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Chapter 32: Problem 13

Complete the following nuclear reactions, assuming that the unknown quantity signified by the question mark is a single entity: a. \(\quad{ }_{20}^{43} \mathrm{Ca}(\alpha, ?)_{21}^{46} \mathrm{Sc}\) b. \(\quad{ }_{4}^{9} \mathrm{Be}(?, n)_{6}^{12} \mathrm{C}\) c. \(\quad{ }_{4}^{9} \operatorname{Be}(p, \alpha)\) ? d. \(?(\alpha, p)_{8}^{17} \mathrm{O}\) e. \(\quad \frac{55}{25} \mathrm{Mn}(n, \gamma)\) ?

Short Answer

Expert verified
a) \( \mathrm{H} \), b) \( \alpha \), c) \( \mathrm{Li} \), d) \( \mathrm{F} \), e) \( \mathrm{Mn} \).

Step by step solution

01

Understanding the Nuclear Reaction Problem

In a nuclear reaction problem, the principle of conservation of nucleons and charge applies. This means that the sum of atomic numbers (bottom numbers, representing protons) and the sum of mass numbers (top numbers, representing total protons and neutrons) must be the same on both sides of the reaction.
02

Solve Part (a)

Given: \( \frac{43}{20} \mathrm{Ca}(\alpha, \, ?)_{21}^{46} \mathrm{Sc} \). Identify the unknown particle using conservation laws. \( \alpha \) is \( \frac{4}{2} \mathrm{He} \), so:\[ \text{Left: Atomic: } 20+2; \text{Mass: } 43+4 \] \[ \text{Right: Atomic: } 21+x; \text{Mass: } 46+y \]Where \( x \) and \( y \) are the atomic and mass number of the unknown particle.Set up the equations:1. \( 20 + 2 = 21 + x \) \( \rightarrow x = 1 \)2. \( 43 + 4 = 46 + y \) \( \rightarrow y = 1 \)The particle is a proton, \( \frac{1}{1} \mathrm{H} \).
03

Solve Part (b)

Given: \({ }_{4}^{9} \operatorname{Be}(?, n)_{6}^{12} C \). Let the unknown particle be \( ^{A}_{Z} X \). Apply conservation laws:\[ \text{Left: Atomic: } 4 + Z; \text{Mass: } 9 + A \] \[ \text{Right: Atomic: } 6 + 0; \text{Mass: } 12 + 1 \] \[1. 4+Z = 6 \Rightarrow Z = 2 \2. 9+A = 13 \Rightarrow A = 4\]The particle is \( ^{4}_{2} \mathrm{He} \), or \( \alpha \).
04

Solve Part (c)

Given: \({ }_{4}^{9} \mathrm{Be}(p, \alpha) ? \). The expression signifies:1. Initial \( \text{Be} \): Atomic: \( 4 \), Mass: \( 9 \)2. Proton \( \text{p:} \): Atomic: \( 1 \), Mass: \( 1 \)3. \( \alpha \) particle: Atomic: \( 2 \), Mass: \( 4 \)Considering the missing participant, solve using:\[ \text{Atomic (Right): } 4 + 1 = 2 + Z \rightarrow Z = 3 \]\[ \text{Mass (Right): } 9 + 1 = 4 + A \rightarrow A = 6 \]The missing particle is \( ^{6}_{3} \mathrm{Li} \).
05

Solve Part (d)

Given: \(?(\alpha, p)_{8}^{17} \mathrm{O} \). The unknown initial particle should satisfy:1. \( \alpha \): Atomic \( = 2 \), Mass \( = 4 \)2. \( \text{p:} \): Atomic \( = 1 \), Mass \( = 1 \)Oxygen \( (\mathrm{O}) \): Atomic \( = 8 \), Mass \( = 17 \)Apply:\[ \text{Left Atomics: } Z = 2 +1 + 8 \rightarrow Z = 9 \]\[ \text{Left Mass: } A = 4 + 1 + 17 \rightarrow A = 20 \]The element is \( ^{20}_{9} \mathrm{F} \), Fluorine.
06

Solve Part (e)

