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Instructor:Dr. Orlando E. Raola

Santa Rosa Junior College

Chemistry 1B

General General ChemistryChemistry

Ch. 24:Nuclear Reactions and Their Applications

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Chapter 24

Nuclear Reactions and Their Applications

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Nuclear Reactions and Their Applications

24.1 Radioactive Decay and Nuclear Stability

24.2 The Kinetics of Radioactive Decay

24.3 Nuclear Transmutation: Induced Changes in Nuclei

24.4 Effects of Nuclear Radiation on Matter

24.5 Applications of Radioisotopes

24.6 Interconversion of Mass and Energy

24.7 Applications of Fission and Fusion

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Table 24.1 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.

Electrons in orbitals are involved as bonds break and form; nuclear particles do not take part.

Protons, neutrons, and other nuclear particles are involved; electrons in orbitals take part much less often.

Reactions are accompanied by relatively small charges in energy and no measurable changes in mass.

Reactions are accompanied by relatively large charges in energy and measurable changes in mass.

Reaction rates are influenced by temperature, concentration, catalyst, and the compound in which an element occurs.

Reaction rates depend on number of nuclei, but are not affected by temperature, catalysts, or, except on rare occasions, the compound in which an element occurs.

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Components of the Nucleus

Most of the mass of the atom is concentrated in the dense, tiny nucleus.

The nucleus comprises the neutrons and protons, collectively called nucleons.The total number of nucleons in a nucleus gives its mass number.

A nuclide is a nucleus with a particular composition. Each isotope of an element has a different nuclide.

A particular nuclide is often designated by its mass number; for example, chlorine-35 and chlorine-37.

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Notation for Nuclides

The relative mass and charge of a particle is described by the notation:

XAZ

A = mass number

Z = charge of the particle

electron e 0-1

proton

neutron n10

p11

Example:

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Radioactivity

Many nuclides are unstable and spontaneously emit radiation, a process termed radioactive decay.

- The intensity of the radiation is not affected by temperature, pressure, or other physical and chemical conditions.

When a nuclide decays, it emits radiation and usually changes into a nuclide of a different element.

There are three natural types of radioactive emission:

Beta particles (β, β-, or ) are high-speed electrons.β 0-1

Alpha particles (α, , or ) are identical to helium-4 nuclei.α 4 2 He 4

2

Gamma rays (γ or ) are very high-energy photons.γ 0 0

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Figure 24.1 How the three types of radioactive emissions behave in an electric field.

The positively charged α particles curve toward the negative plate, the negatively charged β particles curve towards the positive plate, and the γ rays are not affected by the electric field.

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Nuclear Equations

When a nuclide decays, it forms a daughter nuclide of lower energy.The excess energy is carried off by the emitted radiation and the recoiling nucleus of the daughter nuclide.

The decay process is represented by a balanced nuclear equation. Both the total charge and the total mass must be balanced:

Reactants = Products Total A Total Z

Total A Total Z

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Modes of Radioactive Decay

• Alpha (α) decay involves the loss of an α particle from the nucleus.– For each α particle emitted, A decreases by 4 and Z decreases

by 2 in the daughter nuclide.– This is the most common form of decay for a heavy, unstable

nucleus.

• β- decay involves the ejection of a β- particle from the nucleus.– A neutron is converted to a proton, which remains in the

nucleus, and a β- particle is expelled:

– A remains the same in the daughter nuclide but Z increases by 1 unit.

228 222 4 88 86 2Ra → Rn + α

n → p + β 1 0

1 1

0 -1

63 63 0 28 29 -1Ni → Cu + β

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• Positron (β+) emission is the emission of a β+ particle from the nucleus.– The positron is the antiparticle of the electron.– A proton in the nucleus is converted into a neutron, and a

positron is emitted:

– A remains the same in the daughter nuclide but Z increases by 1 unit.

