Download Nuclear Reactions and Their Applications

Nuclear Reactions and Their
Applications
Review
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Atomic Number (Z) – number of protons
Mass Number (A) – sum of protons and neutrons
Z + N = A (mass number of nuclide)
Protons (Z) + Neutrons (N) = Nucleons
Nuclide: a nucleus with a particular composition
Most elements occur in nature as a mixture of isotopes
(characteristic number of protons but different number of
neutrons)
Radioactive Decay
•
Nucleus undergoes decomposition to form a different
nucleus.
Radioactive Stability
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Nuclides with 84 or more protons are unstable.
Light nuclides are stable when Z equals A – Z
(neutron/proton ratio is 1).
For heavier elements the neutron/proton ratio required for
stability is greater than 1 and increases with Z.
Certain combinations of protons and neutrons seem to
confer special stability.
– Even numbers of protons and neutrons are more often
stable than those with odd numbers.
Certain specific numbers of protons or neutrons produce
especially stable nuclides.
– 2, 8, 20, 28, 50, 82, and 126 - magic numbers
It parallels for atoms where 1, 10, 18, 36, 54 and 86 electrons produce chemical stability
Noble gases
Red dots indicate stable nuclides
As the number of protons in a nuclide
increases the proton-to-neutron ratio
Required for stability also increases
Types of Radioactive Decay
•
Alpha production (α):
•
Beta production (β):
•
Gamma ray production (γ):
•
Positron production:
•
Electron capture:
Inner-orbital electron
Three types of radioactive emissions in an electric field.
+
Gamma emission: involves the
radiation of high-energy γ
photons from an excited nucleus.
Gamma rays usually result from
a particle and an anti-particle
annihilation.
0
1
β + −10 e → 2 00γ
Balancing Nuclear Equations
TotalA
TotalZ
Reactants=
TotalA
TotalZ
Products
Write balanced equations for the following nuclear reactions:
(a) Naturally occurring thorium-232 undergoes α decay.
(b) Zirconium-86 undergoes electron capture.
PLAN:
SOLUTION:
Write a skeleton equation; balance the number of neutrons and
charges; solve for the unknown nuclide.
(a)
232
90Th
228
88X
+
4
232
A = 228 and Z = 88
(b) 8640Zr + 0-1e
→
A = 86 and Z = 39
2α
A
90Th
ZX 8640Zr
+ 0-1e
→
228
→
88Ra
86
+
39Y
4
2α
Predicting the Mode of Decay
Neutron-rich nuclides undergo β decay.
Neutron-poor nuclides undergo positron decay or electron capture.
Heavy nuclides undergo α decay.
Which of the following nuclides would you predict to be stable and
which radioactive? Explain.
(a)
18
10Ne
(b)
32
16S
(c)
236
90Th
(a) N/Z = 0.8; there are too few
neutrons to be stable - radioactive
(b) N/Z = 1.0; Z < 20 and N
and Z are even - stable
(d)
123
56Ba
(c) Every nuclide with Z > 83
is radioactive.
(d) N/Z = 1.20; zone of
stability figure shows stability
when N/Z ≥ 1.3. - radioactive
Predicting the Mode of Nuclear Decay
PROBLEM:
Use the atomic mass of the element to predict the mode(s) of decay of
the following radioactive nuclides:
(a)
PLAN:
12
5B
(b)
234
92U
(c)
81
33As
(d)
127
57La
Find the N/Z ratio and compare it to the band stability. Then predict which of
the modes of decay will give a ratio closer to the band.
SOLUTION:
(a) N/Z = 1.4 which is high.
The nuclide will probably undergo β−
decay altering Z to 6 and lowering the
ratio.
(c) N/Z = 1.24 which is in the band of
stability. It will probably undergo β−
decay or positron emission.
(b) The large number of neutrons
makes this a good candidate for α
decay.
(d) N/Z = 1.23 which is too low for this
area of the band. It can increase Z by
positron emission or electron capture.
The 238U decay series.
Rate of Decay
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The rate of decay is proportional to the number of nuclides. This
represents a first-order process.
Half - Life
∆N
Rate =
kN
−
=
∆t
ln
N
= −kt
No
• Time required for the number of nuclides to reach half
the original value.
t1/ 2 =
ln ( 2 ) 0.693
=
k
k
Large k means a short half-life and vice versa
SI unit of decay is the becquerel (Bq) = 1 d/s.
curie (Ci) = number of nuclei disintegrating each second in
1 g of radium-226 = 3.70 x 1010 d/s
The decay of a 10.0-g sample of strontium-90 over time. Note
that the half-life is a constant 28.8 years.
Decrease in the number of 14C nuclei over time.
Carbon–14 Dating
Applying Radiocarbon Dating
PROBLEM:
PLAN:
The charred bones of a sloth in a cave in Chile represent the earliest
evidence of human presence in the southern tip of South America. A
sample of the bone has a specific activity of 5.22 disintegrations per
minute per gram of carbon (d/min•g). If the 12C/14C ratio for living
organisms results in a specific activity of 15.3 d/min•g, how old are the
bones (t1/2 of 14C = 5730 yr)?
Calculate the rate constant using the given half-life. Then use the first-order
rate equation to find the age of the bones.
SOLUTION:
k=
ln 2
=
t1/2
1
t=
ln
k
0.693
= 1.21 x 10-4 yr -1
5730 yr
1
A0
=
ln
1.21 x 10-4 yr -1
At
The bones are about 8900 years old.
