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Complete the notations for the following processes. (a) \({ }^{24} \mathrm{Mg}(d, \alpha) ?\) (e) \(130 \mathrm{Te}(d, 2 n) ?\) (b) \({ }^{26} \mathrm{Mg}(d, p) ?\) (f) \({ }^{55} \mathrm{Mn}(n, \gamma)\) ? (c) \({ }^{40} \operatorname{Ar}(\alpha, p) ?\) \((g){ }^{59} \mathrm{Co}(n, \alpha) ?\) (d) \({ }^{12} \mathrm{C}(d, n) ?\)

Short Answer

Expert verified
(a) ^{22}Na, (e) ^{130}I, (b) ^{27}Mg, (f) ^{56}Mn, (c) ^{43}K, (g) ^{56}Mn, (d) ^{13}N.

Step by step solution

01

Understanding Nuclear Reaction Notation

In nuclear reactions, the notation like \(A(X, Y)Z\) refers to a target nucleus (\(A\)), a bombarding particle (\(X\)), an ejected particle (\(Y\)), and a product nucleus (\(Z\)). We need to find the missing product nucleus in each reaction.
02

Calculate Reaction (a)

Given: \({}^{24}\mathrm{Mg}(d, \alpha) ?\)\. The deuteron (d) has the same mass and proton number as \(^{2}_{1}H\) and alpha (\(\alpha\)) is \(^{4}_{2}He\). Compute the atomic number and mass of missing nucleus: atomic number \(= 12 + 1 - 2 = 11\) and mass number \(= 24 + 2 - 4 = 22\). The product is \(^{22}_{11}Na\).
03

Calculate Reaction (b)

Given: \({}^{26}\mathrm{Mg}(d, p) ?\)\. The proton (p) is \(^{1}_{1}H\). Calculate atomic number: \(12 + 1 - 1 = 12\), mass number: \(26 + 2 - 1 = 27\). The product is \(^{27}_{12}Mg\).
04

Calculate Reaction (c)

Given: \({}^{40}\operatorname{Ar}(\alpha, p) ?\)\. A proton (p) is \(^{1}_{1}H\). Calculate atomic number \(= 18 + 2 - 1 = 19\), and mass number: \(40 + 4 - 1 = 43\). The product is \(^{43}_{19}K\).
05

Calculate Reaction (d)

Given: \({}^{12}\mathrm{C}(d, n) ?\)\. A neutron (n) is \(^{1}_{0}n\). Calculate atomic number: \(6 + 1 - 0 = 7\), mass number: \(12 + 2 - 1 = 13\). The product is \(^{13}_{7}N\).
06

Calculate Reaction (e)

Given: \(130\mathrm{Te}(d, 2n) ?\)\. Atomic number changes: \(52 + 1 = 53\), mass number: \(130 + 2 - 2 = 130\). This results in \(^{130}_{53}I\).
07

Calculate Reaction (f)

Given: \(^{55}\mathrm{Mn}(n, \gamma) ?\)\. In a capture reaction emitting gamma (\(\gamma)\) radiation, the neutron is captured without mass number change: atomic number: \(25\), mass number: \(55 + 1 = 56\). The product is \(^{56}_{25}Mn\).
08

Calculate Reaction (g)

Given: \(^{59}\mathrm{Co}(n, \alpha) ?\)\. The atomic number decreases, and the mass number: \(59 + 1 - 4 = 56\). A cobalt nucleus turns into an iron nucleus: atomic number: \(27 + 0 - 2 = 25\). The product is \(^{56}_{25}Mn\).