• Electron capture occurs when the nucleus interacts with an electron in a low atomic energy level.– A proton is transformed into a neutron:

– The effect on A and Z is the same as for positron emission.

p → n + β 1 1

1 0

0 1

11 11 0 6 5 1C → B + β

p + e → n 1 1

0 -1

0 1

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• Gamma (γ) emission involves the radiation of high-energy γ photons.– Gamma emission usually occurs together with other forms of

radioactive decay.– Several γ photons of different energies can be emitted from an

excited nucleus as it returns to the ground state.– γ emission results in no change in either A or Z since γ rays

have no mass or charge.

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Table 24.2 Modes of Radioactive Decay*

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Table 24.2 Modes of Radioactive Decay*

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Sample Problem 24.1 Writing Equations for Nuclear Reactions

PROBLEM: Write balanced equations for the following nuclear reactions:

(a) Naturally occurring thorium-232 undergoes α decay.

(b) Zirconium-86 undergoes electron capture.

PLAN: We first write a skeleton equation that includes the mass numbers, atomic numbers, and symbols of all the particles on the correct sides of the equation, showing the unknown product particle as X. Then, because the total of mass numbers and the total of charges on the left side and the right side of the equation must be equal, we solve for A and Z, and use Z to determine the identity of X from the periodic table.

AZ

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SOLUTION:

Sample Problem 24.1

(a) Writing the skeleton equation, with the α particle as a product:

232 A 4 90 Z 2Th → X + α

For A, 232 = A + 4, so A= 228.For Z, 90 = Z + 2, so Z = 88.

The daughter nuclide produced in this reaction is Ra.

(b) Writing the skeleton equation, with the captured electron as a reactant:86 0 A40 -1 Z Zr + e → X For A, 86+ 0 = A so A= 86.

For Z, 40 -1 = Z so Z = 39.Element X is yttrium, symbol Y.

232 228 4 90 88 2Th → Ra + α

86 0 8640 -1 39 Zr + e → Y

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Nuclear Stability

Two key factors determine the stability of a nuclide:- the number of neutrons (N), the number of protons (Z), and

their ratio (N/Z), and

- the total mass of the nuclide.

A plot of number of neutrons vs. number of protons for all stable nuclides produces a band of stability that gradually curves above the line for N = Z.

- Lighter nuclides are stable when N = Z.- As Z increases, the N/Z for stable nuclei gradually increases.- All nuclides with Z > 83 are unstable.

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Figure 24.2 A plot of number of neutrons vs. number of protons for the stable nuclides.

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Stability and Nuclear Structure

Protons within the nucleus experience electrostatic repulsive forces, which destabilize the nucleus.

The strong force, which exists between all nucleons, counteracts the weaker repulsive forces.

Nucleons are found in nucleon energy levels, and pairing of the spins of like nucleons leads to greater stability.- Elements with an even Z (number of protons) usually have a

larger number of stable nuclides.

- Over half the stable nuclides have both even N and even Z.

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Table 24.3 Number of Stable Nuclides for Elements 48 to 54*

ElementAtomic No. (Z)

No. of nuclides

Cd 48 8

In 49 2

Sn 50 10

Sb 51 2

Te 52 8

I 53 1

Xe 54 9

* Even Z shown in boldface.

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Table 24.4 An Even-Odd Breakdown of the Stable Nuclides

Z NNo. of nuclides

Even Even 157

Even Odd 53

Odd Even 50

Odd Odd 4

TOTAL 264

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Sample Problem 24.2 Predicting Nuclear Stability

SOLUTION:

PROBLEM: Which of the following nuclides would you predict to be stable and which radioactive? Explain.

PLAN: In order to evaluate the stability of each nuclide, we determine the N and Z values and the N/Z ratio. We can then compare these to the values for stable nuclides. We also note whether Z and N are even or odd.

(a) Ne (b) S (c) Th (d) Ba 18 32 236 123 10 16 90 56

This nuclide has N = (18 – 10) = 8 and Z = 10, so the N/Z ratio is

18 – 10

10= 0.8, which is too low to be stable.

(a) 18Ne is Radioactive.

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Sample Problem 24.2

This nuclide has N = Z = 16, so N/Z = 1.0.With Z < 20 and even values for N and Z, this nuclide is most likely stable.

