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© 2014 Pearson Education, Inc.

Sherril Soman

Grand Valley State University

Lecture Presentation

Chapter 19 放射性與核化學 Radioactivity and Nuclear Chemistry

19.1 Diagnosing Appendicitis 911

19.2 The Discovery of Radioactivity 912 19.3 Types of Radioactivity 913

19.4 The Valley of Stability: Predicting the Type of Radioactivity 918

19.5 Detecting Radioactivity 920

19.6 The Kinetics of Radioactive Decay and Radiometric Dating 921

19.7 The Discovery of Fission: The Atomic Bomb and Nuclear Power 928

19.8 Converting Mass to Energy: Mass Defect and Nuclear Binding Energy 932

19.9 Nuclear Fusion: The Power of the Sun 935

19.10 Nuclear Transmutation and Transuranium Elements 936

19.11 The Effects of Radiation on Life 937

19.12 Radioactivity in Medicine and Other

Applications 940

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He Th

U 234 90 4 2

238

92 → +

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© 2014 Pearson Education, Inc.

19.1 Diagnosing Appendicitis

• Changes in the structure of the nucleus are used in many ways in medicine.

• Nuclear radiation can be used to visualize or test structures in your body to see if they are operating properly.

– For example, labeling atoms so their intake and output can be monitored

• Nuclear radiation can also be used to treat diseases because the radiation is ionizing, allowing it to attack unhealthy tissue.

Nuclear Medicine

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• Antoine-Henri Becquerel designed an experiment to determine if phosphorescent minerals also gave off X-rays.

– Phosphorescence is the long- lived emission of light by atoms or molecules that sometimes occurs after they absorb light.

– X-rays are detected by their ability to penetrate matter and expose a photographic plate.

19.2 The Discovery of Radioactivity

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• Becquerel discovered that certain minerals were constantly producing energy rays that could penetrate matter.

• Becquerel determined that

1. all the minerals that produced these rays contained uranium, and

2. the rays were produced even though the mineral was not exposed to outside energy.

• He called them uranic rays because they were emitted from minerals that contained uranium.

 Like X-rays

 Not related to phosphorescence

Discovery of Radioactivity: Becquerel

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• Marie Curie determined the rays were emitted from specific elements.

• She also discovered new elements by detecting their rays.

 Radium named for its green phosphorescence

 Polonium named for her homeland

• Because these rays were no longer just a property of uranium, she renamed it radioactivity.

Discovery of Radioactivity: Marie Curie

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• Radioactive rays can ionize matter.

Cause uncharged matter to become charged

Basis of Geiger counter and electroscope

• Radioactive rays have high energy .

• Radioactive rays can penetrate matter .

• Radioactive rays cause phosphorescent chemicals to glow .

Basis of scintillation counter

Other Properties of Radioactivity

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What Is Radioactivity?

• Radioactivity is the release of tiny, high- energy particles or gamma rays from

an atom.

• Particles are ejected from the nucleus.

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19.3 Types of Radioactive Decay

• Rutherford discovered three types of rays:

 Alpha (α) rays

Have a charge of +2 and a mass of 4 amu

What we now know to be helium nucleus

 Beta (β) rays

Have a charge of −1 c.u. and negligible mass

Electron-like

 Gamma (γ) rays

Form of light energy (not a particle like α and β)

• In addition, some unstable nuclei emit positrons.

 Like a positively charged electron

• Some unstable nuclei will undergo electrons capture.

 A low energy electron is pulled into the nucleus.

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• The number of neutrons is calculated by subtracting the atomic number from the mass number.

• The nucleus of an isotope is called a nuclide.

– Less than 10% of the known nuclides are nonradioactive; most are radionuclides.

• Each nuclide is identified by a symbol.

– Element – mass number = X – A

Facts about the Nucleus

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Facts about the Nucleus

• Every atom of an element has the same number of protons.

– Atomic number (Z)

• Atoms of the same elements can have different numbers of neutrons.

– Isotopes

– Different atomic masses

• Isotopes are identified by their mass number (A).

– Mass number = number of protons + neutrons

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Nuclear Equations

Alpha Particles, α

Nuclei of He atoms, 4 He 2+ .