15.3
5.22
= 8.89 x 103 yr
PROBLEM: Strontium-90 is a radioactive byproduct of nuclear reactors that behaves
biologically like calcium, the element above it in Group 2A(2). When 90Sr is
ingested by mammals, it is found in their milk and eventually in the bones of those
drinking the milk. If a sample of 90Sr has an activity of 1.2 x 1012 d/s, what are the
activity and the fraction of nuclei that have decayed after 59 yr (t1/2 of 90Sr = 29
yr).
The fraction of nuclei that have decayed is the change in the number of nuclei,
expressed as a fraction of the starting number. The activity of the sample (A) is
proportional to the number of nuclei (N). We are given the A0 and can find At from
the integrated form of the first-order rate equation.
t1/ 2
0.693
−1
ln ( 2 ) 0.693=
k
0.024
yr
=
=
=
29 yr
k
k
ln At = -kt + ln A0
At = 2.9 x
ln At = -(0.024 yr -1)(59 yr) + ln(1.2 x 1012 d/s)
1011 d/s
Fraction
decayed
Fraction
decayed
=
0.76
=
ln At = 26.4
(1.2 x1012 - 2.9 x 1011)
(1.2 x 1012)
Nuclear Transformation
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The change of one element into another.
27
13
1
Al + 24 He → 30
P
+
15
0n
Linear accelerator
Cyclotron accelerator.
249
98
263
Cf + 188 O → 106
Sg + 4 01 n
Nuclear Fission and Fusion
• Fusion – Combining two light nuclei to form a heavier,
more stable nucleus.
• Fission – Splitting a heavy nucleus into two nuclei with
smaller mass numbers.
1
0
n+
235
92
U→
142
56
Ba +
91
36
1
0
Kr + 3 n
18
The Interconversion of Mass and Energy
E = mc2
∆E = ∆mc2
∆E
∆m = 2
c
The mass of the nucleus is less than the
combined masses of its nucleons. The
mass decrease that occurs when
nucleons are united into a nucleus is
called the mass defect.
The mass defect (∆m) can be used to
calculate the nuclear binding energy in
MeV (mega-electron volts).
1 amu = 931.5 x 106 eV = 931.5 MeV
Energy and Mass
When a system gains or loses energy it also gains or loses a
quantity of mass.
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E = mc2
Δm = mass defect
ΔE = change in energy
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If ΔE is negative (exothermic), mass is lost from the
system.
Calculating the Binding Energy per Nucleon
PROBLEM:
Iron-56 is an extremely stable nuclide. Compute the binding energy per
nucleon for 56Fe and compare it with that for 12C (mass of 56Fe atom =
55.934939 amu; mass of 1H atom = 1.007825 amu; mass of neutron =
1.008665 amu).
Find the mass defect, ∆m; multiply that by the MeV equivalent and divide
by the number of nucleons.
PLAN:
SOLUTION:
Mass Defect = [(26 x 1.007825 amu) + (30 x 1.008665 amu)] - 55.934939
∆m = 0.52846 amu
Binding energy =
12C
(0.52846 amu)(931.5 MeV/amu)
56 nucleons
= 8.790 MeV/nucleon
has a binding energy of 7.680 MeV/nucleon, so 56Fe is more stable.
Binding Energy
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The energy required to decompose the nucleus into its
components.
Iron-56 is the most stable nucleus and has a binding energy of
8.79 MeV.
Biological Effects of Radiation
Depend on:
1.
2.
3.
4.
Energy of the radiation
Penetrating ability of the radiation
Ionizing ability of the radiation
Chemical properties of the radiation source
Penetrating power of
radioactive emissions
Penetrating power is inversely
related to the mass and charge
of the emission.
Nuclear changes
cause chemical
changes in
surrounding matter
by excitation and
ionization.
rem (roentgen equivalent for man)
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The energy dose of the radiation and its effectiveness in causing
biologic damage must be taken into account.
Number of rems = (number of rads) × RBE
sievert (Sv), 1 Sv = 0.01 rem
rads = radiation absorbed dose Gray (Gy) SI unit of dose
RBE = relative effectiveness of the
radiation in causing biologic damage
Rad = 0.01 J/kg=0.01 Gy
25
Effects of Short-Term Exposures to Radiation
DNA Damage by High Energy Radiation
Radiation-induced DNA damage can lead to mutagenesis and cancer
PHOTODIMERIZATION: Pyrimidine dimerization causes skin cancer
O
O
O
CH3
N
CH3
CH3
N
N
hv
2
O
O
N
N
R
R
H
H
N
O
R
PHOTOIONIZATION MECHANISMS
Indirect Photoionization Damage: Attack of DNA by OH. in dilute aqueous solutions results
in strand breaks and modification of nucleotide bases (γ radiolysis)
Strand-base +nhν → strand-base+. + e-aq
2H2O + nhν → 2H++OH. + OHOH. + strand-base → strand-base.-OH
strand-base.-OH → ssb and base release
Direct Photoionization Damage: Direct ionization of DNA by high energy radiation
Aqueous base (B1) ionization energies and orbital
diagrams for 5'-dGMP-, 5'-dAMP-, 5'-dCMP-, 5'-dTMP-, 2'deoxythymidine and uridine
P.R. LeBreton, Papadantonakis, G.A. et al Proc. Natl. Acad. Sci. USA (1998) 95, 5550-5555
After consumption of Na 131I, the patient's
thyroid is scanned for radio- activity levels to
determine the efficiency of iodine absorption.