Chapter 25: Problem 80
Of the following nuclides, the one most likely to be radioactive is \((a)^{31} P ;(b)^{66} Z n ;(c)^{35} C l ;(d)^{108} A g\).
Short Answer
Expert verified
Without more specific knowledge or measurements, it's difficult to give a definitive answer. However, judging by neutron to proton ratios, one of $\(^{35}Cl$ or $\(^{31}P$ is more likely to be radioactive. An expert in nuclear chemistry or a more specialized text might provide more insight.
Step by step solution
01
Identify the atomic numbers
The atomic numbers which represent the number of protons in the nuclide's nucleus of each element can be identified from the periodic table. The corresponding atomic numbers are $\(^{31}P (Phosphorus) = 15$, $\(^{66}Zn (Zinc) = 30$, $\(^{35}Cl (Chlorine) = 17$, and $\(^{108}Ag (Silver) = 47$.
02
Compute the number of neutrons
The number of neutrons can be calculated by subtracting the atomic number (number of protons) from the mass number (total number of protons and neutrons). Therefore, the number of neutrons for each nuclide are $\(^{31}P = 31-15=16$, $\(^{66}Zn = 66-30=36$, $\(^{35}Cl = 35-17=18$, and $\(^{108}Ag = 108-47=61$.
03
Determine stability
A nuclide is likely to be stable if the ratio of neutrons to protons is 1:1 for elements with atomic numbers up to 20. For elements with higher atomic numbers, the ratio gradually increases to about 1.5:1. Judging by this criterion, $\(^{66}Zn$ has a neutron to proton ratio of \(36/30=1.2\) which is close to the ideal ratio for elements of its atomic number, while $\(^{108}Ag$ has a neutron to proton ratio of \(61/47=1.3\), which is also close to the ideal ratio for elements of its atomic number. Neither $\(^{35}Cl$ with a neutron to proton ratio of \(18/17 = 1.06\) nor $\(^{31}P$ with a neutron to proton ratio of \(16/15 = 1.07\), have an ideal neutron to proton ratio. Hence, of these, one of $\(^{31}P$ or $\(^{35}Cl$ is more likely to be radioactive.
04
Check atomic numbers
Nuclides with atomic numbers more than 83 tend to be unstable and hence, radioactive. Since all listed isotopes have atomic numbers less than 83, this criterion doesn't help eliminate further nuclides.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Atomic Numbers
The atomic number, symbolized as 'Z', is fundamental to understanding elements and their isotopes. It tells us the number of protons present in the nucleus of an atom, which determines the element's identity. In the context of radioactivity, atomic numbers are crucial as they guide us towards the element's position in the periodic table and its nuclear properties. Every element has a unique atomic number; for instance, Phosphorus (P) has an atomic number of 15. This distinguishes it from other elements like Zinc (Zn), Chlorine (Cl), and Silver (Ag), with atomic numbers of 30, 17, and 47, respectively. In exercises related to radioactivity, recognizing atomic numbers enables students to delve into the behavior and stability of nuclides. Determining whether a nuclide is likely to be stable or radioactive often involves looking at elements with higher atomic numbers, typically greater than 83, which tend to display radioactive characteristics due to their larger, unstable nuclei. However, in cases where the atomic numbers are below this threshold, as with the given nuclides in the textbook exercise, other factors such as the neutron to proton ratio come into play.
Neutron to Proton Ratio
The neutron to proton (n/p) ratio is a key factor in assessing the stability of nuclides. Stable atoms usually have a n/p ratio close to 1:1, especially for lighter elements (atomic numbers up to approximately 20). As atomic numbers increase, a higher ratio is needed for stability, trending towards 1.5:1. For educational purposes, understanding and computing the n/p ratio helps predict nuclide behavior. For example, to find the neutron count in an isotope, students subtract the atomic number (proton count) from the mass number, which is the sum of protons and neutrons together. In the instructional solution provided, we see \(\text{^{\tiny 31}}P\) (16 neutrons, 15 protons) and \(\text{^{\tiny 35}}Cl\) (18 neutrons, 17 protons) have higher n/p ratios than what is typical for their atomic numbers, suggesting a greater likelihood of being radioactive. Meanwhile, \(\text{^{\tiny 66}}Zn\) and \(\text{^{\tiny 108}}Ag\) have n/p ratios more aligned with their respective atomic numbers, indicating a propensity for greater nuclear stability. For students, grappling with the exercise, working through the n/p ratios is not only about doing the math but also understanding the implications of these ratios on the potential radioactivity of nuclides.
Nuclear Stability
Nuclear stability is at the heart of nuclides' potential for radioactivity. A stable nucleus has a balance that prevents it from undergoing radioactive decay under normal conditions. This stability is influenced by several factors but primarily hinges on the neutron to proton ratio. For light nuclides (with lower atomic numbers), a nearly equal number of protons and neutrons typically signifies stability. On the other hand, heavier nuclides (with higher atomic numbers) require more neutrons than protons to remain stable due to the increased electrostatic repulsion between protons. In the scholastic context, comprehending nuclear stability helps students discern why certain isotopes are radioactive. It is this instability that can lead to various types of radioactive decay, such as alpha, beta, or gamma emission, as the nuclide seeks a more stable state. In our textbook example, we don't have any nuclides with an atomic number over 83 (the common threshold for natural radioactivity), so we apply our knowledge of n/p ratios to predict stability. \(\text{^{\tiny 31}}P\) and \(\text{^{\tiny 35}}Cl\) exhibit n/p ratios that stray from the ideal, hinting at a propensity towards instability and potentially making them candidates for radioactivity. For students, it's crucial to note that nuclear stability is a nuanced topic and cannot be determined by a single factor -- it requires an intersection of knowledge about atomic numbers, neutron to proton ratios, and the broader principles governing nuclear forces.