Nuclear Chemistry Notes

Nuclear Chemistry Notes Definitions Nucleons: Subatomic particles in the nucleus : protons and neutrons Radionuclides: Radioactive nuclei. Unstable nuclei that spontaneously emit particles and electromagnetic radiation. Radioisotopes: Atoms containing radioactive nuclei. Half-Life: the length of time required for one-half of the atoms of a radioactive sample to decay. (Number of atoms left) = (( ) n ) * (original # of atoms) n = the number of half lives elapsed. Fission: break up of a large nucleus into two smaller nuclei creating a higher binding energy. Fusion: two or more smaller nuclei combining to form a larger nucleus with a higher binding energy. Synthetic Elements: made by bombarding plutonium with neutrons or other large elements with smaller elements. Nuclear Equations 1. Mass number is conserved in a nuclear change. 2. Electric charge is conserved in a nuclear change. Example of nuclear reaction: 254 99Es + 4 256 2He 101Md + 2 1 0n Types of nuclear reactions: Alpha Decay 238 4 92U 2He + 234 90Th Decay: Beta Emission (Beta-minus) 228 0 88Ra -1e + 228 89Ac Decay: Positron Emission 93 0 44Ru 1e + 93 43Tc K-Capture (capture beta-minus) 83 37Rb + 0 83-1e 36Kr + X-ray + X-ray

Types of Radioactive Decay When unstable nuclei decay, the reactions they undergo generally involve one or more of the following particles listed in the first column. Some facts about these particles may be found in the next 5 columns. In the 7th column you will find an example of each type of decay. Notice that for an alpha particle decay, the equation is balanced with regard to atomic number (92, 90+2) and atomic weight (238, 234+4). This type of balancing is true for all nuclear reactions. In the last column are instructions for how to predict when each type of emission or capture will occur. Fore example, if the atomic number is greater than 83, alpha particle decay is most likely. particle alpha particles beta particles gamma Rays positron emission electron capture What is it? helium nuclei high speed electrons symbol charge mass 2He 4 +2 6.664 or 2 α 4 E-24 g -1e o or - 1 9.11 1β o E-28 g relative penetrating power Example Applies to which particles 1 92U 238 => 90 Th 234 + 2 He 4 Atomic Numbers > 83; the 2 p + 2n 0 loss brings the atom diagonally back to the belt of stability. 100 53I 131 => 54 Xe 131 + -1 e o Isotopes below the belt of stability (high neutron : proton ratios). Causes a When a β-particle is emitted, the at. no. increases by 1. A neutron is converted into a p + and e - : on 1 => 1 p 1 + -1 e o high oγ o 0 0 10000 energy photons Generally accompanies other radioactive radiation because it is the energy lost from other nucleon changes. Gamma radiation is generally not shown in the nuclear equation. positron 1e o +1 9.11 E-28 g 6C 11 => 5 B 11 + 1 e o Causes the atomic number to decrease. It converts a proton to a neutron + positron 1p 1 => o n 1 + 1 e o inner shell electron -1e o 1 9.11 E-28 g 37Rb 81 + -1 e o => 36 Kr 81 The nucleus capture an inner shell electron; thereby converting a p + to a n o 1p 1 + -1 e o => o n 1 loss of 1 neutron and a gain of 1 proton. Isotopes above the belt of stability (low neutron : proton ratios). Causes a loss of 1 proton and a gain of 1 neutron. Isotopes above the belt of stability (low neutron : proton ratios). Causes a loss of 1 proton and a gain of 1 neutron.

