SOLUTIONS FOR THE END-OF-CHAPTER PROBLEMS of

ISOTOPES; PRINCIPLES AND APPLICATIONS (2005) by Gunter Faure and Teresa M. Mensing Department of Geological Sciences The Ohio State University Columbus, Ohio 43210

Third Edition Principles 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 40Ar*/39Ar method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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 electrons because 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: 28Si (92.23 %, 27.976927 amu), 29Si (4.67 %, 28.976494 amu), and 30Si (3.10 %, 29.973770 amu). Atomic weight of Si = 0.9223 × 27.976927 + 0.0467 × 28.97644 + 0.0310 × 29.97377 = 28.085 amu.

2

Chapter 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 amu), (55.934940 amu), of million electron volts (MeV).

(4.002608323 amu),

(115.901743 amu), and

*

*

(24.985837

(232.03805 amu) in units

*

Mass of the proton: 1.00782503 amu Mass 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 energy MeV/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 and draw a smooth curve through the points.(See graph) c. Which element has the highest nuclear stability?

Chapter 1

3

4

Chapter 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).

*

*

*

Molecular weight of Sr (NO3)2 = 87.62 + 2 × 14.00 + 6 ×15.99 = 211.56 Number of moles of Sr =

= 0.0118 = 1.18 ×10-2 mol

b. How many atoms of Sr are present in 2.50 g of Sr nitrate? Number of atoms of Sr = 1.18 × 10-2 × 6.022 × 1023 = 7.11 × 1021 atoms c. How many atoms of 87Sr are present in 2.50 grams of strontium nitrate? (Abundance of 87Sr = 7.0 %)? Number of atoms of

87

Sr

= 7.11 x 1021 × 0.070 = 4.98 × 1020 atoms d. What is the ratio of the number of (Abundance of 86Sr = 9.9 %.)

87

Sr atoms to

Number of atoms of 86Sr = 7.11 × 1021 × 0.099 = 7.04 × 1020 atoms Atomic

ratio =

= 0.7073

86

Sr atoms in 2.50 g of Sr nitrate?

Chapter 1 & 2

Note that the isotope ratio is also equal to the ratio of the abundances of the isotopes expressed as atom percent. 5. The concentration of K2O in a sample of orthoclase is 8.75 % by weight. How many atoms of 40K are present in 1.0 gram of this material? (Atomic weight of K = 39.09, abundance of 40 K = 0.0117 % atomic).

*

*

*

Weight of K2O in 1 g of orthoclase is 0.0875 g. The molecular weight of K2O = 2 ×39.09 + 15.99 = 94.17 The number of moles of K2O =

= 0.000929

The number of moles of K2O = 0.000929 × 2 = 0.00185 The number of atoms of 40K = 0.00185 × 6.022 × 1023 × 0.000117 = 1.309 × 1017 atoms Note that a concentration of 8.75% of K2O is equivalent to an amount of 8.75g of K2O per 100g of sample, or to 0.0875 g of K2O 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.

5

6

Chapter 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 7.029 MeV. 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 is 5.143 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

*

*

*

?

Chapter 2

7

Chapter 2

8

9

Chapter 2 & 3

Recoil energy (ER): ER =

= 0.123 MeV

Total "-decay energy = 6.87 + 0.123 = 6.99 MeV 6.

The most energetic "-particle emitted by the difference in mass between amu).

has a kinetic energy of 5.155 MeV. Estimate

and its daughter in amu. (Mass of

*

*

= 4.0026

*

Total "-decay energy of = 5.155 +

Equivalent mass =

= 5.2427 MeV

= 0.005628 amu

Total mass difference ()m) is )m = mHe+ mdecay = 4.0026 + 0.005628 )m = 4.0082 amu Chapter 3 Radioactive Decay 1.

Calculate the fraction of atoms remaining of 5.0 h.

* Fraction of atoms remaining:

*

(T1/2 = 15.0 h) after a decay interval of

*

Chapter 3

8=

10

= 0.0462 h-1

If t = 5.0, 8t = 0.0462 × 5 = 0.231 e-0.231 = 0.7937 = 0.7937

2.

Plot a straight line in coordinates of lnA and t for a radionuclide (T½ = 2.576 h) given that its initial activity is 4 × 102 dis/s. What is the significance of the slope of this line?

*

*

*

A = A0e-8t lnA = ln A0 - 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 of time of the halflife and of the decay constant of the radionuclide. A0 = 4 × 102 dis/s A0 = 4 × 102 × 60 × 60 = 1.44 × 106 dis/h T½ = 2.576 h, 8 =

= 0.2690 h-1

Let t = 2 h ln A = ln 1.44 × 106 - 0.2690 × 2 = 13.6421 Similarly for other values of t See graph