SOLUTIONS FOR THEEND-OF-CHAPTER

PROBLEMSof

ISOTOPES; PRINCIPLESAND APPLICATIONS (2005)

by

Gunter Faure and Teresa M. MensingDepartment of Geological Sciences

The Ohio State UniversityColumbus, Ohio 43210

Third EditionPrinciples of Isotope Geology

Wiley and Sons, Inc.Hoboken, New Jersey

Table of Contents

1. Principles of atomic physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

2. Decay modes of radionuclides . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

3. Radioactive decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

4. Mass spectrometry and isotope dilution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

5. The Rb-Sr method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

6. The K-Ar method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

7. The Ar*/ Ar method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .40 39 36

8. The K-Ca method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

9. The Sm-Nd Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

10. The U-Pb, Th-Pb, and Pb-Pb methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

11. The common Pb method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

12. The Lu-Hf method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

13. The Re-Os method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

14. The La-Ce method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

15. The La-Ba method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

16. Mixing theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

17. Origin of igneous rocks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

18. Water and sediment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

19. The oceans . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

20. U-Th series disequilibria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129

21. 21. Helium and tritium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136

22. Radiation damage methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144

23. Cosmogenic radionuclides . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148

24. Extinct radionuclides . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156

25. Thermonuclear radionuclides . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161

26. Hydrogen and oxygen . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165

27. Carbon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178

28. Nitrogen . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184

29. Sulfur . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187

30. Boron and other elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192

1

Chapter 1

Principles of Atomic Physics

1. Use Figure 1.1 to answer the following questions:

a. How many neutrons are present in the nucleus of an atom of ?

Answer: 26 - 12 = 14 neutrons

b. What is the mass number of the set of isobars that includes ?

Answer: A = 19

c. How many isotopes of the element beryllium are stable and how many are unstable?

Answer: One isotope of Be is stable; six are unstable

d. What nuclear property do isotones have in common?

Answer: They have the same number of neutrons

e. Why do all of the isotopes of an element have identical chemical properties

Answer: Neutral atoms of the isotopes of an element have the same number of electronsbecause they have the same number of protons.

f. What is the isotopic abundance of ?

Answer: 100%, based on numbers of atoms.

* * *

2. Calculate the atomic weight of silicon: Si (92.23 %, 27.976927 amu), Si (4.67 %,28 29

28.976494 amu), and Si (3.10 %, 29.973770 amu).30

Atomic weight of Si = 0.9223 × 27.976927 + 0.0467 × 28.97644 + 0.0310 × 29.97377 =28.085 amu.

2Chapter 1

Note that in chemistry the atomic weight is considered to be a dimensionless number

because it is a multiple of the mass of .

3a. Calculate the binding energies per nucleon for (4.002608323 amu), (24.985837

amu), (55.934940 amu), (115.901743 amu), and (232.03805 amu) in units

of million electron volts (MeV).

* * *

Mass of the proton: 1.00782503 amuMass of the neutron: 1.00866491 amu

The mass of = 2 × 1.00782503 + 2 × 1.00866491 = 4.0239788 amu

Mass defect = 4.03297988 - 4.002608323= 0.030371557 amu

Binding energy per nucleon:

=

= 7.072 MeV/nucleon:

Similarly, for the other nuclides:

Nuclide Binding energyMeV/nucleon

7.072

8.223

8.790

8.523

7.615

b. Plot the binding energies calculated above versus the atomic number of these nuclides anddraw a smooth curve through the points.(See graph)

c. Which element has the highest nuclear stability?

3Chapter 1

4Chapter 1

Answer: Iron has the highest nuclear stability

4a. Calculate the number of moles of strontium in 2.50 g of strontium nitrate. (Atomic weights:Sr = 87.62, N = 14.00, O = 15.99).

* * *

3 2Molecular weight of Sr (NO )

= 87.62 + 2 × 14.00 + 6 ×15.99 = 211.56

Number of moles of Sr

= = 0.0118 = 1.18 ×10 mol-2

b. How many atoms of Sr are present in 2.50 g of Sr nitrate?

Number of atoms of Sr

= 1.18 × 10 × 6.022 × 10 = 7.11 × 10 atoms21-2 23

c. How many atoms of Sr are present in 2.50 grams of strontium nitrate? (Abundance of Sr87 87

= 7.0 %)?