:Stable n:p ratios: Neutrons are needed to hold protons together in the nucleus by the strong force. At low atomic numbers, the ratio of neutrons to protons in stable isotopes is generally very close to 1. As the atomic number rises, so does the neutron/proton ratio, such that at atomic number 80, stable isotopes have a ratio of about 1.5. This information is tabulated below. atomic number 7 40 50 80 ratio n/p for stable isotopes 1 1.25 1.4 1.5 Predict the type of decay of the isotope 92 U 238 explanation emission preliminary equation balancing to solve for X z > 83 alpha 92U 238 => X + 2 He 4 92U 238 => 90 Th 234 + 2 He 4 Stable nuclear configurations: Some configurations of protons and neutrons are particularly stable just as some configurations of electrons (2, 10, 18, 36, 54, 86) are. proton numbers: 2, 8, 20, 28, 50, 82 neutron number: 2, 8, 20, 28, 50, 82, 126 nuclei with even numbers of protons and neutrons are more stable than those with odd numbers. Nuclear Transmutations: occur when nuclei are struck by neutrons or other nuclei. These reactions are useful in creating new radioisotopes. Notation: The reaction in which N-14 is bombarded with an alpha particle: 7 N 14 + 2 He 4 => 8 O 17 + 1 H 1 is written as 7N 14 (a, p) 8 O 17 in the following order: target nucleus, (bombarding particle, ejected particle), product nucleus. Rates: Half life is the time required for half of the radioisotope to react. Dating: Carbon-14 has been used to determine the age of organic materials. In the upper atmosphere, 7 N 14 captures a neutron to form 6 C 14 7N 14 + 0 n 1 => 6 C 14 + 1 p 1 which supplies a constant source of C-14. However 6 C 14 is radioactive and decays back to 7 N 14 with a half life of 5730 years. 6C 14 => 7 N 14 + -1 e o We assume that the ratio of C-14 to C-12 in the atmosphere has been constant for at least 50000 years. So long as an organism is living, the CO 2 cycle keeps renewing the supply of C-14 keeping the ratio of C-14 to C-12 constant. Upon death, the organism does not replenish its supply of C-14 while existing C-14 decomposes with a half life of 5730 years. The smaller the fraction of C-14 in the organism, the longer ago that organism died. For example, if the organism has half the amount of C-14 that is present in the atmosphere, the organism probably died 5730 years ago.

Mass Defect: It is experimentally observed that the mass of an atom (containing neutrons) is always slightly less than the sum of the masses of its component particles. The difference between the atomic mass and the sum of the masses of its protons, neutrons, and electrons is called the mass defect. isotope Should weigh: Do weigh: Mass Defect: Hydrogen 1.0073 +.00055 = 1.00785 1.0078 none Deuterium 1.0073 +.00055 + 1.0087 = 2.01655 2.0140.0025 Tritium 1.0073 +.00055 + 2(1.0087) = 3.02525 3.01605.0092 The loss in mass is accounted for by Einstein's E = mc 2, which describes conversions between matter (m) and energy (E). When the nucleus is being formed, some matter was converted into energy (called nuclear binding energy). Energy changes in nuclear reactions Given the nuclear reaction: 92 U 238 => 90 Th 234 + 2 He 4, and given the masses of the three atoms (238.0003 amu, 233.9942 amu, 4.0015 amu) calculate the mass and energy change per mole associated with this reaction. [Energy - mass conversion equation: E = m c 2 ; Energy conversion formula: 1.00 J = 1.00 kg m 2 / s 2 ] mass change energy change mass of products - mass of reactants = 233.9942g + 4.0015g - 238.0003g = -.0046 g (exothermic).0046g * [3.00 E8 m/s] 2 [1kg/1000g] = -4.1 E11 kg m 2 /s 2 = -4.1 E11 J Radioactive Emission Worksheet For the following particle, what type of emission is expected, why, and what is the equation describing that emission? 92U 238, 53 I 131, 6 C 11, 37 Rb 81, 88 Ra 226, 84 Po 212, 81 Tl 209, 56 Ba 140 isotope emission equation 92U 238 53I 131 6C 11 37Rb 81 88Ra 226 84Po 212 81Tl 209 56Ba 140

Nuclear Transmutation Notation Worksheet Part 1: Write balanced equations for the following transmutations: in other words, from the information in column "" you should be able to write the answers in column "." 7N 13 (a, p) 8 O 16 4Be 9 (a,n) 6 C 12 13Al 27 (n,a) 11 Na 24 8O 16 (p,a) 7 N 13 11Na 23 (a,p) 12 Mg 26 7N 14 (a, p) 8 O 17 5B 10 (a,n) 7 N 13 Part 2: Complete the following nuclear equations: Note that the equation must balance on each side with respect to atomic number and atomic mass. 7N 13 + 2 He 4 => 8 O 16 +? 1H 3 => 2 He 3 +? 4Be 9 + 2 He 4 => 6 C 12 +? 15P 30 => 14 S 30 +? 83Bi 213 => 2 He 4 +? 11Na 23 + 2 He 4 => 12 Mg 26 +? Ag 106 => Cd 106 +? 5B 10 + 2 He 4 => 7 N 13 +? Part 3: Write the following reactions using transmutation notation. Note that this is the reverse of Part 1. 7N 13 + 2 He 4 => 8 O 16 + 1 H 1 4Be 9 + 2 He 4 => 6 C 12 + o n 1 13Al 27 + 0 n 1 => 11 Na 24 + 2 He 4 8O 16 + 1 H 1 => 7 N 13 + 2 He 4