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Key Concepts

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

Nuclear Physics
Nuclear physics dives into the fascinating world of atomic nuclei. It aims to understand the fundamental particles and forces that govern atomic nuclei. Unlike traditional chemistry, which focuses on electrons and how atoms interact at a surface level, nuclear physics explores much deeper into the core of atoms. This field studies how protons and neutrons bind together to form a nucleus, along with the forces and reactions that can change one element into another. Additionally, it investigates phenomena such as nuclear decay, fission, fusion, and radioactive decay. Through nuclear physics, we gain a deep understanding of energy release in nuclear reactions and how it can be harnessed in various applications, from nuclear power plants to medical imaging techniques. Learning nuclear physics reveals the profound mechanisms that keep the universe ticking, allowing for breakthroughs in both energy production and medical technologies.
Atomic Number
The atomic number of an element is one of its most significant attributes. It represents the number of protons in the nucleus of an atom. This number is crucial because it defines the identity of the element. For example, any atom with an atomic number of 1 is Hydrogen, whereas an atomic number of 6 is Carbon. The atomic number determines the element's position in the periodic table and, consequently, its chemical properties.
  • Defines the type of element
  • Dictates electron configuration
  • Influences chemical behavior
  • Equal to the number of protons in the nucleus
Understanding atomic numbers helps us predict how an element will react chemically and what kind of compounds it will form. When elements undergo nuclear reactions, monitoring changes in atomic numbers is crucial, as this can lead to the formation of new elements entirely.
Mass Number
The mass number gives insight into the total number of particles in an atom's nucleus. It is the sum of protons and neutrons. This number is significant in nuclear reactions as it helps track the conservation of matter. Although elements can have the same atomic number, their mass numbers can differ due to variations in the number of neutrons, leading to different isotopes. For example, Carbon typically has a mass number of 12, however, isotopes like carbon-13 and carbon-14 have additional neutrons, altering their mass numbers.
  • Calculates the total nucleons (protons + neutrons)
  • Varies among isotopes of the same element
  • Helps in identifying isotopic changes in reactions
In nuclear reactions, managing the mass number ensures the reaction is balanced, and no nucleon 'appears' or 'disappears', preserving the law of conservation of mass.
Deuteron
Deuteron is a term you will often see in nuclear physics, as it refers to the nucleus of deuterium. Deuterium, also known as "heavy hydrogen," contains one proton and one neutron. Deuterons are frequently utilized in nuclear reactions due to their stability and mass slightly greater than that of a single proton. In nuclear reactions, deuterons can be used to bombard other nuclei, leading to the formation of new elements. Understanding deuterons is important because:
  • They play a role in nuclear fusion processes.
  • They help scientists create new elements in nuclear laboratories.
  • They serve as a tool in exploring nuclear structure and processes.
Deuterons are crucial in progressing nuclear physics research and influencing modern applications involving nuclear energy and reactions.
Alpha Particle
An alpha particle is one of the most common particles encountered in nuclear physics. It consists of two protons and two neutrons, similar to a helium nucleus. Alpha particles are important in nuclear decay processes, such as alpha decay, where unstable nuclei emit an alpha particle to become more stable. This process decreases the atomic number by 2 and the mass number by 4, effectively transforming the atom into a different element altogether.
  • Highly ionizing due to its charge and mass
  • Limited penetration ability, stopped by a sheet of paper
  • Widely used in radiation therapy and smoke detectors
By understanding the behavior and properties of alpha particles, scientists are able to interpret decay sequences in radioactive materials, offering insights into both the nature of the material and potential applications in technology. Alpha particles play a critical role in the exploration of nuclear energy and have practical applications in fields ranging from medicine to safety devices.

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Most popular questions from this chapter

The half-life of carbon- 14 is \(5.7 \times 10^{3}\) years. What fraction of a sample of \({ }^{14} \mathrm{C}\) will remain unchanged after a period of five half-lives?

The half-life of uranium- 238 is about \(4.5 \times 10^{9}\) years, and its end product is lead-206. We notice that the oldest uranium-bearing rocks on Earth contain about a \(50: 50\) mixture of \({ }^{238} \mathrm{U}\) and \({ }^{206} \mathrm{~Pb}\). Roughly, what is the age of these rocks? Apparently about half the \({ }^{258} \mathrm{U}\) has decayed to \({ }^{206} \mathrm{~Pb}\) during the existence of the rock. Hence, the rock must have been formed about \(4.5\) billion years ago.