This nuclide has Z = 90, and every nuclide with Z < 83 is radioactive.

This nuclide has N = 67 and Z = 56, so N/Z = 1.20. For Z values of 55 to 60, Figure 24.2A shows that N/Z ≥ 1.3, so this nuclide has too few neutrons to be stable

(b) 32S is stable.

(c) 236Th is radioactive.

(d) 123Ba is radioactive.

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Predicting the Mode of Decay

Nuclide type Description Mode of Decay

neutron-rich high N/Z β- decayneutron → proton, lowers N/Z

proton-rich low N/Z β+ emission or e- captureproton → neutron, increases N/Z

heavy nuclides Z > 83 α decayreduces both Z and N

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Sample Problem 24.3 Predicting the Mode of Nuclear Decay

PLAN: If the nuclide is too heavy to be stable (Z > 83), it undergoes α decay. For other cases, we use the Z value to obtain its atomic mass from the periodic table.

If the mass number of the nuclide is higher than the atomic mass, the nuclide has too many neutrons: N is too high and β- decay occurs.

If the mass number is lower than the atomic mass, the nuclide has too many protons: Z is too high and the nuclide decays by β+ emission or e- capture.

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

(a) B (b) U (c) As (d) La 12 234 81 127 5 92 33 57

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SOLUTION:

Sample Problem 24.3

(a) 12B has Z = 5 and its atomic mass is 10.81. The nuclide’s A value of 12 is significantly higher than its atomic mass, so it is neutron rich. It will probably undergo β- decay.

(b) 234U has Z = 92, so it will undergo α decay and decrease its total mass.

(c) 81As has Z = 33 and its atomic mass is 74.92. The A value of 81 is much higher than the atomic mass, so it is neutron rich and will probably undergo β- decay.

(d) 127La has Z = 57 and its atomic mass is 138.9 The A value of 127 is much lower than the atomic mass, so it is proton rich and will probably undergo β+ emission or e- capture.

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Figure 24.3 The 238U decay series.

A parent nuclide may undergo a series of decay steps before a stable daughter nuclide is formed.

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Detection and Measurement of Radioactivity

An ionization counter detects radioactive emissions as they ionize a gas.Ionization produces free electrons and gaseous cations, which are attracted to electrodes and produce an electric current.

A scintillation counter detects radioactive emissions by their ability to excite atoms and cause them to emit light.

Radioactive particles strike a light-emitting substance, which emits photons. The photons strike a cathode and produce an electric current.

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Figure 24.4 Detection of radioactivity by an ionization counter.

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Figure 24.5 A scinatillation “cocktail” in tubes to be placed in the counter.

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Units of Radioactivity

The SI unit of radioactivity is the becquerel (Bq), defined as one disintegration per second (d/s).

The curie (Ci) is a more commonly used unit:

1 Ci = 3.70x1010 d/s

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Rate of Radioactive Decay

Radioactive nuclei decay at a characteristic rate, regardless of the chemical substance in which they occur.

The rate of radioactive decay (A) (also called the activity) is proportional to the number of nuclei present.

A = kN

Radioactive decay follows first-order kinetics, and the rate constant k is called the decay constant.

The larger the value of k, the higher the activity of the substance.

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Half-Life of Radioactive Decay

The half-life of a nuclide is the time taken for half the nuclei in a sample to decay.

- The number of nuclei remaining is halved after each half-life.

- The mass of the parent nuclide decreases while the mass of the daughter nuclide increases

- Activity is halved with each succeeding half-life.

ln = -kt or Nt = N0e-kt and ln = ktNt

N0

Nt

N0

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Figure 24.6 Decrease in the number of 14C nuclei over time.

t1/2 =ln 2

k

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Table 24.5 Decay Constants (k) and Half-Lives (t1/2) of Beryllium Isotopes

Nuclide k t1/2

Be 1.30x10-2/day 53.3 days

Be 1.0x1016/s 6.7x10-17 s

Be Stable

Be 4.3x10-7/yr 1.6x106 yr

Be 5.02x10-2/s 13.8 s

748494

10 411 4

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Sample Problem 24.4 Finding the Number of Radioactive Nuclei

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)?