Number of atoms of Sr87

= 7.11 x 10 × 0.070 = 4.98 × 10 atoms2021

d. What is the ratio of the number of Sr atoms to Sr atoms in 2.50 g of Sr nitrate? 87 86

(Abundance of Sr = 9.9 %.)86

Number of atoms of Sr86

= 7.11 × 10 × 0.099 = 7.04 × 10 atoms21 20

Atomic ratio = = 0.7073

5Chapter 1 & 2

Note that the isotope ratio is also equal to the ratio of the abundances of the isotopesexpressed as atom percent.

2 5. The concentration of K O in a sample of orthoclase is 8.75 % by weight. How many atomsof K are present in 1.0 gram of this material? (Atomic weight of K = 39.09, abundance of40

K = 0.0117 % atomic).40

* * *

2 2Weight of K O in 1 g of orthoclase is 0.0875 g. The molecular weight of K O= 2 ×39.09 + 15.99 = 94.17

2The number of moles of K O = = 0.000929

2The number of moles of K O= 0.000929 × 2 = 0.00185

The number of atoms of K40

= 0.00185 × 6.022 × 10 × 0.00011723

= 1.309 × 10 atoms17

2 2Note that a concentration of 8.75% of K O is equivalent to an amount of 8.75g of K O per

2100g of sample, or to 0.0875 g of K O per 1.0 gram of sample.

Chapter 2

Decay Modes of Radionuclides

1. Complete the following decay equations by providing the appropriate values of Z and A:

a.

Answer:

b.

Answer:

c.

6Chapter 2

Answer:

d.

Answer:

e.

Answer:

f.

Answer:

2. Naturally occurring decays to stable through a series of radioactive daughters.

In this series, four negatrons ($ ) are emitted. How many alpha particles must be emitted to-

produce from ?

Answer:

3. Draw a decay-scheme diagram for that emits a suite of $ particles with an endpoint+

energy of 5.398 MeV followed by a (-ray of 1.6326 MeV. The total decay energy is 7.029MeV.

Answer: See diagram

4. Draw a decay-scheme diagram for that emits a suite of positrons ($ ) having an end+

point energy of 1.81 Me V and a gamma ray of 2.31264 MeV. The total decay energy is5.143 MeV.

Answer: See diagram

5. emits "-particles having a kinetic energy of 6.87 MeV. What is the recoil energy of the

product nucleus and what is the total "-decay energy of ?

* * *

7Chapter 2

8Chapter 2

9Chapter 2 & 3

RRecoil energy (E ):

RE = = 0.123 MeV

Total "-decay energy = 6.87 + 0.123 = 6.99 MeV

6. The most energetic "-particle emitted by has a kinetic energy of 5.155 MeV. Estimate

the difference in mass between and its daughter in amu. (Mass of = 4.0026

amu).

* * *

Total "-decay energy of

= 5.155 + = 5.2427 MeV

Equivalent mass = = 0.005628 amu

Total mass difference ()m) is

He decay)m = m + m = 4.0026 + 0.005628

)m = 4.0082 amu

Chapter 3

Radioactive Decay

1/21. Calculate the fraction of atoms remaining of (T = 15.0 h) after a decay interval of5.0 h.

* * *

Fraction of atoms remaining:

10Chapter 3

8 = = 0.0462 h-1

If t = 5.0, 8t = 0.0462 × 5 = 0.231

e = 0.7937-0.231

= 0.7937

2. Plot a straight line in coordinates of lnA and t for a radionuclide (T½ = 2.576 h) given thatits initial activity is 4 × 10 dis/s. What is the significance of the slope of this line?2

* * *

0A = A e-8t

0lnA = ln A - 8t

The slope of the line is the decay constant of the radionuclide.

Note that the units of time of the stated disintegration rate must be identical to the units oftime of the halflife and of the decay constant of the radionuclide.

0A = 4 × 10 dis/s2

0A = 4 × 10 × 60 × 60 = 1.44 × 10 dis/h2 6

T½ = 2.576 h, 8 = = 0.2690 h-1

Let t = 2 h

ln A = ln 1.44 × 10 - 0.2690 × 26

= 13.6421

Similarly for other values of t

See graph