Chapter 21 Nuclear Chemistry

1. Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 21Nuclear Chemistry John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall, Inc.

2. The Nucleus • Remember that the nucleus is comprised of the two nucleons, protons and neutrons. • The number of protons is the atomic number. • The number of protons and neutrons together is effectively the mass of the atom.

3. Isotopes • Not all atoms of the same element have the same mass due to different numbers of neutrons in those atoms. • There are three naturally occurring isotopes of uranium: • Uranium-233 • Uranium-235 • Uranium-238

4. Radioactivity • It is not uncommon for some nuclides of an element to be unstable, or radioactive. • We refer to these as radionuclides. • There are several ways radionuclides can decay into a different nuclide.

5. Types ofRadioactive Decay

6. 238 92 234 90 4 2 4 2 He U Th He +  Alpha Decay: Loss of an -particle (a helium nucleus)

7. 131 53 131 54 0 −1 0 −1 0 −1 e I Xe e  +  or Beta Decay: Loss of a -particle (a high energy electron)

8. 11 6 11 5 0 1 0 1 e C B e +  Positron Emission: Loss of a positron (a particle that has the same mass as but opposite charge than an electron)

9. 0 0  Gamma Emission: Loss of a -ray (high-energy radiation that almost always accompanies the loss of a nuclear particle)

10. 0 −1 1 1 1 0 p e n +  Electron Capture (K-Capture) Addition of an electron to a proton in the nucleus • As a result, a proton is transformed into a neutron.

11. Measuring Radioactivity • One can use a device like this Geiger counter to measure the amount of activity present in a radioactive sample. • The ionizing radiation creates ions, which conduct a current that is detected by the instrument.

12. Answer: SAMPLE EXERCISE 21.1 continued PRACTICE EXERCISE Which element undergoes alpha decay to form lead-208?

13. Answer: PRACTICE EXERCISE Write a balanced nuclear equation for the reaction in which oxygen-15 undergoes positron emission.

14. Neutron-Proton Ratios • Any element with more than one proton (i.e., anything but hydrogen) will have repulsions between the protons in the nucleus. • A strong nuclear force helps keep the nucleus from flying apart.

15. Neutron-Proton Ratios • Neutrons play a key role stabilizing the nucleus. • Therefore, the ratio of neutrons to protons is an important factor.

16. Neutron-Proton Ratios For smaller nuclei (Z  20) stable nuclei have a neutron-to-proton ratio close to 1:1.

17. Neutron-Proton Ratios As nuclei get larger, it takes a greater number of neutrons to stabilize the nucleus.

18. Stable Nuclei The shaded region in the figure shows what nuclides would be stable, the so-called belt of stability.

19. Stable Nuclei • Nuclei above this belt have too many neutrons. • They tend to decay by emitting beta particles.

20. Stable Nuclei • Nuclei below the belt have too many protons. • They tend to become more stable by positron emission or electron capture.

21. Stable Nuclei • There are no stable nuclei with an atomic number greater than 83. • These nuclei tend to decay by alpha emission.

22. Radioactive Series • Large radioactive nuclei cannot stabilize by undergoing only one nuclear transformation. • They undergo a series of decays until they form a stable nuclide (often a nuclide of lead).

23. Some Trends Nuclei with 2, 8, 20, 28, 50, or 82 protons or 2, 8, 20, 28, 50, 82, or 126 neutrons tend to be more stable than nuclides with a different number of nucleons.

24. Some Trends Nuclei with an even number of protons and neutrons tend to be more stable than nuclides that have odd numbers of these nucleons.

25. Which of the following nuclei are especially stable: Solve: The nucleus (the alpha particle) has a magic number of both protons (2) and neutrons (2) and is very stable. The nucleus also has a magic number of both protons (20) and neutrons (20) and is especially stable. The nucleus does not have a magic number of either protons or neutrons. In fact, it has an odd number of both protons (43) and neutrons (55). There are very few stable nuclei with odd numbers of both protons and neutrons. Indeed, technetium-98 is radioactive. PRACTICE EXERCISE Which of the following nuclei would you expect to exhibit a special stability: SAMPLE EXERCISE 21.4 Predicting Nuclear Stability Solution Analyze: We are asked to identify especially stable nuclei, given their mass numbers and atomic numbers. Plan: We look to see whether the numbers of protons and neutrons correspond to magic numbers.

