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NUCLEAR CHEM
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COMPARISON OF CHEMICAL AND NUCLEAR REACTIONS
Chemical Reactions Nuclear Reactions
One substance is converted into another but atoms never change identity
Atoms of one element typically are converted into atoms of another element
Orbital electrons are involved as bonds break and form; nuclear particles do not take part.
Protons neutrons, and other particles are involved; orbital electrons take part much less often
Reactions are acoompanied by relatively small changes in energy and no measurable changes in mass.
Reactions are coompanied by relatively large changes in energy and measurable changes in mass.
Reactions are influenced by temperature, concentrations catalysts, and the compound in which the element occurs.
Reaction rates depend on number of nuclei but are not affected by temperature, catalysts.
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THREE TYPES OF RADIOACTIVE EMISSIONS IN AN ELECTRIC FIELD
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TYPES OF RADIOACTIVE DECAY: BALANCING NUCLEAR EQUATIONS
Alpha decay - A decreases by 4 and Z decreases by 2. Every element heavier than Pb undergoes decay.
Beta decay - ejection of a particle from the nucleus from the conversion of a neutron into a proton and the expulsion of 0
-1. The product nuclide will have the same A but will be one atomic number higher.Positron decay - a positron (0
-1) is the antiparticle of an electron. A proton in the nucleus is converted into a neutron with the expulsion of the positron. A remains the same but the atomic number decreases.
Electron capture - a nuclear proton is converted into a neutron by the capture of an electron. A remains the same but the atomic number decreases.
Gamma emission - energy release; no change in Z or A.
ProductsReactants TotalATotalZ
TotalATotalZ
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WRITING EQUATIONS FOR NUCLEAR REACTIONS
Write balanced equations for the following nuclear reactions:(a) Naturally occurring thorium-232 undergoes decay.(b) Zirconium-86 undergoes electron capture.
Write a skeleton equation; balance the number of neutrons and charges; solve for the unknown nuclide.
A = 228 and Z = 88 23290Th 228
88Ra + 42α
(a) 23290Th 228
88X + 42α
(b) 8640Zr + 0
-1e AZX
A = 86 and Z = 3986
40Zr + 0-1e 86
39Y
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A PLOT OF NUMBER OF NEUTRONS vs. NUMBER OF PROTONS FOR THE STABLE NUCLIDES.
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NUCLEAR STABILITY AND MODE OF DECAY
Very few stable nuclides exist with N/Z < 1
The N/Z ratio of stable nuclides gradually increases a Z increases.
All nuclides with Z > 83 are unstable
Elements with an even Z usually have a larger number of stable nuclides than elements with an odd Z
Well over half the stable nuclides have both even N and even Z
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PREDICTING THE MODE OF DECAY
Neutron-rich nuclides undergo decay.
Neutron-poor nuclides undergo positron decay or electron capture.
Heavy nuclides undergo decay.
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PREDICTING NUCLEAR STABILITY
Which of the following nuclides would you predict to be stable and which radioactive? Explain.
(a) 1810Ne (b) 32
16S (c) 23690Th (d) 123
56Ba
Stability will depend upon the N/Z ratio, the value of Z, the value of stable N/Z nuclei, and whether N and Z are even or odd.
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Predicting Nuclear Stability
PLAN:
SOLUTION:
Which of the following nuclides would you predict to be stable and which radioactive? Explain.
(a) 1810Ne (b) 32
16S (c) 23690Th (d) 123
56Ba
Stability will depend upon the N/Z ratio, the value of Z, the value of stable N/Z nuclei, and whether N and Z are even or odd.
(a) Radioactive
N/Z = 0.8; there are too few neutrons to be stable.
(b) Stable
N/Z = 1.0; Z < 20 and N and Z are even.
(c) Radioactive
Every nuclide with Z > 83 is radioactive.
(d) Radioactive
N/Z = 1.20; Figure 23.2A shows stability when N/Z ≥ 1.3.
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Predicting the Mode of Nuclear Decay
PLAN:
PROBLEM: Use the atomic mass of the element to predict the mode(s) of decay of the following radioactive nuclides:
(a) 125B (b) 234
92U (c) 8133As (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.
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Predicting the Mode of Nuclear Decay
PLAN:
SOLUTION:
PROBLEM: Use the atomic mass of the element to predict the mode(s) of decay of the following radioactive nuclides:
(a) 125B (b) 234
92U (c) 8133As (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.
(a) N/Z = 1.4 which is high.The nuclide will probably undergo decay altering Z to 6 and lowering the ratio.
(b) The large number of neutrons makes this a good candidate for decay.
(c) N/Z = 1.24 which is in the band of stability. It will probably undergo decay or positron emission.
(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.
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Figure 23.3 The 238U decay series.
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Decay rate (A) = -
SI unit of decay is the becquerel (Bq) = 1 d/s.
curie (Ci)curie (Ci) = = number of nuclei disintegrating each second in 1 g of radium-226 = 3.70 x 1010 d/s
Nuclear decay is a first-order rate process.
Large k means a short half-life and vice versa.
N
t
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Figure 23.4 Decrease in the number of 14C nuclei over time.
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Sample Problem 23.4 Finding the Number of Radioactive Nuclei
PLAN:
SOLUTION:
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 = ln so k = = 0.024 yr -1
ln = ln = kt ln At = -kt + ln A0
ln At = -(0.024 yr -1)(59 yr) + ln(1.2 x 1012 d/s)
ln At = 26.4 At = 2.9 x 1011 d/s
Fraction decayed
= (1.2 x1012 - 2.9 x 1011)
(1.2 x 1012)
Fraction decayed
= 0.76
N0
Nt
A0
At
2
k
0.693
29 yr
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Sample Problem 23.5 Applying Radiocarbon Dating
SOLUTION:
PROBLEM: 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)?PLAN: Calculate the rate constant using the given half-life. Then use the
first-order rate equation to find the age of the bones.
k = = = 1.21 x 10-4 yr -1
t = ln = ln = 8.89 x 103 yr
The bones are about 8900 years old.
ln 2
t1/2
0.693
5730 yrA0
At
1
k
1
1.21 x 10-4 yr -1
15.3
5.22
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Figure 23.5 Schematic of a linear accelerator.
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Figure 23.6 Schematic of a cyclotron accelerator.
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Figure 23.7
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.
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Figure 23.8
The use of radioisotopes to image the thyroid gland.
Figure 23.9
PET and brain activity.
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Figure 23.10 The increased shelf life of irradiated food.
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The Interconversion of Mass and Energy
EE = = mcmc22
EE = = mcmc22
mm = =
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
EEcc22
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Sample Problem 23.6 Calculating the Binding Energy per Nucleon
PLAN:
SOLUTION:
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.
Mass Defect = [(26 x 1.007825 amu) + (30 x 1.008665 amu)] - 55.934939m = 0.52846 amu
Binding energy = = 8.790 MeV/nucleon(0.52846 amu)(931.5 MeV/amu)56 nucleons
12C has a binding energy of 7.680 MeV/nucleon, so 56Fe is more stable.
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Figure 23.11 The variation in binding energy per nucleon.
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Figure 23.12 Induced fission of 235U.
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Figure 23.13 A chain reaction of 235U.
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Figure 23.14 A light-water nuclear reactor.
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Figure 23.15
The tokamak design for magnetic containment of a
fusion plasma.
