CHEM 108 1st Edition Lecture 36 Chapter 21: Nuclear ChemistryNuclear Stability:- The existence of stable nuclei with more than one proton is due to the nuclear force. o The nuclear force is a strong force of attraction between nucleons that acts only at very short distances (about 10-15 m). o This force can more than compensate for the repulsion of electrical charges and thereby give a stable nucleus.- Several factors appear to contribute to the stability of a nucleus.o The shell model of the nucleus is a nuclear model in which protons and neutrons exist in levels, or shells, analogous to the shell structure exhibited in electron configurations. o Experimentally, note that nuclei with certain numbers of protons and neutrons appear to be very stable.o These numbers, called magic numbers, are the numbers of nuclear particles in a completed shell of protons or neutrons. o Because nuclear forces differ from electrical forces, these numbers are not the same as those for electrons in atoms. o For protons, the magic numbers are – 2, 8, 20, 28, 50, and 82 o For neutrons, the magic numbers are – 2, 8, 20, 28, 50, 82, and 126Stability of Nuclei: - Belt of stability: Region on a graph of the number of neutrons vs number of protons that includes all stable nuclei.- Nuclides above the belt of stability: Neutron rich; undergo β-decay. - Nuclides below the belt of stability: Neutron poor; undergo positron-decay or electron captureNuclear Fission: - Nuclear reaction in which a heavy nucleus splits into two lighter nuclei; usually accompanied by release of one or more neutrons. - Chain Reaction: A self-sustaining series of fission reactions in which neutrons released when nuclei split apart initiate additional fission events. - Critical Mass: The minimum amount of fissionable material needed to sustain a chain reaction.- Steps of Fission:o 1. Induced Fission o 2. Products: 141Ba, 92Kr, 3n o 3. Neutrons create more fission events.These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.Units of Radiation: - Becquerel (Bq): SI unit = one decay event per second. - Curie (Ci): 3.7 × 1010 decays per second. - Grey (Gy): SI unit of absorbed radiation = 1 J/kg tissue. - Sievert(Sv): SI unit to express amount of biological damage.o Ionizing Radiation: High-energy products of radioactive decay that can ionize molecules:o Relative Biological Effectiveness (RBE): Factor that accounts for differences in physical damage caused by different types of radiationRadiometric Dating: - Determining the age of an object based on quantity of a radioactive nuclide and/or products of its decay contained in the object. - Based on half-life: Nt / No = 0.5n o Example: 206Pb / 238U: age of geological samples. - 14C / 12C: age of carbon-based samples.- Radiocarbon Dating: Method for establishing the age of a carbon containing object by measuringthe activity of C-14 in the object. o Carbon-14 (t1/2 = 5730 years)Practice problem:A radioactivity counter gives a reading of 350 counts per minute, for a 12.5-mg sample of cobalt(II) chloride, partially enriched with cobalt-60 (half-life = 5.26 yr.). What percentage of the cobalt atoms in this sample are cobalt-60?First we need to get the number of radioactive nuclei. Then we need to calculate the total number of nuclei in the sample. The answer is simply the ratio of the two, multiplied by 100.- rate = k * No k is the radioactive decay constant o N is number of radioactive nucleio Rate is counts per second - rate = 350/60 = 5.83 nuclei/sec- Then convert the half-life from years to seconds. In this case: o Half-life = 5.26*365*24*3600 = 1.66*108 sec. - Then get the decay constant (k) from the formula k= .693/half-life. o k = .693/1.66*108 = 4.17*10-9 /sec. - So what is left is to solve for N. o N = rate / k = 5.83 / 4.17*10-9 = 1.40*109 nuclei. - We are told we have a 12.5 mg sample so we can get the total number of nuclei in this sample bytaking the mass of the sample, dividing the sample by its molar mass and multiplying by Avogadro's constant. o Total nuclei = (12.5*10-3 g) / (129.84 g / mol of CoCl2)* (6.02 *1023 nuclei / mol) = 5.80 *1019 nuclei.- Thus % isotope = N/total nuclei = 1.40*109/5.80*1019 * 100 % = 2.41 *10-9