26. PRACTICE EXERCISE Predict the mode of decay of (a) plutonium-239, (b) indium-120. Answer:(a) decay, (b) decay

27. Nuclear Transformations Nuclear transformations can be induced by accelerating a particle and colliding it with the nuclide.

28. Particle Accelerators These particle accelerators are enormous, having circular tracks with radii that are miles long.

29. Nt N0 = -kt ln Kinetics of Radioactive Decay • Nuclear transmutation is a first-order process. • The kinetics of such a process, you will recall, obey this equation:

30. 0.693 k = t1/2 Kinetics of Radioactive Decay • The half-life of such a process is: • Comparing the amount of a radioactive nuclide present at a given point in time with the amount normally present, one can find the age of an object.

31. Kinetics of Radioactive Decay • A wooden object from an archeological site is subjected to radiocarbon dating. The activity of the sample that is due to 14C is measured to be 11.6 disintegrations per second. The activity of a carbon sample of equal mass from fresh wood is 15.2 disintegrations per second. The half-life of 14C is 5715 yr. What is the age of the archeological sample?

32. 0.693 5715 yr 0.693 k 0.693 k = k = t1/2 = 5715 yr = k 1.21  10−4 yr−1 Kinetics of Radioactive Decay First we need to determine the rate constant, k, for the process.

33. 11.6 15.2 ln = (1.21  10−4 yr−1) t Nt N0 = kt ln ln 0.763 = (1.21  10−4 yr−1) t = t 6310 yr Kinetics of Radioactive Decay Now we can determine t:

34. Solve: Let’s assume that the rock contains 1.000 mg of uranium-238 at present. The amount of uranium- 238 in the rock when it was first formed therefore equals 1.000 mg plus the quantity that decayed to lead-206. We obtain the latter quantity by multiplying the present mass of lead-206 by the ratio of the mass number of uranium to that of lead, into which it has decayed. The total original was thus Using Equation 21.20, we can calculate the decay constant for the process from its half-life: SAMPLE EXERCISE 21.7 Calculating the Age of a Mineral A rock contains 0.257 mg of lead-206 for every milligram of uranium-238. The half-life for the decay of uranium-238 to lead-206 is 4.5  109 yr. How old is the rock? Solution Analyze: We’re told that a rock sample has a certain amount of lead-206 for every unit weight of uranium-238 and asked to estimate the age of the rock. Plan: Presumably the lead-206 is due entirely to radioactive decay of uranium-228 to form lead-206, with a known half-life. To apply first-order kinetics expressions (Equations 21.19 and 21.20) to calculate the time elapsed since the rock was formed, we need first to calculate how much initial uranium-238 there was for every 1 milligram that remains today.

35. Rearranging Equation 21.19 to solve for time, t, and substituting known quantities gives SAMPLE EXERCISE 21.7 continued PRACTICE EXERCISE A wooden object from an archeological site is subjected to radiocarbon dating. The activity of the sample that is due to 14C is measured to be 11.6 disintegrations per second. The activity of a carbon sample of equal mass from fresh wood is 15.2 disintegrations per second. The half-life of 14C is 5715 yr. What is the age of the archeological sample? Answer: 2230 yr

36. Solve: Equation 21.19 is solved for the decay constant, k, and then Equation 21.20 is used to calculate half-life, t1/2: SAMPLE EXERCISE 21.8 Calculations Involving Radioactive Decay If we start with 1.000 g of strontium-90, 0.953 g will remain after 2.00 yr. (a) What is the half-life of strontium-90? (b) How much strontium-90 will remain after 5.00 yr? (c) What is the initial activity of the sample in Bq and in Ci? Solution (a) Analyze: We are asked to calculate a half-life, t1/2, based on data that tell us how much of a radioactive nucleus has decayed in a given period of time and (N0 = 1.000 g, Nt = 0.953 g, and t = 2.00 yr). Plan: We first calculate the rate constant for the decay, k, then use that to compute t1/2. (b) Analyze: We are asked to calculate the amount of a radionuclide remaining after a given period of time. Plan: We need to calculate the amount of strontium at time t, Nt , using the initial quantity, N0 , and the rate constant for decay, k, calculated in part (a).