Low penetrating power, stopped by a sheet of paper.

Copyright © 2011 Pearson Canada Inc.

General Chemistry: Chapter 25 Slide 12 of 45

2

The sum of the mass numbers must be the same on both sides.

The sum of the atomic numbers must be the same on both sides

He Th

U 234 90 4 2

238

92 → +

Sum of mass numbers: 238 = 234 + 4

Sum of atomic numbers: 92 = 90 + 2

EOS

Parent nuclide Daughter nuclide

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Example 19.1 Writing Nuclear Equations for Alpha Decay

Begin with the symbol for Ra-224 on the left side of the equation and the symbol for an alpha particle on the right side.

Equalize the sum of the mass numbers and the sum of the atomic numbers on both sides of the equation by writing the appropriate mass number and atomic number for the unknown daughter nuclide.

Refer to the periodic table to deduce the identity of the unknown daughter nuclide from its atomic number and write its symbol. Since the atomic number is 86, the daughter nuclide is radon (Rn).

For Practice 19.1

Write the nuclear equation for the alpha decay of Po-216.

Solution

Write the nuclear equation for the alpha decay of Ra-224.

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• Occurs when an unstable nucleus emits an electron

• About 10 times more penetrating than α, but only about half the ionizing ability

• When an atom loses a β particle its

 atomic number increases by 1, and

 the mass number remains the same .

Beta Decay

Sum of mass numbers: 228 = 228 + 0 Sum of atomic numbers: 88 = 89 - 1

1 n → 1 p + 0 β + ν

0 1 -1

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Gamma Emission

Highly penetrating energetic photons.

238 92 U → 234 90 Th + 4 2 He 2+

234 90 Th

234 90 Th + γ

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Positrons Emission, β +

Simplest process is the decay of a free proton:

Commonly encountered in artificially produced radioactive nuclei of the lighter elements:

1 p → 1 n + 0 e

1 0 +1

30 P

15

30 14 Si + +1 0 e

Sum of mass numbers: 30 = 30 + 0 Sum of atomic numbers: 15 = 14 + 1

1 p → 1 n + 0 e

1 0 +1

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• Occurs when an inner orbital electron is pulled into the nucleus.

• No particle emission, but atom changes

 Same result as positron emission

• When a proton combines with the electron to make a neutron its

 mass number stays the same, and

 its atomic number decreases by 1.

Electron Capture

Sum of mass numbers: 92 +0 = 92

Sum of atomic numbers: 44 -1 = 43

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© 2014 Pearson Education, Inc.

Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 19.2 Writing Nuclear Equations for Beta Decay, Positron Emission, and Electron Capture

a. In beta decay, the atomic number increases by 1 and the mass number remains unchanged.

The daughter nuclide is element number 98, californium.

b. In positron emission, the atomic number decreases by 1 and the mass number remains unchanged.

The daughter nuclide is element number 7, nitrogen.

c. In electron capture, the atomic number also decreases by 1 and the mass number remains unchanged.

The daughter nuclide is element number 52, tellurium.

Solution

Write the nuclear equation for each type of decay.

a. beta decay in Bk-249 b. positron emission in O-15 c. electron capture in I-111

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Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

For Practice 19.2

a. Write three nuclear equations to represent the nuclear decay sequence that begins with the alpha decay of U-235 followed by a beta decay of the daughter nuclide and then another alpha decay.

b. Write the nuclear equation for the positron emission of Na-22.

c. rite the nuclear equation for electron capture in Kr-76.

Example 19.2 Writing Nuclear Equations for Beta Decay, Positron Emission, and Electron Capture

Continued

For More Practice 19.2

Potassium-40 decays to produce Ar-40. What is the method of decay? Write the nuclear equation for this decay.

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A X

Z

Mass Number

Element Symbol

Atomic Number

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For Z = 1 ⇒ 20, stable N/Z ≈ 1.

19.4 The Valley of Stability: Predicting

the Type of Radioactivity

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For Z = 1 ⇒ 20, stable N/Z ≈ 1.

For Z = 20 ⇒ 40,

stable N/Z approaches 1.25.

