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Lecture 17 - Neutron ­Proton Ra-o •  Any element...

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Unformatted text preview: Neutron ­Proton Ra-o •  Any element with more than one proton will have repulsions between the protons in the nucleus. •  Nuclear force helps keep the nucleus from pulling apart. •  Neutrons play a key role stabilizing the nucleus. •  The ra-o of neutrons to protons is an important factor. Neutron ­Proton Ra-o For smaller nuclei (Z ≤ 20) stable nuclei have a neutron ­to ­ proton ra-o close to 1:1. Neutron ­Proton Ra-o As nuclei get larger, it takes a greater number of neutrons to stabilize the nucleus. Stable Nuclei The shaded region in the figure shows what nuclides would be stable, the so ­called belt of stability. Stable Nuclei •  Nuclei above this belt have too many neutrons. •  They tend to decay by emiKng beta par-cles. Stable Nuclei •  Nuclei below the belt have too many protons. •  They tend to become more stable by positron emission or electron capture. Stable Nuclei •  There are no stable nuclei with an atomic number greater than 83. •  These nuclei tend to decay by alpha emission. A plot of number of neutrons vs. number of protons for the stable nuclides Nuclear Stability and Mode of Decay • Very few stable nuclides exist with N/Z < 1. • The N/Z ra-o of stable nuclides gradually increases a Z increases. • All nuclides with Z > 83 are unstable. • Elements with an even Z usually have a larger number of stable nuclides than elements with an odd Z. • Well over half the stable nuclides have both even N and even Z. Predic<ng the Mode of Decay • Neutron ­rich nuclides undergo β decay. • Neutron ­poor nuclides undergo positron decay or electron capture. • Heavy nuclides undergo α decay. Sample Problem 23.2 PROBLEM: Predic<ng Nuclear Stability Which of the following nuclides would you predict to be stable and which radioac-ve? Explain. (a) 1810Ne PLAN: (b) 3216S (c) 23690Th (d) 12356Ba Stability will depend upon the N/Z ra-o, the value of Z, the value of stable N/Z nuclei, and whether N and Z are even or odd. SOLUTION: (a) Radioac-ve (b) Stable N/Z = 0.8; there are too few neutrons to be stable. N/Z = 1.0; Z < 20 and N and Z are even. (c) Radioac-ve (d) Radioac-ve Every nuclide with Z > 83 is radioac-ve. N/Z = 1.20; Figure 23.2A shows stability when N/Z ≥ 1.3. Sample Problem 23.3 PROBLEM: Use the atomic mass of the element to predict the mode(s) of decay of the following radioac-ve nuclides: (a) 125B PLAN: Predic<ng the Mode of Nuclear Decay (b) 23492U (c) 8133As (d) 12757La Find the N/Z ra-o and compare it to the band stability. Then predict which of the modes of decay will give a ra-o closer to the band. SOLUTION: (a) N/Z = 1.4 which is high. The nuclide will probably undergo β- decay altering Z to 6 and lowering the ra-o. (c) N/Z = 1.24 which is in the band of stability. It will probably undergo β- decay or positron emission. (b) The large number of neutrons makes this a good candidate for α decay. (d) N/Z = 1.23 which is too low for this area of the band. It can increase Z by positron emission or electron capture. The 238U decay series. Decay rate (A) =  ­ ΔN Δt SI unit of decay is the becquerel (Bq) = 1 d/s. Nuclear decay is a first ­order rate process. Large k means a short half ­life and vice versa. Decrease in the number of 14C nuclei over <me Sample Problem 23.4 Finding the Number of Radioac<ve Nuclei PROBLEM: Stron-um ­90 is a radioac-ve byproduct of nuclear reactors that behaves biologically like calcium, the element above it in Group 2A(2). When 90Sr is ingested by mammals, it is found in their milk and eventually in the bones of those drinking the milk. If a sample of 90Sr has an ac-vity of 1.2 x 1012 d/s, what are the ac-vity and the frac-on of nuclei that have decayed aeer 59 yr (t1/2 of 90Sr = 29 yr). PLAN: The frac-on of nuclei that have decayed is the change in the number of nuclei, expressed as a frac-on of the star-ng number. The ac-vity of the sample (A) is propor-onal to the number of nuclei (N). We are given the A0 and can find At from the integrated form of the first ­order rate equa-on. 0.693 2 SOLUTION: t1/2 = ln so k = = 0.024 yr  ­1 29 yr k N0 A0 (1.2 x1012  ­ 2.9 x 1011) ln = ln = kt ln At =  ­kt + ln A0 Frac-on Nt At = decayed (1.2 x 1012) ln At =  ­(0.024 yr  ­1)(59 yr) + ln(1.2 x 1012 d/s) ln At = 26.4 At = 2.9 x 1011 d/s Frac-on decayed = 0.76 Sample Problem 23.5 PROBLEM: PLAN: Applying Radiocarbon Da<ng The charred bones of a sloth in a cave in Chile represent the earliest evidence of human presence in the southern -p of South America. A sample of the bone has a specific ac-vity of 5.22 disintegra-ons per minute per gram of carbon (d/min•g). If the 12C/14C ra-o for living organisms results in a specific ac-vity of 15.3 d/min•g, how old are the bones (t1/2 of 14C = 5730 yr)? Calculate the rate constant using the given half ­life. Then use the first ­order rate equa-on to find the age of the bones. SOLUTION: ln 2 0.693 k = = = 1.21 x 10 ­4 yr  ­1 t1/2 5730 yr 1 1 A0 15.3 t = ln =  ­ 4 l ­n 1 = 8.89 x 103 yr k 1.21 x 10 yr 5.22 At The bones are about 8900 years old. ...
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