCourtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
P = N
e-captureor
e+ emission
stable nuclei
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120
100
80
60
40
20
0N
eutr
on
s (A
-Z)
P = N
0 20 40 60 80 100 120Protons (Z)
stable nuclei
e-captureor
e+ emission
120
100
80
60
40
20
0
Neu
tro
ns
(A-Z
)
P = N
0 20 40 60 80 100 120Protons (Z)
stable nuclei
• Why nuclides decay…– need stable ratio of neutrons to protons
Nuclear DecayNuclear Decay
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Discovery of the Neutron
James Chadwick bombarded beryllium-9 with alpha particles, carbon-12 atoms were formed, and neutrons were emitted.
n10
+He42
+Be94 C12
6
Dorin, Demmin, Gabel, Chemistry The Study of Matter 3rd Edition, page 764
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Types of Radiation
Type Symbol Charge Mass (amu)
Alpha particle2+ 4.015062
Beta particle1- 0.0005486
Positron 1+ 0.0005486
Gamma ray0 0
01
01
He42
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Alpha, Beta, Gamma Rays
Lead block
Radioactivesubstance
Electrically chargedplates
Photographicplate
rays
rays
rays
(negative charge)
(positive charge)
(no charge)
(+)
(-)
Aligningslot
(detecting screen)
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Alpha, Beta, Positron Emission
Examples of Nuclear Decay Processes
emission(alpha)
emission(beta)
emission(positron)
Th He U 23490
42
23892
Ra He Th 22688
42
23090
Rn He Ra 22286
42
22688
Al Mg 2713
01-
2712 e
Cl S 3517
01-
3516 e
Ca K 4020
01-
4019 e
N O 147
01
148 e
S Cl 3216
01
3217 e
N O 147
01
148 e
Although beta emission involves electrons, those electrons come from the nucleus. Within the nucleus, a neutron decays into a proton and an electron. The electron is emitted, leaving behind a proton to
replace the neutron, thus transforming the element. (A neutrino is also produced and emitted in the process.)
Herron, Frank, Sarquis, Sarquis, Schrader, Kulka, Chemistry, Heath Publishing,1996, page 275
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Nuclear Decay
• Alpha Emission
He Th U 42
23490
23892
parentnuclide
daughternuclide
alphaparticle
Numbers must balance!!
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
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Nuclear Decay
• Beta Emission
e Xe I 0-1
13154
13153
electron• Positron Emission
e Ar K 01
3818
3819
positronCourtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
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Nuclear Decay
• Electron Capture
Pd e Ag 10646
0-1
10647
electron• Gamma Emission
– Usually follows other types of decay.
• Transmutation – One element becomes another.
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
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Half-Lives of Isotopes
Isotope Half-Live Radiation emitted
Half-Lives and Radiation of Some Naturally Occurring Radioisotopes
Carbon-14 5.73 x 103 years
Potassium-40 1.25 x 109 years
Thorium-234 24.1 days
Radon-222 3.8 days
Radium-226 1.6 x 103 years
Thorium-230 7.54 x 104 years
Uranium-235 7.0 x 108 years
Uranium-238 4.46 x 109 years
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Half-Life Plot
Timberlake, Chemistry 7th Edition, page 104
Am
ount
of
odin
e-13
1 (g
)
20
15
10
5
0
40 48 560 8
1 half-life
16
2 half-lives
24
3 half-lives
32
4 half-lives etc…
Time (days)
Half-life of iodine-131 is 8 days
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Half-LifeHalf-life (t½)
– Time required for half the atoms of a radioactive nuclide to decay.
– Shorter half-life = less stable.1/1
1/2
1/4
1/8
1/160R
atio
of
Rem
ain
ing
Po
tass
ium
-40
Ato
ms
to O
rig
inal
Po
tass
ium
-40
Ato
ms
0 1 half-life1.3
1 half-lives2.6
3 half-lives3.9
1 half-lives5.2
Time (billions of yearsTime (billions of years))
Newly formed rockPotassium
Argon
Calcium
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Half-LifeHalf-life (t½)
– Time required for half the atoms of a radioactive nuclide to decay.
– Shorter half-life = less stable.
1/1
1/2
1/4
1/8
1/16
0
Rat
io o
f R
emai
nin
g P
ota
ssiu
m-4
0 A
tom
sto
Ori
gin
al P
ota
ssiu
m-4
0 A
tom
s
0 1 half-life1.3
1 half-lives2.6
3 half-lives3.9
1 half-lives5.2
Time (billions of yearsTime (billions of years))
Newly formed rock
Potassium
Argon
Calcium
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Nuclear Fusion
Sun
+ +
Fourhydrogen
nuclei(protons)
Two betaparticles
(electrons)
Oneheliumnucleus
He e2 H4 4
2
0
1-
1
1 + Energy
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Conservation of Mass…mass is converted into energy
Hydrogen (H2) H = 1.008 amuHelium (He) He = 4.004 amu
FUSIONFUSION
2 H2 1 He + ENERGY
1.008 amux 44.0032 amu = 4.004 amu + 0.028 amu
This relationship was discovered by Albert EinsteinE = mcE = mc22
Energy= (mass) (speed of light)2
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Mass Defect
• Difference between the mass of an atom and the mass of its individual particles.
4.00260 amu 4.03298 amu
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
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The Energy of FusionThe fusion reaction releases an enormous amount of energy relative to themass of the nuclei that are joined in the reaction. Such an enormous amountof energy is released because some of the mass of the original nuclei is con-verted to energy. The amount of energy that is released by this conversioncan be calculated using Einstein's relativity equation E = mc2.
Suppose that, at some point in the future, controlled nuclear fusion becomes possible. You are a scientist experimenting with fusion and you want to determine the energy yield in joules produced by the fusion of one mole of deuterium (H-2) with one mole of tritium (H-3), as shown in the following equation:
n He H H 10
42
31
21
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n He H H 10
42
31
21
First, you must calculate the mass that is "lost" in the fusion reaction. Theatomic masses of the reactants and products are as follows: deuterium (2.01345 amu), tritium (3.01550 amu), helium-4 (4.00150 amu), and a neutron (1.00867 amu).
2.01345 amu 3.01550 amu 4.00150 amu 1.00867 amu
5.01017 amu5.02895 amu
Mass defect:
5.02895 amu 5.01017 amu
-0.01878 amu