For Z = 40 ⇒ 80,

stable N/Z approaches 1.5.

N/Z too small

positron emission or electron

capture

N/Z too large

Z=83

For Z > 83,

there are no stable nuclei.

β decay

19.4 The Valley of Stability: Predicting

the Type of Radioactivity

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N/Z Ratio, N=中子, Z=質子

• The ratio of neutrons : protons is an important measure of the stability of the nucleus.

• If the N/Z ratio is too high, neutrons are converted to protons via β decay.

N/Z=中子/質子比太高,表示中子太多,所以中子放出β decay 變成質子

• If the N/Z ratio is too low, protons are converted to neutrons via positron emission or electron capture.

• N/Z=中子/質子比太低,表示質子子太多,所以質子放出

positron emission, or electron capture變成中子

– Or via α decay, though not as efficiently

1 p → 1 n + 0 e

1 0 +1

1 n → 1 p + 0 β + ν

0 1 -1

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© 2014 Pearson Education, Inc.

Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 19.3 Predicting the Type of Radioactive Decay

Predict whether each nuclide is more likely to decay via beta decay or positron emission.

a. Mg-28 b. Mg-22 c. Mo-102

a. Magnesium-28 has 16neutrons and 12protons, so N/Z = 1.33. However, for Z = 12, you can see from Figure 19.5 that stable nuclei should have an N/Z of about 1. Alternatively, you can see from the periodic table that the atomic mass of magnesium is 24.31. Therefore, a nuclide with a mass number of 28 is too heavy to be stable because the N/Z ratio is too high and Mg-28 undergoes beta decay, resulting in the conversion of a neutron to a proton.

b. Magnesium-22 has 10 neutrons and 12 protons, so N/Z = 0.83 (too low).

Alternatively you can see from the periodic table that the atomic mass of magnesium is 24.31. A nuclide with a mass number of 22 is too light; the N/Z ratio is too low. Therefore, Mg-22 undergoes positron emission, resulting in the conversion of a proton to a neutron. (Electron capture would accomplish the same thing as positron emission, but in Mg-22, positron emission is the only decay mode observed.)

Solution

Figure 19.5Stable and Unstable Nuclei A plot of N (the number of neutrons) versus Z (the number of protons) for all known stable nuclei—represented by green dots on this graph—shows that these nuclei cluster together in a region known as the valley (or island) of stability. Nuclei with an N/Z ratio that is too high tend to undergo beta decay. Nuclei with an N/Z ratio that is too low tend to undergo positron emission or electron capture.

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© 2014 Pearson Education, Inc.

Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 19.3 Predicting the Type of Radioactive Decay

Continued

For Practice 19.3

Predict whether each nuclide is more likely to decay via beta decay or positron emission.

a. Pb-192 b. Pb-212 c. Xe-114

c. Molybdenum-102 has 60 neutrons and 42 protons, so N/Z = 1.43. However, for Z = 42, you can see from Figure 19.5 that stable nuclei should have an N/Z ratio of about 1.3. Alternatively you can see from the periodic table that the atomic mass of molybdenum is 95.94. A nuclide with a mass number of 102 is too heavy to be stable; the N/Z ratio is too high.

Therefore, Mo-102 undergoes beta decay, resulting in the conversion of a neutron to a proton.

Figure 19.5Stable and Unstable Nuclei A plot of N (the number of neutrons) versus Z (the number of protons) for all known stable nuclei—represented by green dots on this graph—shows that these nuclei cluster together in a region known as the valley (or island) of stability. Nuclei with an N/Z ratio that is too high tend to undergo beta decay. Nuclei with an N/Z ratio that is too low tend to undergo positron emission or electron capture.

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• Besides the N/Z ratio, the actual numbers of protons and neutrons affect stability.

• Most stable nuclei have even numbers of protons and neutrons.

• Only a few have odd numbers of protons and neutrons.

• If the total number of nucleons adds to a magic number, the nucleus is more stable.

– Same principle as stability of the noble gas electron configuration

– Most stable when N or Z = 2, 8, 20, 28, 50, 82; or N = 126

Magic Numbers

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U-238 Decay Series

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Decay Series

• In nature, often one radioactive nuclide changes into another radioactive nuclide.

