Study Guide Chapter 20 PDF

Preview

Summary

Study Guide Chapter 20...

Description

Chapter 20 Nuclear Chemistry 20.1 Nuclei and Nuclear Reactions 20.2 Nuclear Stability 

Patterns of Nuclear Stability

 Nuclear Binding Energy 20.3 Natural Radioactivity 

Kinetics of Radioactive Decay

 Dating Based on Radioactive Decay 20.4 Nuclear Transmutation 20.5 Nuclear Fission 20.6 Nuclear Fusion 20.7 Uses of Isotopes 

Chemical Analysis

 Isotopes in Medicine 20.8 Biological Effects of Radiation

20.1

Nuclei and Nuclear Reactions

This chapter emphasizes changes that occur within the nucleus of an atom. To begin, we will discuss two types of nuclear reactions: radioactive de cay and nuclear transmutation. Radioactive decay, or radioactivity, is described as the spontaneous emission of particles and/or radiation by unstable atomic nuclei. These processes often result in the formation of a new element. The kinds of particles emitted from various nuclei are shown in Table 20.1. Table 20.1 Particles From Radioactive Decay _____________________________________________________ Particle Mass (amu) Charge Symbol _____________________________________________________ alpha

4.0

+2

beta

0.0005

–1

positron

0.0005

+1

gamma

0

0

4 2

4

or 2 He

0 0 –1  or –1 e 0 0 +1  or +1 e 0 0

_____________________________________________________ A nucleus can also undergo change by nuclear transmutation. In this process, one nucleus reacts with another nucleus, an elementary particle, or a photon (gamma particle) to produce one or more new nuclei. Radioactive decay and nuclear transmutation processes are described by nuclear equations. These equations use isotopic and elementary particle symbols to represent the reactants and products of nuclear reactions. For example,

394

Chapter 20: Nuclear Chemistry

in the first nuclear transmutation ever observed (in 1919) alpha particles were used to bombard nitrogen-14 nuclei. The observed products were oxygen-17 nuclei and protons. The nuclear equation is: 14 7N

4

+ 2 He  

17 8O

1

+ 1H

The balancing rules for nuclear equations are given below and are applied to the above equation: 1. 2.

The sum of the mass numbers of the reactants must equal the sum of the mass numbers of the products (conservation of mass number). 14 + 4 = 18 = 17 + 1 The sum of the nuclear charges of the reactants must equal the sum of the nuclear charges of the products (conservation of atomic number). 7 + 2 = 9 = 8 + 1

EXAMPLE 20.1: NUCLEAR EQUATIONS Complete the following nuclear equations. Label the nuclear reaction as radioactive decay or nuclear transmutation. a. b.

14 7N 226

1

 + 0 n 

Ra  

4 2

14 6C

+ ___

+ ___

Solution a.

According to rule 1, the sum of the mass numbers of the reactants must equal the sum of the mass numbers of the products. 14 + 1 = 15 = 14 + A. Therefore, the unknown product will have a mass number of 1. According to rule 2, the sums of the nuclear charges on both sides of the equation must be the same. 7 + 0 = 7 = 6 + Z. Therefore, the nuclear charge of the unknown product must be 1, making it a proton. 14 7N

1

+ 0 n  

14 6C

1

+ 1p

This reaction is a nuclear transmutation b.

Note that the atomic number of radium is missing. The periodic table indicates that Ra is element number 88. Balancing the mass numbers first, we find that the unknown product must have a mass number of 222. Balancing the nuclear charges next, we find that the atomic number of the unknown must be 86. Element number 86 is radon. 226 88 Ra

 

4 2

+

222 86 Rn

This reaction is a radioactive decay. Table 20.1 of the text lists a number of comparisons between chemical and nuclear processes. Keep in mind that in chemical reactions the number of atoms of each element is conserved. Only changes in chemical bonding occur. However, in nuclear reactions the compositions of the atomic nuclei are altered, and so elements are converted from one to another. The forces that operate within the nucleus are the strong and the weak forces. The string force, as the name suggests, is the strongest of the four forces that occur in Nature. It is much stronger than gravitational or

395

Chapter 20: Nuclear Chemistry

electromagnetic forces but operates only over very small distances. The energy changes that accompany nuclear reactions are, therefore, much greater than the ones that occur in chemical reactions. ____________________________________________________________________________________________ PRACTICE EXERCISES 1. Write symbols for alpha particles, beta particles, and gamma rays. 2.

