Chapter 23

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Chapter 23

Nuclear Reactions and Their Applications

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Nuclear Reactions and Their Applications

23.1 Radioactive Decay and Nuclear Stability

23.2 The Kinetics of Radioactive Decay

23.3 Nuclear Transmutation: Induced Changes in Nuclei

23.4 The Effects of Nuclear Radiation on Matter

23.5 Applications of Radioisotopes

23.6 The Interconversion of Mass and Energy

23.7 Applications of Fission and Fusion

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Table 23.1

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Figure 23.1The behavior of three types of radioactive emissions in

an electric field.

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Types of Radioactive Decay: Balancing Nuclear Equations

Total ATotal Z Reactants = Total A

Total Z Products

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 Z but will be one atomic number higher.

Positron decay - a positron (01) is the antiparticle of an electron. A

proton in the nucleus is converted into a neutron with the expulsion of the positron. Z remains the same but the atomic number decreases.

Electron capture - a nuclear proton is converted into a neutron by the capture of an electron. Z remains the same but the atomic number decreases.

Gamma emission - energy release; no change in Z or A.

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Table 23.2

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Sample Problem 23.1 Writing Equations for Nuclear Reactions

PLAN:

SOLUTION:

PROBLEM: Write balanced equations for the following nuclear reactions:

(a) Naturally occurring thorium-232 undergoes decay.

(b) Chlorine-36 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 + 42He

(a) 23290Th 228

88Ra + 42He

(b) 3617Cl + 0

-1e AZX

A = 36 and Z = 16 3617Cl + 0

-1e 3616S

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Figure 23.2 A plot of neutrons vs. 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.

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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Table 23.3

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Sample Problem 23.2 Predicting Nuclear Stability

PLAN:

SOLUTION:

PROBLEM: Which of the following nuclides would you predcit 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; the diagram on shows stability when N/Z ≥ 1.3.

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Sample Problem 23.3 Predicting the Mode of Nuclear Decay

PLAN:

SOLUTION:

PROBLEM: Predict the nature of the nuclear change(s) each of the following radioactive nuclides is likely to undergo:

(a) 125B (b) 234

92U (c) 7433As (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) = N/t

SI unit of decay is the becquerel (Bq) = 1d/s.

curie (Ci) =curie (Ci) =

number of nuclei disintegrating each second in 1g of radium-226 =

3.70x1010d/s

Nuclear decay is a first-order rate process.

Large k means a short half-life and vice versa.

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Figure 23.4 Decrease in the number of 14C nuclei over time.

Table 23.4

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Sample Problem 23.4 Finding the Number of Radioactive Nuclei

PLAN:

SOLUTION:

PROBLEM: Strontium-90 is a radioactive by-product 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 1.2x1012 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 A from the integrated form of the first-order rate equation.

t1/2 = ln2/k so k = 0.693/29 yr = 0.024 yr-1

ln N0/Nt = ln A0/At = kt ln At = -kt + ln A0

ln At = -(0.024yr-1)(59yr) + ln(1.2x1012d/s)

ln At = 26.4 At = 2.9x1011d/s

Fraction decayed

=(1.2x1012-2.9x1011)

(1.2x1012)

Fraction decayed

= 0.76

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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 ratio of 12C:14C in 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 = ln 2/t1/2 = 0.693/5730yr = 1.21x10-4yr-1

t = 1/k ln A0/At = 1/(1.21x10-4yr-1) ln (15.3/5.22) = 8.89x103 yr

The bones are about 8900 years old.

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Figure 23.5 A linear accelerator.

The linear accelerator operated by Standford University, California

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Figure 23.6 The cyclotron accelerator.

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Table 23.5

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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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Table 23.6

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Figure 24.9

R C OH

O

R' OH18

+ R C O

O

R'18

+ H O H

R C OH

O

R' OH18

+ R C O

O

R' + H O H18

R C OH

O

H OR'+ R C O

O

R' + H O HH+

Which reactant contributes which group to the ester?

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Table 23.7

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Figure 23.10

PET and brain activity.

normal Alzheimer’s

Figure 23.9

The use of radioisotopes to image the thyroid

gland.

asymmetric scan indicates disease

normal

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Figure 23.11 The increased shelf life of irradiated food.

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The Interconversion of Mass and Energy

E = mcE = mc22

E = E = mcmc22

m = m = E / cE / c22

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.

1 amu = 931.5x106 eV = 931.5 MeV

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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)] - 55.934939

m = 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 23.12 The variation in binding energy per nucleon.

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Figure 23.13 Induced fission of 235U.

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Figure 23.14 A chain reaction of 235U.

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Figure 23.15

Schematic of a light-water nuclear reactor.

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Figure 23.16

The tokamak design for magnetic containment of a

fusion plasma.

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Chapter 23