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Unformatted text preview: Neutron
Proton Rao • 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 rao of neutrons to protons is an important factor. Neutron
Proton Rao For smaller nuclei (Z ≤ 20) stable nuclei have a neutron
to
proton rao close to 1:1. Neutron
Proton Rao 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 parcles. 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 rao 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 radioacve? Explain. (a) 1810Ne PLAN: (b) 3216S (c) 23690Th (d) 12356Ba Stability will depend upon the N/Z rao, the value of Z, the value of stable N/Z nuclei, and whether N and Z are even or odd. SOLUTION: (a) Radioacve (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) Radioacve (d) Radioacve Every nuclide with Z > 83 is radioacve. 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 radioacve nuclides: (a) 125B PLAN: Predic<ng the Mode of Nuclear Decay (b) 23492U (c) 8133As (d) 12757La Find the N/Z rao and compare it to the band stability. Then predict which of the modes of decay will give a rao 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 rao. (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: Stronum
90 is a radioacve 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 acvity of 1.2 x 1012 d/s, what are the acvity and the fracon of nuclei that have decayed aeer 59 yr (t1/2 of 90Sr = 29 yr). PLAN: The fracon of nuclei that have decayed is the change in the number of nuclei, expressed as a fracon of the starng number. The acvity of the sample (A) is proporonal to the number of nuclei (N). We are given the A0 and can ﬁnd At from the integrated form of the ﬁrst
order rate equaon. 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 Fracon 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 Fracon 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 acvity of 5.22 disintegraons per minute per gram of carbon (d/min•g). If the 12C/14C rao for living organisms results in a speciﬁc acvity 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 equaon 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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This note was uploaded on 02/18/2012 for the course CHEM 6C taught by Professor Hoeger during the Spring '08 term at UCSD.
 Spring '08
 HOEGER
 Nucleus, Proton

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