How many protons (p), neutrons (n), and electrons (e) does an atom of the radioisotope phosphorus-32 contain?

3.

Complete the following nuclear reactions: a.

14 7N 210 84

b.

b.

c.

4 2 He

d.

15 7N

1

+ 0 n   Po   +

9 4 Be

4 2

14 6C

+ _____

+ ____

 

  ____ +

12 6C

+ ____

0 –1 

____________________________________________________________________________________________ 20.2

Nuclear Stability

Patterns of Nuclear Stability Little is known about the forces that hold a nucleus together. However, some interesting facts emerge if we examine the numbers of protons and neutrons found in those nuclei that are stable. Nuclei can be classified according to whether they contain even or odd numbers of protons and neutrons. The number of stable isotopes of each of the four types of nuclei classified in this way are shown in Table 20.2 in your textbook. The following rules are useful in predicting nuclear stability: 1.

Nuclei with even numbers of both protons and neutrons are generally more stable than those with odd numbers of these particles. Nuclei that contain certain specific numbers of protons and neutrons ensure an extra degree of stability. These so-called magic numbers for protons and for neutrons are 2, 8, 20, 28, 50, 82, and 126.

2.

Nuclei with even numbers of protons or neutrons are generally more stable than those with odd numbers of these particles.

3.

All isotopes of elements after bismuth (Z = 83) are radioactive.

EXAMPLE 20.2A: NUCLEAR STABILITY Using the stability rules, rank the following isotopes in order of increasing nuclear stability. 39 20 Ca

40 20 Ca

30 15 P

Solution Phosphorus-30 should be the least stable because it has odd numbers of both protons and neutrons. Calcium-39 has an even number of protons (20) and an odd number of neutrons. With an even number of protons and a "magic

396

Chapter 20: Nuclear Chemistry

number" at that, it should be more stable than P-30. Of the three isotopes, Calcium-40 should be the most stable. It has an even number of protons and of neutrons. Both numbers are "magic numbers." 30 15 P

<

39 20 Ca

40 20 Ca

<

The principal factor for determining whether a nucleus is stable is the neutron to proton ratio. Figure 20.1 in your textbook shows a plot of the number of neutrons versus the number of protons in various isotopes. The stable nuclei are located in an area of the graph known as the belt of stability. In this figure, we see that at low atomic numbers, stable nuclei possess a neutron to proton ratio of about 1.0. Above Z = 20 the number of neutrons always exceeds the number of protons in stable isotopes. The n : p ratio increases to about 1.5 at the upper end of the belt of stability. If you were given the symbol of a radioisotope, without any experience, it would be impossible to tell its mode of decay. But with knowledge of the belt of stability, you can make accurate predictions of the expected mode of decay. Isotopes with too many neutrons lie above the belt of stability. The nuclei of these isotopes decay in such a way as to lower their n : p ratio. For example, one neutron may decay into a proton and a beta particle. 1 0n

 

1 1p

+

0 –1 

The proton remains in the nucleus, and the beta particle is emitted from the atom. The loss of a neutron and the gain of a proton produces a new isotope with two important properties. It has a lower n : p ratio , and thus is more likely to be stable. Also, the daughter product has an atomic number that is one greater than the decaying isotope, due to the additional proton. Consider the decay of carbon-14 for example. (Carbon-14 is continually produced in the upper atmosphere by the interaction of cosmic rays with nitrogen.) Carbon-14 has a higher n : p ratio than either of carbon's stable isotopes (C-12 and C-13), and decays by beta decay. 14 6C

Note that the product isotope,

14 7N

 

14 7N

+

0 –1 

, is one atomic number greater than carbon. Also, it is stable. Its n : p ratio is 1.0.

Isotopes with too many protons have low n : p ratios and lie below the belt of stability. These isotopes tend to decay by positron emission because this process produces a new isotope with a higher n : p ratio. During positron emission a proton emits a positron,

0 +1

, and becomes a neutron. 1 1p

 

1 0n

0

+ +1 

The neutron remains in the nucleus, and the positron is ejected from the atom. Thus, newly produced nucleus will contain one less proton and one more neutron than the parent nucleus. The n : p ratio increases due to positron decay.

