Determining Nuclide Symbol In Nuclear Decay Reactions
Nuclear equations represent the transformations that occur during radioactive decay. In this article, we will delve into the process of identifying the original nuclide symbol in a nuclear reaction. Specifically, we'll focus on the given nuclear equation:
This equation illustrates a beta decay process where an unknown nuclide, represented by X, decays into Actinium-228 (${}{89}^{228}Ac$) and a beta particle (${}{-1}^0\beta$). To determine the nuclide symbol of X, we must understand the fundamental principles governing nuclear reactions and how they relate to the conservation of mass number and atomic number.
Understanding Beta Decay and Nuclear Equations
Before we proceed, let's clarify the concept of beta decay and the conventions used in nuclear equations. Beta decay is a type of radioactive decay in which an unstable nucleus emits a beta particle, which is essentially a high-energy electron or positron. In the case of electron emission (beta-minus decay), a neutron within the nucleus transforms into a proton, and an electron is ejected. This process increases the atomic number by one while the mass number remains constant. Conversely, in positron emission (beta-plus decay), a proton transforms into a neutron, and a positron is emitted, decreasing the atomic number by one while the mass number stays the same. Nuclear equations are symbolic representations of nuclear reactions, where the symbols of the nuclides involved are written with their mass numbers as superscripts and atomic numbers as subscripts. The arrow indicates the direction of the reaction, with the original nuclide on the left and the products on the right. Conservation laws are paramount in nuclear reactions. The total mass number (the sum of protons and neutrons) and the total atomic number (the number of protons) must be conserved on both sides of the equation. This principle forms the bedrock of our method for identifying the unknown nuclide X.
Applying Conservation Laws to Determine the Unknown Nuclide
In the given nuclear equation,
we can apply the conservation laws to deduce the mass number and atomic number of the unknown nuclide X. Let's denote the mass number of X as A and the atomic number as Z. According to the conservation of mass number, the total mass number on the left side of the equation must equal the total mass number on the right side. Therefore,
A = 228 + 0 = 228
This tells us that the mass number of the original nuclide X is 228. Similarly, the conservation of atomic number dictates that the total atomic number on the left side must equal the total atomic number on the right side. Hence,
Z = 89 + (-1) = 88
This reveals that the atomic number of the original nuclide X is 90. Now that we have both the mass number (228) and the atomic number (90) of X, we can identify the element by referring to the periodic table. The element with atomic number 90 is Thorium (Th). Therefore, the nuclide symbol for X is ${}_{90}^{228}Th$. This detailed step-by-step application of conservation laws ensures an accurate determination of the unknown nuclide, highlighting the critical role these principles play in understanding nuclear reactions. By carefully balancing the mass and atomic numbers, we successfully unraveled the identity of the parent nuclide in this beta decay process. This methodical approach can be applied to various nuclear equations, making it a fundamental tool in nuclear chemistry.
Identifying the Correct Option
Based on our calculations, the nuclide symbol of X is ${}_{90}^{228}Th$. Now, let's examine the given options:
A. ${}{90}^{230}Th$ B. ${}{89}^{229}Ac$ C. ${}_{90}^{228}Th$
Comparing our result with the options, we can clearly see that option C, ${}{90}^{228}Th$, matches our calculated nuclide symbol for X. Therefore, option C is the correct answer. Option A (${}{90}^{230}Th$) has the correct atomic number but an incorrect mass number. Option B (${}_{89}^{229}Ac$) has an incorrect atomic number and mass number. This process of elimination, coupled with our accurate determination of the nuclide symbol, solidifies our confidence in the correct answer. Understanding the nuances of each option and how they deviate from the calculated values is crucial in mastering nuclear chemistry problems. By systematically analyzing the given information and applying the principles of conservation laws, we can effectively navigate through the options and pinpoint the accurate solution.
Why the Other Options Are Incorrect
To further solidify our understanding, let's discuss why the other options are incorrect. Option A, ${}{90}^{230}Th$, represents an isotope of Thorium with a mass number of 230. While it has the correct atomic number (90), indicating that it is Thorium, the mass number is incorrect. Our calculations based on the conservation of mass number clearly showed that the mass number of X should be 228. Therefore, this option is incorrect. Option B, ${}{89}^{229}Ac$, represents an isotope of Actinium with a mass number of 229. This option is incorrect for two reasons. First, the atomic number is 89, which corresponds to Actinium, not the element we are looking for. Second, the mass number is 229, which does not match our calculated mass number of 228. This detailed examination of incorrect options not only reinforces the correct answer but also enhances our understanding of the underlying principles. By identifying the specific errors in each option, we gain a deeper appreciation for the precision required in nuclear chemistry calculations. This thorough analysis is a valuable tool in avoiding common mistakes and building a strong foundation in the subject.
Conclusion
In conclusion, by applying the principles of conservation of mass number and atomic number, we successfully determined that the nuclide symbol of X in the nuclear equation
is ${}_{90}^{228}Th$. This exercise underscores the importance of understanding beta decay and the rules governing nuclear reactions. Mastering these concepts is crucial for tackling more complex problems in nuclear chemistry and related fields. The systematic approach we used, involving the application of conservation laws and a careful comparison of options, provides a reliable method for solving similar problems. Furthermore, the detailed explanation of why the incorrect options are wrong enhances our understanding and reinforces the correct solution. Nuclear chemistry is a fascinating field with numerous applications, and a solid grasp of fundamental principles like those discussed here is essential for success. By continuously practicing and applying these concepts, we can build our proficiency and confidently navigate the intricacies of nuclear reactions.
Importance of Understanding Nuclear Equations
Understanding nuclear equations is paramount for several reasons. Firstly, nuclear equations provide a concise and accurate way to represent nuclear reactions, which are fundamental processes in various scientific disciplines. From nuclear medicine to astrophysics, nuclear reactions play a crucial role, and being able to interpret and manipulate nuclear equations is essential for professionals in these fields. Secondly, the principles underlying nuclear equations, such as the conservation of mass number and atomic number, are fundamental laws of nature. Grasping these principles allows us to predict the outcomes of nuclear reactions and understand the behavior of radioactive materials. Thirdly, nuclear equations are instrumental in understanding the decay processes of radioactive isotopes, which have significant implications for nuclear waste management, environmental science, and dating techniques like carbon-14 dating. The ability to balance nuclear equations and identify unknown nuclides is a critical skill in these contexts. Finally, the study of nuclear equations fosters a deeper appreciation for the structure of the atom and the forces that govern the nucleus. This understanding is not only valuable in scientific pursuits but also in making informed decisions about societal issues related to nuclear technology and energy. In summary, the significance of understanding nuclear equations extends far beyond the classroom, impacting various aspects of science, technology, and society.
Final Answer
The final answer is (C) ${}_{90}^{228}Th$.