 That is, the daughter nuclide is also radioactive.

• All atoms with Z > 83 are radioactive.

• All of the radioactive nuclides that are produced one

after the other until a stable nuclide is made is called

a decay series.

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Natural Radioactivity

• There are small amounts of radioactive minerals in the air, ground, and water.

• They are even in the food you eat!

• The radiation you are exposed to from natural

sources is called background radiation.

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Particles emitted by radioactive nuclei have a lot of energy and therefore can be readily detected.

•Radioactive rays can expose light-protected photographic film.

•We may use photographic film to detect the presence of radioactive rays—film badge

dosimeters.

19.5 Detecting Radioactivity

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• Radioactive rays cause air to become ionized.

• An electroscope detects radiation by its ability to penetrate the flask and ionize the air inside.

• A Geiger-Müller counter works by counting electrons generated when Ar gas atoms are ionized by radioactive rays.

Detecting Radioactivity

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• Radioactive rays cause certain chemicals to give off a flash of light when they strike the chemical.

• A scintillation counter( 閃爍偵測器) is able to count the number of flashes per minute.

Detecting Radioactivity

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• The rate of change in the amount of radioactivity is constant, and is different for each radioactive “isotope.”

 Change in radioactivity measured with Geiger counter

 Counts per minute

• Each radionuclide had a particular length of time it required to lose half its radioactivity—a constant half-life.

 We know that processes with a constant half-life follow first order kinetic rate laws.

• The rate of radioactive change was not affected by temperature.

 In other words, radioactivity is not a chemical reaction!

Rate of Radioactive Decay

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19.6 The Kinetics of Radioactive Decay and Radiometric Dating

ln N t

N 0 = -kt

The rate of disintegration of a radioactive material – called the activity, A, or the decay rate –

is directly proportional to the number of atoms.

A: decay rate or activity k: decay constant

N: amount

Rate of radioactive decay = A = kN ln A t

A 0 = -kt

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• The shorter the half-life, the larger the value of λ and the faster the decay proceeds.

• Radioactive decay follows first-order kinetics (Ch. 13.4):

N 0 is the amount of isotope initially N t is the amount of isotope at time t k: decay constant k

Radioactive Decay Rates (cont’d)

N t

ln –––– = – kt N 0

0.693

t

1/2 = –––––

k ln –––– = – 1 kt

1/2

2

the half-life

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Half-Lives of Various Nuclides

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Half of the radioactive atoms decay each half-life.

Half-Life

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Kinetics of Radioactive Decay

• Radioactive decay is a first order process.

• N t = number of radioactive nuclei at time, t

• N 0 = initial number of radioactive nuclei

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Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 19.4 Radioactive Decay Kinetics

You are given the initial mass of Pu-236 in a sample and asked to find the mass after 5.00 years.

Sort

Plutonium-236 is an alpha emitter with a half-life of 2.86 years. If a sample initially contains 1.35 mg of Pu-236, what mass of Pu-236 is present after 5.00 years?

Strategize

Use the integrated rate law (Equation 19.3) to solve this problem. You must determine the value of the rate constant (k) from the half-life expression (Equation 19.1).

Use the value of the rate constant, the initial mass of Pu-236, and the time along with integrated rate law to find the final mass of Pu-236. Since the mass of the Pu-236 (mPu-236) is directly proportional to the number of atoms (N), and since the integrated rate law contains the ratio (Nt/N0), the initial and final masses can be substituted for the initial and final number of atoms.

Given: mPu-236(initial) = 1.35 mg;

t = 5.00 yr; t1/2 = 2.86 yr Find: mPu-236 (final)

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© 2014 Pearson Education, Inc.

Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 19.4 Radioactive Decay Kinetics

Conceptual Plan Continued

Follow your plan. Begin by determining the rate constant from the half-life.

Solve the integrated rate law for Ntand substitute the values of the rate constant, the initial mass of Pu-236, and the time into the solved equation. Calculate the final mass of Pu-236.

