Nuclear Chemistry
Chapter 23
1
Radioactivity
• Emission of subatomic particles or high-
energy electromagnetic radiation by nuclei
• Such atoms/isotopes said to be
radioactive
2
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
4
5
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
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
Types of decay: alpha decay
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Alpha () decay
Alpha () particle: helium-4 bereft of 2e= 42He (don’t write He+2)
Parent nuclide  daughter nuclide + He-4
238 U  234 Th + 4 He
92
90
2
Daughter nuclide = parent nuclide atomic
# minus 2
Sum of atomic #’s & mass #’s must be = on
both sides of nuclear equation!
8
9
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
10
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
11
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)
12
13
Gamma ray emission
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•
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Gamma () ray emission
Electromagnetic radiation
High-energy photons
0 
0
No charge, no mass
Usually emitted in conjunction with other
radiation types
Lowest ionizing power, highest penetrating
power
– Requires several inches lead shielding
14
Positron emission
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Positron = antiparticle of e same mass, opposite charge
(Collision with e- causes -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-
15
Electron capture
• Particle absorbed by, instead of ejected from, an
•
•
•
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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
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
17
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
18
Valley of stability
• In determining nuclear stability,
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•
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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
19
Predict type of radioactive decay
• 1. Mg-28
• 2. Mg-22
• 3. Mo-102
20
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
21
Radioactive decay series
22
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
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
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
25
Radioactive decay kinetics
• All radioactive nuclei
decay via 1st-order
kinetics
0.693
t1/2 =
k
Ratef =kN f & Ratei =kN i
Ratef
–  rate of decay  to #
Nf
Rate f
k
of nuclei present


• Rate = kN
N i Ratei
Ratei
• Half-life = time
k
N f Rate f
taken for ½ of parent
Thus,

nuclides to decay to
Ni
Ratei
daughter nuclides
Nf
Rf
ln
= -kt & ln
= -kt
Ni
Ri
26
Decay of Rn-220
27
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
28
Solution
0.693
t1 
 2.86 yrs
2
k
1
k  0.242 yrs
Nf
ln
= -kt
Ni
ln
mg
0.100
mg
1.35
t  10.8 yrs
L  0.242 yrs 1  t
L
29
Radiometric dating: radiocarbon
dating
• Devised in 1949 by Libby at U of
• There is an approximately
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Chicago
Age of artifacts, etc., revealed by
presence of C-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 emission:
14 C  14 N + 0 e; t
6
7
-1
1/2 = 5730
years
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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
30
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.
31
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
ln
min  g carbon
min  g carbon
Rf
= -kt
Ri
4.50 dis
min  g carbon
15.3 dis
min  g carbon
t  1.01104 yrs
 1.2110 4 yrs 1  t
32
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
33
Example
• A meteor contains 0.556 g Pb-206 (to
every 1.00 g U-238). Determine its age.
34
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
ln
 1.5  1010 yrs 1  t
1.64 g U-238
L
t  3.3 109 yrs
35
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?
36
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


206gPb-206 1mol Pb-206 mol U-238
Thus, N i  0.506 g  1.00 g  1.51g U-238
So, 0.438g 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
ln
1.51 g U-238
L
t  2.7  109 yrs
37
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 selfsustaining rxn
38
Nuclear power
• In America, about 20% electricity generated by
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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
Nuclear power plant
• Controlled fission through
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U fuel rods (3.5% U-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
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
43
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!
44
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
Some more stuff
• Nuclear physicists use eV or MeV (mega eV) instead of
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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
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
Solution
Mass defect = 4.03298 amu - 4.00260 amu = 0.03038 amu
931.5MeV
nuclear binding energy = 0.03038 amu 
 28.30 MeV
1 amu
Since He-4 has 4 nucleons (2 protons + 2 neutrons):
28.30 MeV
binding energy per nucleon =
 7.075 MeV per nucleon
4 nucleons
48
Problem
• Calculate the mass defect and nuclear
binding energy per nucleon (in MeV) for C16
– Consider C-16 being made from 6
1 n
0
• C-16 mass = 16.014701 amu
1 H
1
& 10
49
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
nuclea binding energy per nucleon =
 6.9213 MeV/nucleon
16 nucleons
0.11888 amu 
50
Curve of binding energy
• Measure of stability of
nucleus (binding
energy/nucleon)
– Reaches max at Fe-56
51
Fusion
•
2
1H +
3
1H 
4
2He +
1
0n
– Ten times more energy
per gram than fission
52
Transmutation
• Transforming one element into another
• In 1919, Rutherford bombarded N-17 to make
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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
53
Linear accelerator
• Charged-particle
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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
Cyclotron
• Similar alternating
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voltage used
But applied between
two semicircular
halves of cyclotron
Particle spirals due to
magnets
– Hits target
55
Radiation on life
•
1.
2.
3.
3 divisions
Acute radiation
Increased cancer risk
Genetic effects
56
The first
• Quickly dividing cells at greatest risk:
– Intestinal lining
– Immune response cells
• Likelihood of death depends on dose &
duration
57
The second
• Cancer = uncontrolled cell growth leading
to tumors
– Dose?
• Unknown
• Cancer is a murky illness
58
The third
• Causes genetic defects  teratogenic
59
How to measure radiation
exposure?
• Decay events exposure
•
– Curie (Ci)
• 3.7 x 1010 decay events/second
– Based on one gram of Ra-226 decay events/second
Or, amount of energy absorbed by body tissue
– Gray (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)
• 1 Bq = 1 decay/second
– 1 Ci = 3.7 × 1010 Bq
• All measure radiation but none account for amount of damage to life
60
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
61
Average American 360 mrem/yr
62
Good site: let’s take a look
• http://www.deq.idaho.gov/inl_oversight/ra
diation/radiation_guide.cfm
63
More facts
• 20 rem
– Decreased white blood cell
count after instantaneous
exposure
• 100-400 rem
– Vomiting, diarrhea, lesions,
cancer-risk increase
• 500-1000
– Death w/in 2 months
• 1000-2000
– Death w/in 2 weeks
• Above 2000
– Death w/in hours
64
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
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
66
PET
67
Radiotherapy
• Using radiation to treat cancer
• Depending on duration/dose can develop
symptoms of radiation sickness
– Vomiting, diarrhea, skin burns, hair loss
68
Other applications
• Irradiating foods
• Nuking bugs like fruit flies and screwworm flies
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