PLAN: 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), and we are given A0. We can find At from the integrated form of the first-order rate equation, in which t is 59 yr. We need the value of k, which we can calculate from the given t1/2.

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Sample Problem 24.4

SOLUTION:

fraction decayed = A0 – At

A0

=t1/2 =

ln 2

kso k =

ln 2

t1/2

0.693

29 yr= 0.024 yr-1

Nt

N0

ln = lnAt

A0

= kt or lnA0 – lnAt = kt

so lnAt = -kt + lnA0 = -(0.024 yr-1 x 59 yr) + ln(1.2x1012 d/s) = -1.4 + 27.81 lnAt = 26.4 At = e26.4 = 2.9x1011 d/s

=1.2x1012 d/s – 2.9x1011 d/s

1.2x1012 d/s = 0.76

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Radioisotopic Dating

• Radioisotopes can be used to determine the ages of certain objects.

• Radiocarbon dating measures the relative amounts of 14C and 12C in materials of biological origin.–The ratio of 14C/12C remains the same for all living organisms.–Once the organism dies, the amount of 14C starts to decrease as it

decays to form 14N.–Since 14C decays at a predictable rate, measuring the amount

present indicates the time that has passed since the organism died.

• 40K/40Ar ratios can be used to determine the age of certain rocks.

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Figure 24.7 Ages of several objects determined by radiocarbon dating.

t = 1

k

A0

At

ln

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Sample Problem 24.5 Applying Radiocarbon Dating

PROBLEM: The charred bones of a sloth in a cave in Chile represent the earliest evidence of human presence at 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 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: We calculate k from the given half-life. Then use the first-order rate equation to find the age of the bones, using the given activities of the bones and of a living organism.

SOLUTION:

k = ln 2

t1/2

= 0.693

5730 yr= 1.21x10-4 yr-1

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The bones are about 8900 years old.

Sample Problem 24.5

t = 1

k

A0

At

ln ln= 1

1.21x10-4 yr-1

15.3 d/min·g

5.22 d/min·g

= 8.89x103 yr

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Nuclear Transmutation

Nuclear transmutation is the induced conversion of the nucleus of one element into the nucleus of another.

This is achieved by high-energy bombardment of nuclei in a particle accelerator.

14 4 1 17 7 2 1 8N + α → p + O

Nuclear transmutation reactions can be described using a specific short-hand notation:

reactant nucleus (particle in, particle(s) out) product nucleus

The above reaction can be written as: 14N (α, p) 17O.

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Figure 24.8 Schematic diagram of a linear accelerator.

The linear accelerator uses a series of tubes with alternating voltage. A particle is accelerated from one tube to the next by repulsion.

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Figure 24.9 Schematic diagram of a cyclotron accelerator.

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Table 24.6 Formation of some Transuranium Nuclides*

Reaction Half-life of Product

239 94Pu + 2 n → Am + β

1 0

0-1

241 95 432 yr

239 94Pu + α → Cm + n

42

1 0

242 96

241 95Am + α → Bk + 2 n

42

1 0

243 97

242 96 Cm + α → Cf + n

42

1 0

245 98

253 99Es + α → Md + n

42

1 0

256101

253 99 Am + O → Lr + 5 n

188

1 0

256101

163 days

4.5 h

45 min

28 s

76 min

* Like chemical reactions, nuclear reactions may occur in several steps.

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Effects of Nuclear Radiation on Matter

Radioactive emissions collide with surrounding matter, dislodging electrons and causing ionization. Each such event produces a cation and a free electron.

The number of cation-electron pairs is directly related to the energy of the incoming ionizing radiation.

Ionizing radiation has a destructive effect on living tissue. The danger of a particular radionuclide depends on- the type of radiation,- its half-life, and- its biological behavior.

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Units of Radiation

The gray is the SI unit for energy absorption. 1 Gy = 1 J absorbed per kg of body tissue.