Solve

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Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 19.4 Radioactive Decay Kinetics

Continued Check

The units of the answer (mg) are correct. The magnitude of the answer (0.402 mg) is about one-third of the original mass (1.35 mg), which seems reasonable given that the amount of time is between one and two half-lives. (One half-life would result in one-half of the original mass and two half-lives would result in one- fourth of the original mass.)

For Practice 19.4

How long will it take for the 1.35 mg sample of Pu-236 in Example 19.4 to decay to 0.100 mg?

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 C-14 radioactive with half-life = 5730 years

While still living, C-14/C-12 is constant because the organism replenishes its supply of carbon.

 CO2 in air is the ultimate source of all C in an organism. 活體中C-14/C-12 為一定值

Once the organism dies the C-14/C-12 ratio decreases.死亡後C-14/C-12減少

By measuring the C-14/C-12 ratio in a once living artifact and comparing it to the C-14/C-12 ratio in a living organism, we can tell how long ago the organism was alive.可由C-14/C-12比判斷年限或死亡多久

The limit for this technique is 50,000 years old.

 About 9 half-lives, after which radioactivity from C-14 will be below the background radiation

 最多只能判斷到5萬年前,因為經過9次半衰期C-14變成與背景值差不多

Radiocarbon Dating

死海古卷,2000年前

有些人聲稱裹屍布是耶穌的原殮布,奇蹟 般地印有

他的形象。 1988年,羅馬天主教會選擇 三個獨立的實驗室進行

放射性碳測年的壽衣。該實驗室參加布的 樣品和測量

的碳14含量。這三個獨立的實驗室都得出 了類似的結果,在

壽衣是從做亞麻起源於大約公元1325

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Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 19.5 Radiocarbon Dating

Sort

A skull believed to belong to an ancient human being has a carbon-14 decay rate of 4.50 disintegrations per minute per gram of carbon (4.50 dis/min · g C). If living organisms have a decay rate of 15.3 dis/min · g C, how old is the skull? (The decay rate is directly proportional to the amount of carbon-14 present.)

You are given the current rate of decay for the skull and the assumed initial rate. You are asked to find the age of the skull, which is the time that passed in order for the rate to have reached its current value.

Given: ratet= 4.50 dis/min · g C;

rate0= 15.3 dis/min · g C;

Find: t

Strategize

Use the expression for half-life (Equation 19.1) to find the rate constant (k) from the half-life for C-14, which is 5730 yr (Table 19.3).

Use the value of the rate constant and the initial and current rates to find t from the integrated rate law (Equation 19.4).

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© 2014 Pearson Education, Inc.

Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 19.5 Radiocarbon Dating

Continued

Conceptual Plan

Follow your plan. Begin by finding the rate constant from the half-life.

Substitute the rate constant and the initial and current rates into the integrated rate law and solve for t.

Solve

Solution

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Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Check

The units of the answer (yr) are correct. The magnitude of the answer is about 10,000 years, which is a little less than two half-lives. This value is reasonable given that two half-lives would result in a decay rate of about 3.8 dis/min g C.

Example 19.5 Radiocarbon Dating

Continued

For Practice 19.5

A researcher claims that an ancient scroll originated from Greek scholars in about 500 b.c. A measure of its carbon-14 decay rate gives a value that is 89% of that found in living organisms. How old is the scroll and could it be authentic?

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超過5萬年怎麼判斷?The Age of Earth (and Moon)

Copyright © 2011 Pearson Canada Inc.

Slide 47 of 45 General Chemistry: Chapter 25

238 92 U 206 82 Pb + 8 4 2 He 2+ + 6 -1 0 β

A lunar rock that has been radiometrically dated to be about 4.6 billion years old.

14 steps

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Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 19.6 Using Uranium/Lead Dating to Estimate the Age of a Rock

Sort

A meteor contains 0.556 g of Pb-206 to every 1.00 g of U-238. Assuming that the meteor did not contain any Pb-206 at the time of its formation, determine the age of the meteor. Uranium-238 decays to lead-206 with a half-life of 4.5 billion years.

You are given the current masses of Pb-206 and U-238 in a rock and asked to find its age. You are also given the half-life of U-238.