Technetium-99 \(\left({ }_{43}^{99} \mathrm{Tc}\right)\) has an excited state that decays by emission of a gamma ray. The half-life of the excited state is \(360 \mathrm{~min}\). What is the activity, in curies, of \(1.00 \mathrm{mg}\) of this excited isotope? Because we have the half-life \(\left(t_{1 / 2}\right)\) we can determine the decay constant since \(\lambda t_{1 / 2}=0.693 .\) The activity of a sample is \(\lambda N\). In this case, $$ \lambda=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{21600 \mathrm{~s}}=3.21 \times 10^{-5} \mathrm{~s}^{-1} $$ We also know that \(99.0 \mathrm{~kg}\) of Tc contains \(6.02 \times 10^{26}\) atoms. A mass \(m\) will therefore contain \([m /(99.0 \mathrm{~kg})]\left(6.02 \times 10^{26}\right)\) atoms. In our case, \(m=1.00 \times 10^{-6} \mathrm{~kg}\), and so $$ \begin{aligned} \text { Activity } &=\lambda N=\left(3.21 \times 10^{-5} \mathrm{~s}^{-1}\right)\left(\frac{1.00 \times 10^{-6} \mathrm{~kg}}{99.0 \mathrm{~kg}}\right)\left(6.02 \times 10^{26}\right) \\ &=1.95 \times 10^{14} \mathrm{~s}^{-1}=1.95 \times 10^{14} \mathrm{~Bq} \end{aligned} $$

The radius of a carbon nucleus is about \(3 \times 10^{-15} \mathrm{~m}\) and its mass is \(12 \mathrm{u}\). Find the average density of the nuclear material. How many more times dense than water is this? $$ \begin{aligned} \rho &=\frac{m}{V}=\frac{m}{4 \pi r^{3} / 3}=\frac{(12 \mathrm{u})\left(1.66 \times 10^{-27} \mathrm{~kg} / \mathrm{u}\right)}{4 \pi\left(3 \times 10^{-15} \mathrm{~m}\right)^{3} / 3}=1.8 \times 10^{17} \mathrm{~kg} / \mathrm{m}^{3} \\\ \frac{\rho}{\rho_{\text {water }}} &=\frac{1.8 \times 10^{17} \mathrm{~kg} / \mathrm{m}^{3}}{1000 \mathrm{~kg} / \mathrm{m}^{3}}=2 \times 10^{14} \end{aligned} $$

What is the binding energy of the atom \({ }^{12} \mathrm{C} ?\) One atom of \({ }^{12} \mathrm{C}\) consists of 6 protons, 6 electrons, and 6 neutrons. The mass of the uncombined protons and electrons is the same as that of \(\operatorname{six}{ }^{1} \mathrm{H}\) atoms (if we ignore the very small binding energy of each proton-electron pair). The component particles may thus be considered as six \({ }^{1} \mathrm{H}\) atoms and six neutrons. A mass balance may be computed as follows. $$ \begin{array}{ll} \text { Mass of six }{ }^{1} \mathrm{H} \text { atoms }=6 \times 1.0078 \mathrm{u} & =6.0468 \mathrm{u} \\ \text { Mass of six neutrons }=6 \times 1.0087 \mathrm{u} & =6.0522 \mathrm{u} \\\ \text { Total mass of component particles } & =12.0990 \mathrm{u} \\ \text { Mass of }{ }^{12} \mathrm{C} \text { atom } & =\underline{12.0000 \mathrm{u}} \\ \text { Loss in mass on forming }{ }^{12} \mathrm{C} & =0.0990 \mathrm{u} \\ \text { Binding energy }=(931 \times 0.0990) \mathrm{MeV} & =92 \mathrm{MeV} \end{array} $$

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