397

Chapter 20: Nuclear Chemistry

Electron capture accomplishes the same end, that is, a higher n : p ratio. Some nuclei decay by capturing an orbital electron of the atom. 1 1p

+

0 –1 e

 

1 0n

Lanthanum-138, a naturally occurring isotope with an abundance of 0.089 percent, decays by electron capture. 138 57 La

+

0 –1 e

 

138 56Ba

Electron capture is accompanied by X-ray emission, represented in the above reaction as h (the equation for the energy of a photon.) EXAMPLE 20.2B: TYPES OF RADIOACTIVE DECAY The only stable isotope of sodium is sodium-20. What type of radioactivity would you expect from sodium-25? Solution Sodium-23 has 11 protons and 12 neutrons and is in the belt of stability. Sodium-25 must have two more neutrons, 0

and so has a higher n : p ratio than the stable isotope. Sodium-25 will decay by –1 emission. Nuclear Binding Energy. One of the important consequences of Einstein's theory of relativity was the discovery of the interconvertability of mass and energy. The total energy content (E) of a system of mass m is given by Einstein's theory: E = mc2 8 where c is the velocity of light (3.0  10 m/s). Therefore, the mass of a nucleus is a direct measure of its energy content. It was discovered in the 1930s that the measured mass of a nucleus is always smaller than the sum of the separate masses of its constituent nucleons (nuclear particles.) This difference in mass is called the mass defect.

When the mass defect is expressed as energy by applying Einstein's equation, it is called the binding energy of the nucleus. The binding energy is the energy required to break up a nucleus into its component, protons and neutrons. The binding energy provides a quantitative measure of nuclear stability. The greater its binding energy, the more stable a nucleus is toward decomposition. In terms of the

17 8O

nucleus for instance, the binding energy (BE) is the energy required for the process: 17 8O

1

1

+ BE   8 1p + 9 0 n

The mass defect, ∆m, is equal to the total mass of the products minus the total mass of the reactants. 17

∆m = [8(proton mass) + 9(neutron mass)] – ( 8 O nuclear mass) Here, we can substitute atomic masses, which include the electrons for the nuclear masses.

398

Chapter 20: Nuclear Chemistry

1

17

∆m = [8( 1 H atomic mass) + 9(neutron mass)] – ( 8 O atomic mass) This works because the mass of 8 electrons in the 8 hydrogen atoms is canceled by the mass of the 8 electrons in the oxygen atom. 1

8( 1 H atomic mass) = 8(proton mass) + 8(electron mass) (

17 8O

atom mass) = (

17 8O

nuclear mass) + 8(electron mass)

Continue by using the atomic masses given in the text . 17

1 ∆m = [8( 1 H atomic mass) + 9(neutron mass)] – ( 8 O atomic mass)

∆m = [8(1.007825 amu) + 9(1.008665 amu)] – (16.999131 amu) = 0.141454 amu The eight protons and nine neutrons have more mass than the oxygen-17 nucleus. The binding energy is: ∆E = (∆m)c2

E  (0.141454amu)  (1.6606 10 27 kg

amu

) (3.0 108 m ) s

E  2.1141 1011 J

(Recall that 1 J = 1 kg (

m2 s2

)

In comparing the stabilities of any two different nuclei, we must take into account the different numbers of nucleons per nucleus. A satisfactory comparison of nuclear stabilities can be made by using the binding energy per nucleon, that is, the binding energy of each nucleus divided by the total number of nucleons in the nucleus. This is one of the most important properties of a nucleus. When the BE per nucleon is plotted as a function of the atomic mass, we get the curve of binding energy as shown in Figure 20.2 of the text. Note that at first, it rises rapidly with increasing atomic mass, reaching a maximum at mass 56. Above mass 56, the binding energy drops slowly as atomic mass increases. EXAMPLE 20.2C: NUCLEAR BINDING ENERGY Calculate the nuclear binding energy of the light isotope of helium, helium-3. The atomic mass of 32 He is 3.01603 amu. Solution The binding energy is the energy required for the process 3 2 He

1

1

  2 1p + 0 n

399

Chapter 20: Nuclear Chemistry

where, ∆m = [2(proton mass) + (neutron mass)] – 32 He (nuclear mass) The mass difference is calculated using atomic masses: 3

1

∆m = [2( 1 H atomic mass) + (neutron mass)] – (2 He atomic mass) ∆m = [2(1.007825 amu) + 1.008665 amu] – 3.01603 amu –3 ∆m = 8.29  10 amu