Given: mU-238 = 1.00 g; mPb-206 = 0.556 g;

t1/2 = 4.5 × 109yr Find: t

Strategize

Use the integrated rate law (Equation 19.3) to solve this problem. To do so, you must first determine the value of the rate constant (k) from the half-life expression (Equation 19.1).

Before substituting into the integrated rate law, you also need the ratio of the current amount of U-238 to the original amount (Nt/N0). The current mass of uranium is simply 1.00 g. The initial mass includes the current mass (1.00 g) plus the mass that has decayed into lead-206, which can be determined from the current mass of Pb-206.

Use the value of the rate constant and the initial and current amounts of U-238 along with the integrated rate law to find t.

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Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 19.6 Using Mass Percent Composition as a Conversion Factor

Continued

Conceptual Plan

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Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 19.6 Using Mass Percent Composition as a Conversion Factor

Continued Solve

Follow your plan. Begin by finding the rate constant from the half-life.

Solution

Determine the mass in grams of U-238 that is required to form the given mass of Pb-206.

Substitute the rate constant and the initial and current masses of U-238 into the integrated rate law and solve for t. (The initial mass of U-238 is the sum of the current mass and the mass that is required to form the given mass of Pb-206.)

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© 2014 Pearson Education, Inc.

Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 19.6 Using Mass Percent Composition as a Conversion Factor

Continued Solution

Check

The units of the answer (yr) are correct. The magnitude of the answer is about 3.2 billion years, which is less than one half-life. This value is reasonable given that less than half of the uranium in the meteor has decayed into lead.

For Practice 19.6

A rock contains a Pb-206 to U-238 mass ratio of 0.145 : 1.00. Assuming that the rock did not contain any Pb-206 at the time of its formation, determine its age.

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19.7 The Discovery of Fission: The Atomic Bomb and Nuclear Power

Fission 核分裂

– The large nucleus splits into two smaller nuclei.

Fusion 核融合

– Small nuclei can be

accelerated to smash together to make a larger nucleus.

Both fission and fusion release enormous amounts of energy.

 Fusion releases more energy per gram than fission.

Nonradioactive Nuclear Changes

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Nuclear Fission 核分裂

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• Use about 50 kg of fuel to generate

enough electricity for 1 million people

• No air pollution

• Use about 2 million kg of fuel to generate

enough electricity for 1 million people

• Produce NO 2 and SO x that add to acid rain

• Produce CO 2 that adds to the

greenhouse effect

Nuclear Power Plants versus Coal-Burning

Power Plants

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• Core meltdown

 Water loss from core; heat melts core

 China Syndrome

 Chernobyl

• Waste disposal

 Waste highly radioactive

 Reprocessing; underground storage?

 Federal High Level Radioactive Waste Storage Facility at Yucca Mountain, Nevada

• Transporting waste

• How do we deal with nuclear power plants that are no longer safe to operate?

 Yankee Rowe in Massachusetts

Concerns about Nuclear Power

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Where Does the Energy from Fission Come From?

In chemical reactions ΔE is too small to notice m.

In nuclear reactions ΔE is large enough to see m.

1 MeV = 1.6022×10 -13 J

If m = 1.0 u then ΔE =1.4924×10 -10 J or 931.5 MeV

E = mc 2

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• When a nucleus forms, some of the mass of the separate nucleons is converted into energy.

• The difference in mass between the separate

nucleons and the combined nucleus is called the mass defect.

• The energy that is released when the nucleus forms is called the binding energy.

– 1 MeV = 1.602 × 10 −13 J

– 1 amu of mass defect = 931.5 MeV

– The greater the binding energy per nucleon, the more stable the nucleus.

19.8 Converting Mass to Energy: Mass Defect and Nuclear Binding Energy

Mass Defect and Binding Energy

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Mass Defect and Binding Energy

Mass lost (m) = 236.05258 amu – 235.86769 amu

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Mass Defect

Nuclear binding energy in 4 2 He

(4.0320-4.0015)u * (1.66 *10-27kg/1u)

=1.8 8*10-29kg E = mc2

= (1.8 8*10-29kg) x (3x108)2m2s-2

= 1.65 x 10-12J

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Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 19.7 Mass Defect and Nuclear Binding Energy

Solution

Calculate the mass defect and nuclear binding energy per nucleon (in MeV) for C-16, a radioactive isotope of carbon with a mass of 16.014701 amu.