11Na 23 + 2 He 4 => 12 Mg 26 + 1 H 1 7N 14 + 2 He 4 => 8 O 17 + 1 H 1 5B 10 + 2 He 4 => 7 N 13 + o n 1 s calculating mass defect 1: What is the mass defect for 3 Li 6, the lithium isotope with 3 neutrons and an atomic weight of 6.01513? 2. Calculate the mass defect for 17 Cl 35 whose mass is 34.9689 AMU. 3. The actual mass of 29 Cu 63 is 62.9298 AMU. What is its mass defect? 4. The actual mass of 35 Br 81 is 80.9163 AMU. What is its mass defect? 5. The actual mass of 52 Te 120 is 119.9045 AMU. What is its mass defect? answers with solutions isotope mass of protons mass of neutrons mass of electrons calc mass of atom act mass of atom mass defect 3Li 6 17Cl 35 29Cu 63 35Br 81 52Te 120 s calculating energies associated with mass defects Calculate the energies associated with the mass defects for one mole of the atoms from the five problems above using E = mc 2. 1 2 3 4 5 answers with solutions How much energy is lost or gained when a mole of cobalt-60 undergoes beta decay: 27Co 60 => -1 e 0 + 28 Ni 60. The masses of 27 Co 60 and 28 Ni 60 are 59.9338 AMU and 59.9308 AMU. mass change mass change per mole energy change

Answers Radioactive Emission Worksheet For the following particle, what type of emission is expected, why, and what is the equation describing that emission? 92U 238, 53 I 131, 6 C 11, 37 Rb 81, 88 Ra 226, 84 Po 212, 81 Tl 209, 56 Ba 140 isotope explanation emission equation 92U 238 z > 83 alpha 92U 238 => 90 Th 234 + 2 He 4 53I 131 ratio = 78/53=1.47 high beta 53I 131 => 54 Xe 131 + -1 e o 6C 11 37Rb 81 ratio = 5/6 = 0.83 low ratio = 44/37 = 1.19 low positron emission or electron capture positron emission or electron capture 6C 11 => 5 B 11 + 1 e o 6C 11 + -1e o => 5 B 11 37Rb 81 => 36 Kr 81 + 1 e o 37Rb 81 + -1 e o => 36 Kr 81 88Ra 226 z > 83 alpha 88Ra 226 => 86 Rn 222 + 2 He 4 84Po 212 z > 83 alpha 84Po 212 => 82 Pb 208 + 2 He 4 81Tl 209 ratio = 128/1.58 high beta 81Tl 209 => 82 Pb 209 + -1 e o 56Ba 140 ratio = 84/56 = 1.5 high beta 56Ba 140 => 57 La 140 + -1 e o Nuclear Transmutation Notation Worksheet Part 1: Write balanced equations for the following transmutations: in other words, from the information in column "" you should be able to write the answers in column "." 7N 13 (a, p) 8 O 16 7N 13 + 2 He 4 => 8 O 16 + 1 H 1 4Be 9 (a,n) 6 C 12 4Be 9 + 2 He 4 => 6 C 12 + o n 1 13Al 27 (n,a) 11 Na 24 13Al 27 + 0 n 1 => 11 Na 24 + 2 He 4 8O 16 (p,a) 7 N 13 8O 16 + 1 H 1 => 7 N 13 + 2 He 4 11Na 23 (a,p) 12 Mg 26 11Na 23 + 2 He 4 => 12 Mg 26 + 1 H 1 7N 14 (a, p) 8 O 17 7N 14 + 2 He 4 => 8 O 17 + 1 H 1 5B 10 (a,n) 7 N 13 5B 10 + 2 He 4 => 7 N 13 + o n 1 Part 2: Complete the following nuclear equations: Note that the equation must balance on each side with respect to atomic number and atomic mass. 7N 13 + 2 He 4 => 8 O 16 +? 7N 13 + 2 He 4 => 8 O 16 + 1 H 1 1H 3 => 2 He 3 +? 1H 3 => 2 He 3 + -1 e 0 4Be 9 + 2 He 4 => 6 C 12 +? 4Be 9 + 2 He 4 => 6 C 12 + o n 1 15P 30 => 14 S 30 +? 15P 30 => 14 S 30 + 1 e 0 (positron) 83Bi 213 => 2 He 4 +? 83Bi 213 => 2 He 4 + 81 Tl 209 11Na 23 + 2 He 4 => 12 Mg 26 +? 11Na 23 + 2 He 4 => 12 Mg 26 + 1 H 1 Ag 106 => Cd 106 +? 47Ag 106 => 48 Cd 106 + -1 e 0 5B 10 + 2 He 4 => 7 N 13 +? 5B 10 + 2 He 4 => 7 N 13 + o n 1 Part 3: Write the following reactions using transmutation notation. Note that this is the reverse of Part 1. 7N 13 + 2 He 4 => 8 O 16 + 1 H 1 7N 13 (a, p) 8 O 16

