1. 1 Nuclear Chemistry Chapter 20 Glenn T. Seaborg 1912-1999.* Transuranium elements. Pierre and Marie Curie. 1859-1906,* 1867-1934.** Discovered radium; defined “radioactivity.”

2. 2 Basics: Radioactivity Nuclear Equations Nucleons: particles in the nucleus: p + : proton n: neutron. Mass number A: the number of p + + n. Atomic number Z: the number of p +. Symbol: A z X; e.g. 14 6 C is “carbon-14” Isotopes: have the same number of p + and different numbers of n (and therefore, different mass) In nuclear equations, the total number of nucleons is conserved: 238 92 U  234 90 Th + 4 2 He

3. 3 Radioactivity There are three types of radiation which we consider: –  -Radiation is the loss of 4 2 He from the nucleus, –  -Radiation is the loss of an electron from the nucleus (electrons represented as either 0 -1 e or 0 -1 β) –  -Radiation is the loss of high-energy photon from the nucleus. Example of α emission : HeThU 4 2 234 90 238 92   Example of β emission : Example of γ emission : Tc 99m 43  0 hνhνTc 99 43  

4. 4 Radioactivity - Separating the types of radiation (-) (++) Charge 2+ 1- 0 Mass(g) 6.64x10 -24 9.11x10 -28 0 Rel. mass 7,300 1 0 Rel. penetration 1 100 10,000 4 2 He nucleus electron high energy photons α β γ _

5. 5 Radioactivity Complete the following nuclear reactions :

6. 6 Patterns of Nuclear Stability Neutron-to-Proton Ratio Neutron/proton ratio increases as atoms become larger Above 83 Bi, all nuclei are unstable and belt of stability ends.

7. 7 238 U Series Stable Radioactive Series

8. 8

9. 9 Nuclear Transmutations Using Charged Particles - cyclotron

10. 10 Rates of Radioactive Decay Calculations Based on Half-Life Radioactive decay is a first order process: Rate = kN In radioactive decay the constant, k, is called the decay constant, and N is number of nuclei. The rate of decay is called activity (disintegrations per unit time). If N 0 is the initial number of nuclei and N t is the number of nuclei at time t, then with half-life t 1/2 =0.693/k

11. 11 Rates of Radioactive Decay 5.0 2.5 1.25

12. 12 Rates of Radioactive Decay Dating Carbon-14 is a radioactive isotope of carbon and is used to determine the ages of organic compounds. We assume the ratio of 12 C to 14 C has been constant over time. For us to detect 14 C, the object must be less than 50,000 years old. The half-life of 14 C is 5,730 years. Its abundance is <1% (the most common isotope of carbon is C-12)

13. 13 Rates of Radioactive Decay 14 C is created in upper atmosphere by bombardment of nitrogen with cosmic neutrons: 14 C itself decays to stable 14 N by β emission: with a half-life of t 1/2 = 5715 yr The amount of 14 C in the environment is constant.

14. 14 How does 14 C dating work? When an organism dies, it no longer takes in carbon compounds but its 14 C continues to decay. 5715 years (or one half-life) after the death of the organism (or 1 half-life), the relative amount of 14 C is half that found in living matter. 11,430 years after the death of the organism (two half-lives), the relative amount of 14 C is ¼ that found in living matter. First order rate law: and t 1/2 =.693/k Since [ 14 C] is proportional to radiation emitted (in counts/min or cpm), the law become:

15. 15 Rates of Radioactive Decay- 14 C Dating 21.37 Artifact has 14 C activity of 24.9 counts/m, compared to current count of 32.5 counts/m for a standard. What is the age of the artifact? Given: t 1/2 = 5715 year. k=.693/t 1/2 =.693/5715 yr = 1.21x10 -4 yr -1 Use first order expression 2200 yr = t

16. 16 40 K – 40 Ar Dating 40 K is a strange beast – with a half life of 1.3 x 10 9 yr, it has two simultaneous modes of decay. 88.8% decays by electron-emission to give Ca-40: Ca 40 20 K 40 19  e 0 1   11.2% decays by electron-capture (of one of the orbital electrons) to give Ar-40: 40 K constitutes 0.01% of the natural abundance of potassium in the earth’s crust.

17. 17 Problem: a mineral is found with a 40 K/ 40 Ar mass ratio of 3/1. How old is the mineral? Answer: For the purposes of calculation, let’s assume a femtogram total mass of 40 K and 40 Ar. The respective masses of K-40 and Ar-40 will be 0.75 fg K-40 and 0.25 fg Ar-40. Since the mode of decay giving Ar-40 is 11.2%, the mass of K-40 which decayed is 0.25/0.112 = 2.23 fg. Hence, the original mass of K-40 was 0.75 + 2.23 = 2.98 fg. (5.33 x 10 -10 yr -1 ) t 0.75 2.98 ln  t = 2.58 x 10 9 yr k = 0.693/t 1/2

18. 18Half-lives Isotope Half-life Type of decay U-238 4.5 x 10 9 yr Alpha U-235 7.0 x 10 8 yr Alpha Th-232 1.4 x 10 10 yr Alpha K-40 1.3 x 10 9 yr Beta-capture or -emission C-14 5715 yr Beta Rn-222 3.825 days Alpha Tc-99 210,000 yr Beta

19. 19 Detection of Radioactivity Matter is ionized by radiation. Geiger counter determines the amount of ionization by detecting an electric current. A thin window is penetrated by the radiation and causes the ionization of Ar gas. The ionized gas carried a charge and so current is produced. The current pulse generated when the radiation enters is amplified and counted.

20. 20 Energy Changes in Nuclear Reactions Einstein showed that mass and energy are proportional: E = mc 2 If a system loses mass it loses energy (exothermic). If a system gains mass it gains energy (endothermic). Since c 2 is a large number (8.99  10 16 m 2 /s 2 ) small changes in mass cause large changes in energy. Mass and energy changed in nuclear reactions are much greater than in chemical reactions.

21. 21 Energy Changes in Nuclear Reactions Consider reaction : 238 92 U  234 90 Th + 4 2 He –Molar masses: 238.0003 g  233.9942 g + 4.0015 g. –The change in mass during reaction is 233.9942 g + 4.0015 g - 238.0003 g = -0.0046 g. =  m –The process is exothermic because the system has lost mass. –To calculate the energy change per mole of 238 92 U: This is a very large number!!

22. 22 Nuclear Fission Splitting of heavy nuclei is exothermic for large mass numbers. Consider a neutron bombarding a 235 U nucleus:

23. 23 Nuclear Fusion Light nuclei can fuse to form heavier nuclei. Most reactions in the Sun are fusion. Fusion processes occurring in sun: Sun currently is about 75% H and 25% He. Another 15b yrs or so to go before sun “burns” up all its H.

24. 24 Nuclear Fusion Fusion of tritium and deuterium requires about 40,000,000K: 2 1 H + 3 1 H  4 2 He + 1 0 n These temperatures can be achieved in a nuclear bomb or a tokamak. A tokamak is a magnetic bottle: strong magnetic fields contained a high temperature plasma so the plasma does not come into contact with the walls. (No known material can survive the temperatures for fusion.) To date, about 3,000,000 K has been achieved in a tokamak.