Calculate the mass defect as the difference between the mass of one C-16 atom and the sum of the masses of 6 hydrogen atoms and 10 neutrons.

Calculate the nuclear binding energy by converting the mass defect (in amu) into MeV. (Use 1 amu = 931.5 MeV.)

Determine the nuclear binding energy per nucleon by dividing by the number of nucleons in the nucleus.

For Practice 19.7

Calculate the mass defect and nuclear binding energy per nucleon (in MeV) for U-238, which has a mass of 238.050784 amu.

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• Fusion is the combining of light nuclei to make a heavier, more stable nuclide.

• The sun uses the fusion of hydrogen isotopes to make helium as a power source.

• It requires high input of energy to initiate the process.

– Because need to overcome repulsion of positive nuclei

• It produces 10 times the energy per gram as fission.

• It produces no radioactive byproducts.

• Unfortunately, the only currently working application is the H-bomb.

19.9 Nuclear Fusion: The Power of the Sun

Nuclear Fusion

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Fusion

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Tokamak Fusion Reactor

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19.10 Nuclear Transmutation and Transuranium Elements

• High energy particles can be smashed into target nuclei, resulting in the production of new nuclei.

• The particles may be radiation from another

radionuclide, or charged particles that are accelerated.

 Rutherford made O-17 bombarding N-14 with alpha rays from radium.

 Cf-244 is made by bombarding U-238 with C-12 in a particle accelerator.

Making New Elements: Artificial Transmutation

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Artificial Transmutation

• Bombardment of one nucleus with

another causing new atoms to be made

 Can also bombard with neutrons

• Reaction done in a particle accelerator

 Linear

 Cyclotron

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Target

- + - + - + - + - + - + - + - + - + - + - + -

Source

+

+ - + - + - + - + - + - + - + - + - + - + - +

+ +

Linear Accelerator

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Cyclotron

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19.11 The Effects of Radiation on Life

• Radiation has high energy—enough energy to

knock electrons from molecules and break bonds.

– Ionizing radiation

• Energy transferred to cells can damage biological

molecules and cause malfunction of the cells.

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Acute Effects of Radiation

• High levels of radiation over a short period of time kill large numbers of cells.

– From a nuclear blast or exposed reactor core

• It causes weakened immune system and lower ability to absorb nutrients from food.

– May result in death, usually from infection

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• Low doses of radiation over a period of time show an increased risk for the development of cancer.

– Radiation damages DNA that may not get repaired properly.

• Low doses over time may damage

reproductive organs, which may lead to sterilization.

• Damage to reproductive cells may lead to genetic defects in offspring.

Chronic Effects

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• The curie (Ci) is an exposure of 3.7 × 10 10 events per second.

– No matter the kind of radiation

• The gray (Gy) measures the amount of energy absorbed by body tissue from radiation.

– 1 Gy = 1 J/kg body tissue

• The rad also measures the amount of energy absorbed by body tissue from radiation.

– 1 rad = 0.01 Gy

Measuring Radiation Exposure

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• A correction factor is used to account for a number of factors that affect the result of the exposure.

• This biological effectiveness factor is the RBE, and the result is the dose in rems.

– Rads × RBE = rems

– Rem = roentgen equivalent man

Measuring Radiation Exposure

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1. More energetic radiation has a larger effect.

2. More ionizing radiation penetrates human tissue more deeply.

– Gamma >> Beta > Alpha

3. More ionizing radiation has a larger effect.

– Alpha > Beta > Gamma

4. Characteristics of the radionuclide are as follows:

The radioactive half-life of the radionuclide The biological half-life of the element

The physical state of the radioactive material

Factors That Determine the Biological

Effects of Radiation

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Biological Effects of Radiation

• The amount of danger to humans of

radiation is measured in the unit rems.

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• Radiotracers

Certain organs absorb most or all of a particular element.

You can measure the amount absorbed by using tagged isotopes of the element and a Geiger counter.