37. Solve: Again using Equation 21.19, with k = 0.0241 yr–1, we have Nt /N0is calculated from ln(Nt /N0) = –0.120 using the ex or INV LN function of a calculator: Because N0 = 1.000 g, we have Solve: The number of disintegrations per atom per second is given by the rate constant, k. SAMPLE EXERCISE 21.8 continued (c) Analyze: We are asked to calculate the activity of the sample in becquerels and curies. Plan: We must calculate the number of disintegrations per second per atom, then multiply by the number of atoms in the sample.

38. To obtain the total number of disintegrations per second, we calculate the number of atoms in the sample. We multiply this quantity by k, where we express k as the number of disintegrations per atom per second, to obtain the number of disintegrations per second: Because a Bq is one disintegration per second, the activity is just 5.1  1012 Bq. The activity in Ci is given by SAMPLE EXERCISE 21.8 continued Bq = 1 disintegration/sec Ci = 3.7 x 10^10 disintegration/sec We have used only two significant figures in products of these calculations because we don’t know the atomic weight of 90S to more than two significant figures without looking it up in a special source. PRACTICE EXERCISE A sample to be used for medical imaging is labeled with 18F, which has a half-life of 110 min. What percentage of the original activity in the sample remains after 300 min? Answer: 15.1%

39. How much energy is lost or gained when a mole of cobalt-60 undergoes beta decay: The mass of the atom is 59.933819 amu, and that of a atom is 59.930788 amu. Solve: A atom has 27 electrons. The mass of an electron is 5.4858  10–4 amu. (See the list of fundamental constants in the back inside cover). We subtract the mass of the 27 electrons from the mass of the atom to find the mass of the nucleus: Likewise, for the mass of the nucleus is The mass change in the nuclear reaction is the total mass of the products minus the mass of the reactant: Thus, when a mole of cobalt-60 decays, SAMPLE EXERCISE 21.9 Calculating Mass Change in a Nuclear Reaction Solution Analyze: We are asked to calculate the energy change in a nuclear reaction. Plan: We must first calculate the mass change in the process. We are given atomic masses, but we need the masses of the nuclei in the reaction. We calculate these by taking account of the masses of the electrons that contribute to the atomic masses.

40. Because the mass decreases (m < 0), energy is released (E < 0). The quantity of energy released per mole of cobalt-60 is calculated using Equation 21.22: PRACTICE EXERCISE Positron emission from occurs with release of 2.87  1011 J per mole of 11C. What is the mass change per mole of 11C in this nuclear reaction? SAMPLE EXERCISE 21.9 continued Answer: –3.19  10–3g

41. Energy in Nuclear Reactions • There is a tremendous amount of energy stored in nuclei. • Einstein’s famous equation, E = mc2, relates directly to the calculation of this energy.

42. Energy in Nuclear Reactions • In the types of chemical reactions we have encountered previously, the amount of mass converted to energy has been minimal. • However, these energies are many thousands of times greater in nuclear reactions.

43. Energy in Nuclear Reactions For example, the mass change for the decay of 1 mol of uranium-238 is −0.0046 g. The change in energy, E, is then E = (m) c2 E= (−4.6  10−6 kg)(3.00  108 m/s)2 E= −4.1  1011 J

44. Nuclear Fission • How does one tap all that energy? • Nuclear fission is the type of reaction carried out in nuclear reactors.

45. Nuclear Fission • Bombardment of the radioactive nuclide with a neutron starts the process. • Neutrons released in the transmutation strike other nuclei, causing their decay and the production of more neutrons.

46. Nuclear Fission This process continues in what we call a nuclear chain reaction.

47. Nuclear Fission If there are not enough radioactive nuclides in the path of the ejected neutrons, the chain reaction will die out.

48. Nuclear Fission Therefore, there must be a certain minimum amount of fissionable material present for the chain reaction to be sustained: Critical Mass.

49. Nuclear Reactors In nuclear reactors the heat generated by the reaction is used to produce steam that turns a turbine connected to a generator.

50. Nuclear Reactors • The reaction is kept in check by the use of control rods. • These block the paths of some neutrons, keeping the system from reaching a dangerous supercritical mass.