The rad is more widely used: 1 rad = 0.01 J/kg or 0.01 Gy.

The rem is the unit of radiation dosage equivalent to a given amount of tissue damage in a human.

no. of rems = no. of rads x RBE

The RBE is the relative biological effectiveness factor. The rem allows us to assess actual tissue damage by taking into account the strength of the radiation, the exposure time, and the type of tissue.

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Figure 24.10 Penetrating power of radioactive emissions.

Penetrating power is inversely related to the mass, charge, and energy of the emission.

The effect of radiation on living tissue depends on both the penetrating power and the ionizing ability of the radiation.

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Molecular Interactions with Radiation

The interaction of ionizing radiation with molecules causes the loss of an electron from a bond or a lone pair.

This results in the formation of free radicals, molecular or atomic species with one or more unpaired electrons.

Free radicals are unstable and extremely reactive.

H2O + H2O + e-

H2O + H2O H3O+ + OH and e- + H2O H + OH-

H + RCH2RCH CHR' CHR'

Double bonds in membrane lipids are very susceptible to attack by free-radicals:

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Sources of Ionizing Radiation

• There are several natural sources of background radiation.

• Cosmic radiation increases with altitude.• Radon is a radioactive product of uranium and thorium

decay.– Rn contributes to 15% of annual lung cancer deaths.

• Radioactive 40K is present in water and various food sources.

• Radioactive 14C occurs in atmospheric CO2.

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Figure 24.11 US radon distribution.

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Table 24.7 Typical Radiation Doses from Natural and Artificial Sources

Source of Radiation Average Adult Exposure

Natural

Cosmic radiation 30-50 mrem/yr

Radiation from the groundFrom clay soil and rocksIn wooden housesIn brick housesIn concrete (cinder block) houses

~25-170 mrem/yr10-20 mrem/yr60-70 mrem/yr60-160 mrem/yr

Radiation from the air (mainly radon)Outdoors, average valueIn wooden housesIn brick housesIn concrete (cinder block) houses

20 mrem/yr 70 mrem/yr130 mrem/yr260 mrem/yr

Internal radiation from minerals in tap water and daily intake of food.(40K, 14C, Ra) ~ 40 mrem/yr

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Table 24.7 Typical Radiation Doses from Natural and Artificial Sources

Source of Radiation Average Adult Exposure

Artificial

Diagnostic x-ray methodsLung (local)Kidney (local)Dental (does to the skin)

0.04-0.2 rad/film1.5-3 rad/film≤ 1 rad/filmLocally ≤ 10,000 rad

Therapeutic radiation treatment

Other SourcesJet flight (4 h)Nuclear testingNuclear power industry

~1 mrem< 4 mrem/yr< 1 mrem/yr

Total average value 100-200 mrem/yr

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Figure 24.12 Two models of radiation risk.

The linear response model proposes that radiation effects a`ccumulate over time regardless of dose.The S-shaped response model implies there is a threshold above which the effects are more significant.

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Table 24.8 Acute Effects of a Single Dose on Whole-Body Irradiation

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Radioactive Tracers

• The isotopes of an element exhibit very similar chemical and physical behavior. – A small amount of radioactive isotope mixed with the stable

isotope will undergo the same chemical reactions, and can act as a tracer.

• Radioactive tracers are used– to study reaction pathways,– to track physiological functions,– to trace material flow,– to identify the components of a substance from a very small

sample, and– to diagnose a wide variety of medical conditions.

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Figure 24.13 The use of radioisotopes to image the thyroid gland.

This 131I scan shows an asymmetric image that is indicative of disease.

A 99Tc scan of a healthy thyroid.

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Table 24.9 Some Radioisotopes Used as Medical Tracers

Isotope Body Part or Process

11C, 18F, 13N, 15O PET studies of brain, heart60Co, 192Ir Cancer therapy64Cu Metabolism of copper59Fe Blood flow, spleen67Ga Tumor imaging123I, 131I Thyroid111In Brain, colon42K Blood flow81mKr Lung99mTc Heart, thyroid, liver, lung, bone201Tl Heart muscle90Y Cancer, arthritis

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Figure 24.14 PET and brain activity.

