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Chapter 32: Problem 17

Complete the following nuclear reactions, assuming that the unknown quantity signified by the question mark is a single entity: (a) \(^{34}_{18} \mathrm{Ar}(n, \alpha) ?\) (b) \(^{82}_{34} \operatorname{Se}(?, n) ^{82}_{35} \mathrm{Br}\) (c) \(^{58}_{28} \mathrm{Ni}\left(\begin{array}{l}{40} \\ {18}\end{array} \mathrm{Ar}, ?\right) ^{57}_{27} \mathrm{Co}\) (d) \(?(\gamma, \alpha)_{8}^{16} \mathrm{O}\)

Short Answer

Expert verified
(a) \(^{31}_{16}\mathrm{S}\), (b) \(^{1}_{1}\mathrm{H}\), (c) \(^{41}_{19}\mathrm{K}\), (d) \(^{20}_{10}\mathrm{Ne}\).

Step by step solution

01

Understand the Reaction for Part (a)

The given nuclear reaction is \(^{34}_{18} \mathrm{Ar}(n, \alpha) ?\). This means a neutron \((n)\) is bombarding \(^{34}_{18} \mathrm{Ar}\), resulting in the emission of an alpha particle \((\alpha = ^4_2 \mathrm{He})\) and the production of an unknown nucleus. We need to balance the atomic and mass numbers.
02

Balance Equation for Part (a)

The initial atom is \(^{34}_{18}\mathrm{Ar}\) with an added neutron \(^{1}_{0}\mathrm{n}\). This gives a total of \(35\) for the mass number and \(18\) for the atomic number before the reaction. After emitting an alpha \((\alpha)\) particle \(^{4}_{2}\mathrm{He}\), the remaining nucleus should have \(31\) as its mass number \((35-4)\) and \(16\) as its atomic number \((18-2)\). The resulting nucleus is \(^{31}_{16}\mathrm{S}\).
03

Understand the Reaction for Part (b)

The given reaction is \(^{82}_{34}\operatorname{Se}(?, n) ^{82}_{35}\mathrm{Br}\). Here, an unknown particle \(?)\) reacts with \(^{82}_{34}\operatorname{Se}\) to produce \(^{82}_{35}\mathrm{Br}\) while releasing a neutron \((n)\). We need to find this unknown particle.
04

Balance Equation for Part (b)

The mass number of \(^{82}_{34}\operatorname{Se}\) doesn't change. \(^{82}_{35}\mathrm{Br}\) implies an increase in the atomic number by 1, indicating that the unknown must carry a charge of \(+1\) without altering the mass number. Therefore, the unknown particle is a proton \(^{1}_{1}\mathrm{H}\).
05

Understand the Reaction for Part (c)

The given reaction is \(^{58}_{28} \mathrm{Ni}\left(\begin{array}{l}{40} \ {18}\end{array}\mathrm{Ar}, ?\right) ^{57}_{27} \mathrm{Co}\). \(^{40}_{18}\mathrm{Ar}\) hits \(^{58}_{28}\mathrm{Ni}\) and produces \(^{57}_{27}\mathrm{Co}\) plus an unknown particle.
06

Balance Equation for Part (c)

The total mass number is initially \(98\) \((58+40)\) and the atomic number is \(46\) \((28+18)\). After the reaction forming \(^{57}_{27}\mathrm{Co}\), the resulting mass is \(57 + m\) and atomic number is \(27 + z\). Balancing gives \(m = 41\) and \(z = 19\). Therefore, the unknown particle is \(^{41}_{19}\mathrm{K}\).
07

Understand the Reaction for Part (d)

The unknown nucleus reacts with a gamma photon to release an \(\alpha\) particle and form \(^{16}_{8}\mathrm{O}\). We want to solve for the unknown nucleus.
08

Balance Equation for Part (d)

\(^{16}_{8}\mathrm{O}\) and \(^{4}_{2}\mathrm{He}\) have a total combined mass and atomic numbers of \(20\) and \(10\) respectively. Therefore, the unknown must also have a mass number of \(20\) and an atomic number of \(10\). This is \(^{20}_{10}\mathrm{Ne}\).

