1. Ch 19 Nuclear Chemistry

2.  Know how nuclear equations are balanced: The same sums of both mass and atomic numbers appear on both sides of the equation  Ex: Cl-35 is bombarded by a neutron and H-1 is created along with another product. What is the identity of the second product?  35 Cl + 1 n → 1 H + ? so atomic # of the unknown must be 16 (S) and the mass # must be 35 170 1

3.  Know the five naturally occurring decay modes: ◦ Alpha emission ◦ Beta emission ◦ Gamma emission ◦ Positron Emission ◦ Electron capture

4.  Alpha Emission ◦ An alpha particle is a helium nucleus with two protons and 2 neutrons and is represented by α or 4 He ◦ Example: 222 Rn → 218 Po + 4 He ◦ Note that alpha radiation is fairly weak 2 8684 2

5.  Beta Emission ◦ The emission of an electron ◦ Given by 0 β or 0 e ◦ This electron comes from the nucleus, not the electron cloud! ◦ The Beta particle results from the conversion of a neutron into a proton and an electron 1 n → 1 p + 0 e Ex: 63 Ni → 63 Cu + 0 e 0 1 28 29

6.  Gamma Emission ◦ A high energy, short wavelength photon similar to X-rays ◦ Given by symbol γ ◦ Accompanies many other types of decay but is not shown in the balanced nuclear equation ◦ Penetration power/risk due to exposure: ◦ alpha < beta < gamma

7.  Positron Emission  The emission of an electron with a positive charge  Represented by 0 β or 0 e  Results from the conversion of a proton to a neutron and a positron 1 p → 1 n + 0 e  Ex: 40 K → 40 Ar + 0 e +1 1 1 0 19181

8.  Electron Capture ◦ The capture of an electron from the energy level closest to the nucleus by a proton inside the nucleus, thus creating a neutron ◦ 0 e + 1 p → 1 n ◦ Electrons from higher energy levels cascade down to fill successive vacated lower energy levels and in the process release energy in the X-ray range of the electromagnetic spectrum ◦ Ex: 204 Po + 0 e → 204 Bi + X-rays 10 84 83

9.  Know that nuclear stability is best related to the neuron-to-proton ratio (n/p) ◦ The n/p ratio starts at about 1/1 for light isotopes and ends at about 1.5/1 for heavier isotopes with atomic numbers up to 83. ◦ All isotopes of atomic number greater than 84 are unstable and will commonly undergo alpha decay. ◦ Below atomic number 84:  neutron poor isotopes will probably undergo positron emission or electron capture  neutron rich isotopes will probably undergo beta emission

10.  Know that the half-life, t ½, of a radioactive isotope is the amount of time it takes for one-half of the sample to decay.  Know how to use the appropriate equations to calculate amounts of an isotope remaining at any given time or to calculate the half-life  ***We reviewed first order decay in Ch 13 Chemical Kinetics  Key equations: t ½ = 0.693/k ln[A] t – ln[A] 0 = -kt

11.  Know how to use Einstein’s equation E =mc 2 to calculate the amount of energy produced from a mass defect (the amount of matter that was converted into energy) ◦ Ex: When one mole of U-238 decays to Th-234, 5 x 10 -6 kg of matter is converted to energy (the mass defect). How much energy is released? ◦ E =mc 2 ◦ E = (5 x 10 -6 kg )(3.00 x 10 8 m/s) 2 ◦ E = 5 x 10 11 kgm 2 /s 2 or 5 x 10 11 J

12.  Mass Defect  Conservation of matter does not necessarily occur in nuclear reactions! Some matter can be converted to energy in the form of heat and light, gamma radiation, or kinetic energy of the expelled particle

13.  When 226 Ra decays, it emits 2 alpha particles, then a beta particle, followed by an alpha particle. The resulting isotope is:  (A) 212 Bi  (B) 222 Rn  (C) 214 Pb  (D) 214 Bi  (E) 212 At 88 83 86 82 83 85

14.  The formation of 230 Th from 234 U occurs by:  (A) electron capture  (B) α decay  (C) β decay  (D) positron decay  (E) γ decay 90 92

15.  What is the missing product in the following nuclear reaction? 236 U → 4 1 n + 136 I +  (A) 90 Y  (B) 96 Sr  (C) 96 Y  (D) 98 Zr  (E) 98 Nb 92 053 39 38 40 41

16.  If 75% of a sample of pure tritium decays in 24.6 years, what is the half-life of tritium?  (A) 24.6 years  (B) 18.4 years  (C) 12.3 years  (D) 6.15 years  (E) 3.07 years  ln0.25 – ln1 = -k(24.6 years) k = 0.056 years -1  then t = 0.693/k = 12.3 years