Tagged = radioisotope that can then be detected and measured

Use radioisotope with a short half-life

Use radioisotope that is low ionizing

Beta or gamma

19.12 Radioactivity in Medicine and Other Applications

Medical Uses of Radioisotopes, Diagnosis

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A Bone Scan

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PET scan

Positron emission tomography

F-18 tagged glucose

F-18 is a positron emitter.

Brain scan and function

Medical Uses of Radioisotopes,

Diagnosis

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• Cancer treatment

Cancer cells are more sensitive to radiation than healthy cells; radiation can be used to kill cancer cells without doing significant damage.

Brachytherapy

Place radioisotope directly at site of cancer.

Teletherapy

Use gamma radiation from Co-60 outside to penetrate inside

IMRT

Radiopharmaceutical therapy

Use radioisotopes that concentrate in one area of the body

Medical Uses of Radioisotopes,

Treatment – Radiotherapy

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Gamma Ray Treatment

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Nonmedical Uses of Radioactive Isotopes

• Smoke detectors

 Am-241

 Smoke blocks ionized air; breaks circuit

• Insect control

 Sterilize males

• Food preservation

• Radioactive tracers

 Follow progress of a “tagged”

atom in a reaction

• Chemical analysis

 Neutron activation analysis

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• END

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• Authenticating art object

 Many older pigments and ceramics were made from minerals with small amounts of

radioisotopes.

• Crime scene investigation

• Measure thickness or condition of industrial materials

 Corrosion

 Track flow through process

 Gauges in high temp processes

 Weld defects in pipelines

 Road thickness

Nonmedical Uses of Radioactive Isotopes

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• Agribusiness

 Develop disease-resistant crops

 Trace fertilizer use

• Computer disks to enhance data integrity.

• Nonstick pan coatings

 Initiates polymerization

• Photocopiers to help keep paper from jamming

• Sterilize cosmetics, hair products, contact lens solutions, and other personal hygiene products

Nonmedical Uses of Radioactive Isotopes

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++++++++++++

---

α β γ

Rutherford’s Experiment

8-1

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α β γ

0.01 mm 1 mm 100 mm Pieces of Lead

Penetrating Ability of Radioactive Rays

8-1

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• Very small volume compared to volume of the atom

• Essentially entire mass of atom

• Very dense

• Composed of protons and neutrons that are tightly held together

– The particles that make up the nucleus are called nucleons.

Facts about the Nucleus

8-1

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Radioactivity

• Radioactive nuclei spontaneously decompose into smaller nuclei—

radioactive decay.

– We say that radioactive nuclei are unstable.

– Decomposing involves the nuclide emitting a particle and/or energy.

• The parent nuclide is the nucleus that is undergoing radioactive decay.

• The daughter nuclide is the new nucleus that is made.

• All nuclides with 84 or more protons are radioactive.

8-1

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Important Atomic Symbols

8-1

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• Rutherford discovered that during the

radioactive process, atoms of one element are changed into atoms of a different

element—transmutation.

 Showing that statement three of Dalton’s atomic theory is not valid all the time, only for chemical reactions

• For one element to change into another, the number of protons in the nucleus must change!

Transmutation

8-1

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• Chemical reactions involve changes in the electronic structure of the atom.

– Atoms gain, lose, or share electrons.

– No change in the nuclei occurs.

• Nuclear reactions involve changes in the structure of the nucleus.

– When the number of protons in the nucleus changes, the atom becomes a different

element.

Chemical Processes versus Nuclear Processes

8-1

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• The design is used in the United States (GE, Westinghouse).

• Water is both the coolant and moderator.

• Water in the core kept under pressure to keep it from boiling.

• Fuel is enriched uranium.

 Subcritical

• Containment dome is made of concrete.

Pressurized Light Water Reactor

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PLWR

Core

Containment building

Turbine

Condenser

Cold

water

Boiler

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PLWR – Core

Cold water

Fuel rods Hot water Control

rods

The control rods are made of neutron

absorbing material.

This allows the rate of neutron flow through the reactor to be

controlled.

Because the neutrons are required to

continue the chain

reaction, the control

rods control the rate of

nuclear fission.

參考文獻

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