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Key Concepts

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

Atomic Structure
Atomic structure is the foundation of understanding nuclear reactions. It is essential to know that every atom consists of a nucleus containing protons and neutrons, with electrons orbiting around it. The number of protons, known as the atomic number, defines the element. Neutrons contribute to the mass of the atom, together forming the mass number when added to the number of protons. In nuclear chemistry, changes to the nucleus occur, which can alter both the atomic and mass numbers. For example, in a nuclear reaction, if a neutron is added to an atom, the mass number increases by one, but the atomic number remains unchanged. These structural changes to the nucleus are the basis for nuclear equations.
Particle Interactions
Particle interactions play a critical role in nuclear reactions. Different particles such as neutrons, protons, and alpha particles (which are helium nuclei with two protons and two neutrons) can interact with atomic nuclei, resulting in nuclear transformations.
  • Neutrons: These neutral particles can penetrate deeply into the nucleus, causing fission or transforming the nucleus by forming an isotope.
  • Protons: Positively charged, protons can increase the atomic number when they interact with nuclei, often resulting in a new element.
  • Alpha particles: Emitted in some radioactive decays, they can cause the nucleus to lose protons and neutrons, thus transmuting into a different element.
Understanding these interactions helps in balancing nuclear equations since they define the changes in mass and atomic numbers.
Nuclear Equations
Nuclear equations are a symbolic representation of nuclear reactions, showing how particles interact. Much like chemical equations, nuclear equations must be balanced. This means that the total mass and atomic numbers must be the same on both sides of the equation. The balancing of nuclear equations follows a process:
  • Identify the known particles, such as isotopes, neutrons, protons, etc.
  • Calculate the total mass and atomic numbers of the reactants and products.
  • Add or determine the unknown entity to ensure the sums are equivalent on both sides.
By doing this, you determine the identity of unknown particles and predict the result of nuclear reactions. For instance, in a reaction where an element transforms upon neutron absorption, or releases an alpha particle, you carefully calculate the differences in mass and atomic numbers to complete the equation.

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Most popular questions from this chapter

During a nuclear reaction, an unknown particle is absorbed by a copper \(^63_{29}\) Cu nucleus, and the reaction products are \(^{62}_{29} \mathrm{Cu}\) , a neutron, and a proton. What are the name, atomic number, and nucleon number of the nucleus formed temporarily when the copper \(^{63}_{29} \mathrm{Cu}\) nucleus absorbs the unknown particle?

Determine the atomic number \(Z\) , the atomic mass number \(A\) , and the element \(X\) for the unknown species \(\frac{A}{2} X\) in the following reaction for the fission of uranium \(_{92}^{235} \mathrm{U} :\) \(_{0}^{1} n+ ^{235}_{92} \mathrm{U} \longrightarrow^{133}_{51} \mathrm{Sb} +^{A}_{Z} X+4_{0}^{1} n\) Consult the periodic table on the inside of the back cover of the text as needed.

If each fission reaction of a \(^{235}_{92} \mathrm{U}\) nucleus releases about \(2.0 \times 10^{2}\) MeV of energy, determine the energy (in joules) released by the complete fissioning of 1.0 gram of \(^{235}_{92} \mathrm{U}\) (b) How many grams of \(^{235}_{92} \mathrm{U}\) would be consumed in one year to supply the energy needs of a household that uses 30.0 kWh of energy per day, on the average?

A nuclear power plant is 25\(\%\) efficient, meaning that only 25\(\%\) of the power it generates goes into producing electricity. The remaining 75\(\%\) is wasted as heat. The plant generates \(8.0 \times 10^{8}\) watts of electric power. If each fission releases \(2.0 \times 10^{2}\) MeV of energy, how many kilograms of \(^{235}_{92} \mathrm{U}\) are fissioned per year?

When a nuclear reactor is in a critical state, the neutrons released in each fission trigger an average of exactly one additional fission. If the average number of additional fissions triggered rises above one, the reactor enters a supercritical state in which the fission rate and the power output grow very rapidly. A reactor in a critical state has a power output of 25 kW. The reactor then enters a supercritical state in which each fission triggers an average of 1.01 additional fissions. The average time for the neutrons released by one generation of fissions to trigger the next generation of fissions is \(1.2 \times 10^{-8}\) s. How much time elapses before the power output from a single generation of fissions grows to 3300 \(\mathrm{MW}\) (roughly the normal output of a commercial reactor)?
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