4Be 9 + 2 He 4 => 6 C 12 + o n 1 4Be 9 (a,n) 6 C 12 13Al 27 + 0 n 1 => 11 Na 24 + 2 He 4 13Al 27 (n,a) 11 Na 24 8O 16 + 1 H 1 => 7 N 13 + 2 He 4 8O 16 (p,a) 7 N 13 11Na 23 + 2 He 4 => 12 Mg 26 + 1 H 1 11Na 23 (a,p) 12 Mg 26 7N 14 + 2 He 4 => 8 O 17 + 1 H 1 7N 14 (a, p) 8 O 17 5B 10 + 2 He 4 => 7 N 13 + o n 1 5B 10 (a,n) 7 N 13 s calculating mass defect 1: What is the mass defect for 3 Li 6, the lithium isotope with 3 neutrons and an atomic weight of 6.01513? 2. Calculate the mass defect for 17 Cl 35 whose mass is 34.9689 AMU. 3. The actual mass of 29 Cu 63 is 62.9298 AMU. What is its mass defect? 4. The actual mass of 35 Br 81 is 80.9163 AMU. What is its mass defect? 5. The actual mass of 52 Te 120 is 119.9045 AMU. What is its mass defect? isotope mass of protons mass of neutrons 3Li 6 3 * 1.0073 = 3.0219 35 17 * 1.0073 17Cl = 17.124 63 29 * 1.0073 29Cu = 29.212 81 35 * 1.0073 35Br = 35.256 120 52 * 1.0073 52Te = 52.380 3 * 1.0087= 3.0261 18 * 1.0087=18.157 34 * 1.0087= 34.296 46 * 1.0087= 46.400 68 * 1.0087=68.592 answers with solutions mass of electrons 3 *.00055 =.0017 17 *.00055 =.0093 29 *.00055 =.016 35 *.00055 =.019 52 *.00055 =.029 calc mass of atom act mass of atom mass defect 6.0497 6.01513.0346 35.290 34.9689.321 63.523 62.9298.594 81.675 80.9163.759 121.001 119.9045 1.096 s calculating energies associated with mass defects Calculate the energies associated with the mass defects for one mole of the atoms from the five problems above using E = mc 2. answers with solutions 1 2 3 4 5.0345 AMU* 6.02E23 atoms * [1g/6.023E23AMU] [3.00E8m/s] 2 *[1kg/1000g] = 3.11E12 kg m 2 /s 2 = 3.11E12 J.321 AMU* 6.02E23 atoms * [1g/6.023E23AMU] [3.00E8m/s] 2 *[1kg/1000g] = 2.89E13 kg m 2 /s 2 = 2.89E13 J.594AMU* 6.02E23 atoms * [1g/6.023E23AMU] [3.00E8m/s] 2 *[1kg/1000g] = 5.35E13 kg m 2 /s 2 = 5.35E13 J.759AMU* 6.02E23 atoms * [1g/6.023E23AMU] [3.00E8m/s] 2 *[1kg/1000g] = 6.84E13 kg m 2 /s 2 = 6.84E13 J 1.095AMU* 6.02E23 atoms * [1g/6.023E23AMU] [3.00E8m/s] 2 *[1kg/1000g] = 9.87E13 kg m 2 /s 2 = 9.87E13 J How much energy is lost or gained when a mole of cobalt-60 undergoes beta decay: 27Co 60 => -1 e 0 + 28 Ni 60. The masses of 27 Co 60 and 28 Ni 60 are 59.9338 AMU and 59.9308 AMU. mass change mass change per mole = mass of Ni - mass of Co = 59.9308 AMU - 59.9338 AMU = -.0030 AMU. = -.0030 g energy change = mc 2 = -.0030 g (3.00 E8 m/s) 2 (1 kg/1000g) = -2.7 E 11 kg m 2 / s 2 = -2.7 E 11 J