Using Einstein's equation: 2 8 ∆E = (∆m)c2 = (8.29  10–3 amu)  (3.00  10 m/s)

= 7.46  1014 amu m2/s2

= 7.46  1014 amu m2/s2 

1.6606 10 27 kg 1J  1 amu 1 kg m 2 /s2

The binding energy is: ∆E = 1.24  10–12 J/atom Each 23 He atom contains three nucleons. The binding energy per nucleon is:

BE per nucleon =

1.24  10  12 J/atom 3 nucleons/atom

The binding energy per nucleon is: ∆E = 4.13  10–13 J/nucleon Comment By combining the above conversion factors into one constant, the number of steps in future calculations can be lessened. 2 8 ? J/amu = (3.00  10 m/s) 

1J 1.6606×10 -27 kg  1 amu 1 kg m 2 /s 2

= 1.49  10–10 J/amu

400

Chapter 20: Nuclear Chemistry

This constant is a useful factor relating energy to mass in amu. Applying this to the mass defect (0.14145 amu) calculated earlier in the discussion for

17 8O

, the binding energy is 2.10  10–11 J/atom.

________________________________________________________________________________________ PRACTICE EXERCISES 4. For each pair of nuclei, predict which one is the more stable. a.

3 2 He

or 2 He .

4

b.

26 13 Al

or

27 13 Al

5.

When a nucleus decays by positron emission, the atomic number of the decay product will be (higher or lower) than the original radioisotope?

6.

Fluorine has only one stable isotope, fluorine-19. a. The nucleus of fluorine-18 lies below the belt of stability. Write an equation for the decay of fluorine-18. b. The nucleus of fluorine-21 lies above the belt of stability. Write an equation for the decay of fluorine-21.

7.

How many atomic mass units are in one kilogram?

8.

Which has more mass, the nucleus of aluminum–27 or 13 protons and 14 neutrons?

9.

What is the binding energy of 13 Al ? Calculate its binding energy per nucleon. The atomic mass of aluminum-

27

27 is 26.981541 amu. 10. How much energy is released when one Po-214 atom decays by alpha emission? 214 84 Po

 

4 2

+

210 82 Pb

Given the atomic masses: Pb-214 = 213.99519 amu, Po-210 = 209.98286 amu, and He-4 = 4.00260 amu. ________________________________________________________________________________________ 20.3

Natural Radioactivity

A number of isotopes exist in nature that have an n : p ratio that places them outside the belt of stability. These isotopes, called radioisotopes, occur naturally on Earth and give rise to natural radioactivity. Uranium, thorium, radon, potassium-40, carbon-12, and tritium (H-3) are naturally occurring radioisotopes. For example, the radioactive decay of Uranium-238, which is fairly abundant in Earth's crust, begins a sequence of decay reactions that ultimately changes U-238 to a stable isotope of lead. The uranium series is shown in Figure 20.3 in your textbook. Kinetics of Radioactive Decay Radioactive decay rates obey first-order kinetics.

401

Chapter 20: Nuclear Chemistry

decay rate = number of atoms disintegrating per unit time = kN where k is the first-order rate constant and N is the number of atoms of the particular radioisotope present in the sample being studied. Recall that the half-life is related to the rate constant by:

t 12 =

0.693 k

The integrated first-order equation is

ln

N = –kt N0

where N is the number of atoms of the radioisotope present in the sample after time t has elapsed, and N0 is the number of atoms of the radioisotope present initially. If N0, N, and t are known, we can calculate the rate constant, k. The purpose of radioactive dating is to determine the age of geological and archaeological samples and specimens. The age (t in the calculation) of certain rocks, for instance, can be estimated from analysis of the number of atoms of a particular radioisotope present now (N), as compared to the number present when the rock was formed originally (N0). Rearranging the integrated first-order equation, the age t is given by:

t=–

1 N ln  N0

The value of the initial number of atoms N0 is the sum, N + D, where D is the number of daughter nuclei resulting from the decay of atoms of the radioisotope. The original number of atoms of a radioisotope present in a rock sample is equal to the number N remaining at time t, plus the number of daughter atoms (D). EXAMPLE 20.3: RADIOACTIVE DATING The rubidium-87/strontium-87 method of dating rocks was used to analyze lunar samples. The half-life of Rb-87 is 4.9  1010 yr. 87 3...