Given: \(\frac{55}{25} \mathrm{Mn}(n, \gamma) ? \). \(\gamma\) refers to a gamma photon with no atomic or mass numbers, simplifying our equations:1. Atomic Number: \( 25 \left(\text{Mn} \right) + 0\left(\text{n}\right) = Z\), meaning the product also has an atomic number of 25.2. Mass Number: \( 55 + 1 = 56 \)Thus, the answer is \( \frac{56}{25} \mathrm{Mn} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

conservation of nucleons
In nuclear reactions, the principle of conservation of nucleons is a fundamental concept. It states that the total number of nucleons (which includes both protons and neutrons) must be the same before and after a reaction. This can be understood by looking at mass numbers, which sum up the nucleons. For instance, consider a reaction - The number of nucleons on the left side of the reaction must equal the number on the right side, represented by their mass numbers. - If you start with a nucleus and an alpha particle on the left, their combined mass must equal the mass numbers on the right side of the equation. This principle allows physicists and chemists to predict the outcome of nuclear reactions, ensuring the total nuclear count remains constant, much like balancing a chemical reaction equation.
conservation of charge
Another crucial aspect in understanding nuclear reactions is the conservation of charge. In every reaction, the total charge, represented by atomic numbers, should remain constant both before and after the process. This means that the sum of atomic numbers on one side of the equation must equal that on the other side. - The atomic number signifies the number of protons in a nucleus, which directly corresponds to the charge since each proton carries a positive charge. - For example, if a reaction begins with a nucleus and an alpha particle, the total charge (sum of atomic numbers) must also reflect these components on the reaction's right side. This preserves the neutrality of atomic interactions and ensures that the creation of new isotopes or particles doesn't disrupt the balance of charges.
isotopes
Isotopes are a fascinating feature of elements that occur when atoms have the same number of protons but different numbers of neutrons. This results in atoms of the same element but with different mass numbers. Understanding isotopes is essential in nuclear reactions as they form the basis for many reactions and their outcomes. - Each isotope of an element can have distinct properties and stability, which affect the types of nuclear reactions it may undergo or produce. - For instance, in the nuclear reactions problem you encountered, isotopes can transform into other isotopes or even different elements based on the conservation laws. When working with isotopes in reactions, one key focus is knowing how these various forms interact or convert, directly impacting the reaction results and applications in fields like radiochemistry and nuclear energy.
alpha particles
Alpha particles play a significant role in nuclear reactions. They are made up of 2 protons and 2 neutrons, equivalent to a helium nucleus, - and they are represented with the symbol \( ^{4}_{2} \text{He} \).Because they carry a relatively large mass and charge, - alpha particles can significantly impact the behavior and results of nuclear reactions. They are commonly emitted in radioactive decay processes and can interact with other elements to form new ones.In the context of completing nuclear reactions, identifying the presence and influence of alpha particles is crucial. They contribute to both the conservation of nucleons and charge, altering the equation to maintain balance. Their presence is a key factor in determining the identity of unknown particles or isotopes formed during nuclear reactions.

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Most popular questions from this chapter

When a 235 92 \(\mathrm{U}(235.043924 \mathrm{u})\) nucleus fissions, about \(200 \mathrm{MeV}\) of energy is released. What is the ratio of this energy to the rest energy of the uranium nucleus?

The energy consumed in one year in the United States is about \(9.3 \times 10^{19} \mathrm{~J}\). With each Q35 U fission, about \(2.0 \times 10^{2} \mathrm{MeV}\) of energy is released. How many kilograms of \({ }_{92}^{235} \mathrm{U}\) would be needed to generate this energy if all the nuclei fissioned?

The energy released by each fission within the core of a nuclear reactor is \(2.0 \times 10^{2}\) \(\mathrm{MeV}\). The number of fissions occurring each second is \(2.0 \times 10^{19}\). Determine the power (in watts) that the reactor generates.

Identify the unknown species \({ }_{Z}^{A} \mathrm{X}\) in the nuclear reaction \({ }_{11}^{22} \mathrm{Na}(d, \alpha){ }_{Z}^{A} \mathrm{X}\), where \(d\) stands for the deuterium isotope \({ }_{1}^{2} \mathrm{H}\) of hydrogen.

concept Questions Identical body tissues are being irradiated by two types of radiation (e.g., X-rays and protons). (a) If the absorbed doses are the same, which type of radiation, the one with the larger \(\mathrm{RBE}\) or the smaller \(\mathrm{RBE}\), produces the greater biological effect on the tissue? (b) If the two types of radiation have the same \(\mathrm{RBE}\), which type, the one that generates the greater absorbed dose or the smaller absorbed dose, produces the greater biological effect? Provide a reason for each answer.Problem Someone stands near a radioactive source and receives doses of the following types of radiation: \(\gamma\) rays \((20 \mathrm{mrad}, \mathrm{RBE}=1)\), electrons \((30 \mathrm{mrad}, \mathrm{RBE}=1)\), protons \((5\) mrad, \(\mathrm{RBE}=10\) ), and slow neutrons \((5 \mathrm{mrad}, \mathrm{RBE}=2\) ). Rank the types of radiation, highest first, as to which produces the largest biologically equivalent dose. Verify that your answers are consistent with your answers to the Concept Questions.
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