These PET scans show brain activity in a normal person (left) and in a patient with Alzheimer’s disease (right). Red and yellow indicate relatively high activity within a region.

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Other Applications of Ionizing Radiation

• Radiation therapy– Cancer cells divide more rapidly than normal cells, and are

therefore susceptible to radioisotopes that interfere with cell division.

• Destruction of microbes– Irradiation of food increases its shelf life by killing

microorganisms that cause rotting or spoilage.

• Insect control• Power for spacecraft instruments

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Figure 24.15 The increased shelf life of irradiated food.

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The Interconversion of Mass and Energy

The total quantity of mass-energy in the universe is constant.Any reaction that releases or absorbs energy also loses or gains mass.

E = mc2 or ΔE = Δmc2 so Δm = ΔE

c2

In a chemical reaction, the energy changes in breaking or forming bonds is relatively small, so mass changes are negligible.

In a nuclear reaction, the energy changes are enormous and the mass changes are easily measurable.

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Nuclear Binding Energy

The mass of the nucleus is less than the combined masses of its nucleons.- Mass always decreases when nucleons form a nucleus, and the “lost” mass is released as energy.- Energy is required to break a nucleus into individual nucleons.

The nuclear binding energy is the energy required to break 1 mol of nuclei into individual nucleons.- Binding energy is expressed using the electron volt (eV).

Nucleus + nuclear binding energy → nucleons

1 u = 931.5 x 106 eV = 931.5 MeV

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Sample Problem 24.6 Calculating the Binding Energy per Nucleon

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 u; mass of 1H atom = 1.007825 u; mass of neutron = 1.008665 u).

PLAN: Iron-56 has 26 protons and 20 neutrons. We calculate the mass difference, Δm, when the nucleus forms by subtracting the given mass of one 56Fe atom from the sum of the masses of 26 1H atoms and 30 neutrons. To find the binding energy per nucleon, we multiply Δm by the equivalent in MeV and divide by the number of nucleons.

Δm = [(26 x mass 1H atom) + (30 x mass neutron)] – mass 56Fe atom= [(26 x 1.007825 u) + (30 x 1.008665 u)] - 55.934939= 0.52856 u

Calculating the mass difference:

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Sample Problem 24.6

Binding energy per nucleon = 0.52846 u x 931.5 MeV/u

56 nucleons

= 8.790 MeV/nucleon

An 56Fe nucleus would require more energy per nucleon to break up into its nucleons than would 12C, so 56Fe is more stable than 12C.

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Figure 24.16 The variation in binding energy per nucleon.

The greater the binding energy per nucleon, the more stable the nuclide.

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Fission or Fusion

The binding energy per nucleon peaks at elements with mass number A ≈ 60.- Nuclides become more stable with increasing number up to around 60 nucleons, after which stability decreases.

There are two ways nuclides can increase their binding energy per nucleon:

A heavier nucleus can split into lighter ones by undergoing fission.

Lighter nuclei can combine to form a heavier nucleus in a process called fusion.

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Figure 24.17 Fission of 235U caused by neutron bombardment.

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Nuclear Fission

Nuclear fission involves the splitting of large nuclei into smaller nuclei, using neutron bombardment to start the process.

Fission releases energy and generates more high-energy neutrons, which cause further fission to occur.

The fission process becomes self-sustaining by a chain reaction. The mass required to achieve this is called the critical mass.

The energy from nuclear fission can be harnessed and converted to other forms of energy.

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Figure 24.18 A chain reaction involving fission of 235U.

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Figure 24.19 An atomic bomb based on 235U.

An atomic bomb uses an uncontrolled chain reaction to produce a powerful explosion.

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Figure 24.20 A light-water nuclear reactor.

The dome-shaped structure is the containment shell for the nuclear reactor.

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Figure 24.20 A light-water nuclear reactor.

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Figure 24.21 The tokamak design for magnetic containment of a fusion plasma.

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Figure B24.1 Element synthesis I the life cycle of a star.

Chemical Connections