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

Short Answer

Expert verified
(a) \(^{22}\text{Na}\), (b) \(^{27}\text{Mg}\), (d) \(^{13}\text{N}\), (e) \(^{130}\text{I}\), (f) \(^{56}\text{Mn}\), (g) \(^{55}\text{Fe}\).

Step by step solution

01

Analyze the Reaction (a)

The reaction is \[^{24} \rm{Mg}(d, \alpha) ?\]where \(d\) is a deuteron (a type of hydrogen nucleus, \(^2_1\rm{H}\)), and \(\alpha\) is an alpha particle (\(^4_2\rm{He}\)). The target nucleus \(^A_Z \rm{X}\) undergoes a transformation, emitting an alpha particle and a resulting nucleus.
02

Determine Unknown Nucleus for (a)

Using conservation of mass and charge, the initial and final reactions can be equated.Initial components: \(^2_1\rm{H} + \ ^{24}_{12}\rm{Mg}\)Final components: \(^4_2\rm{He} + \ ^{X}_{Y}\rm{Element}\)By balancing mass:\[24 + 2 = 4 + A \Rightarrow A = 22\]By balancing charge:\[12 + 1 = 2 + Z \Rightarrow Z = 11\]The resulting element is \(^{22}_{11}\rm{Na}\).
03

Analyze the Reaction (b)

The reaction is \[^{26} \rm{Mg}(d, p) ?\]where a proton \(p\) is emitted. Apply conservation principles. Initial components: \(^2_1\rm{H} + \ ^{26}_{12}\rm{Mg}\).
04

Determine Unknown Nucleus for (b)

Using the reaction\[^{26} \rm{Mg}(d, p)\ ^{27} \text{Mg} \],balathe resulting mass and charge to solve shows that after the reaction the resulting nucleus is: By conservation of mass:\[26 + 2 = 1 + A \Rightarrow A = 27\]By conservation of charge:\[12 + 1 = 1 + Z \Rightarrow Z = 12\]The resulting nucleus is \(^{27}_{12}\rm{Mg}\).
05

Analyze the Reaction (c)

This reaction is incomplete: \[^{40} ext{Ar}(p) ? \]without details on the emission particle. Additional data needed for a complete solution, but can make assumptions with common particles like neutron or gamma.
06

General Solution Approach (c)

Many reactions with missing emission products can result in stable or unstable isotopes ranging typically resulting from common particle emissions like (n), (\(\gamma\)), causing minimal isotope shift. Await instructor's direction for desired approach.
07

Analyze the Reaction (d)

Given the reaction:\[^{12} \rm{C}(d, n) ?\]Conservation of mass and charge give us the reaction products. Consider each step with a particle, neutron \(n\), emission.
08

Determine Unknown Nucleus for (d)

From the given:\[^{12}_{6} \text{C} + ^{2}_{1} ext{H} \longrightarrow n + ^{13}_7 ext{N}\]Mass and charge consistency:\[12 + 2 = 1 + A \Rightarrow A = 13\]Charge balancing:\[6 + 1 = 0 + Z \Longrightarrow Z = 7\]
09

Analyze the Reaction (e)

This reaction involves:\[^{130} \rm{Te}(d, 2n) ?\]Deuteron joins telure, emits two neutrons.
10

Determine Unknown Nucleus for (e)

Initial mass and charge assessment gives\[A: 130 + 2 - (2 * 1) = 130\]\[Z: 52 + 1 = Z, \], where \(\text{element} \euler_{52} = ^{130}_{53}\text{I}\)
11

Considerations for Reactions (f) and (g)

Analogous to reactions involving neutron capture and alpha emission. Analysis identifies the reactions as capturing neutron or emitting alpha.
12

Specify the Element for Reactions (f) and (g)

(f) \(n, \gamma)\): Capture occurs leaving mass as an excited element, 56 màss and no loss Z: 25 :\(^{56}_{25}\text{Mn}\) with mass increase by one. (g) \(\alpha \): \([A= 59 + 1; Z=27 + 3] = ^{55}_{24} ext{Fe}\)
13

Evaluate Based on Concepts

Generalize step descriptions or attempt reverting to conceptual logical directions, else assumptions change reactants or products.
14

Review Results for Completion

(a) \(^{22}\rm{Na}\), (b) \(^{27}\rm{Mg}\), (d) \(^{13}\text{N}\), (e) \(^{130}\text{I}\), (f) \(^{56}\text{Mn}\), (g) \(^{55}\text{Fe}\).

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Key Concepts

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

Conservation of Mass
In nuclear reactions, the principle of conservation of mass manifests as the conservation of mass and atomic numbers. This principle states that the total mass and total charge before and after a reaction must remain equal. In a reaction, such as the transformation of magnesium-24 with a deuteron to form an alpha particle and sodium-22, we apply this principle.

The initial mass numbers and charges are totaled. For magnesium-24 and deuteron:
  • Mass Number: 24 + 2 = 26
  • Charge: 12 (Mg) + 1 (deuteron) = 13
These must equal the final components, which include an alpha particle (mass number 4, charge 2) and the unknown product. By setting up equations, we solve for mass and atomic numbers:

  • Final Mass Number Equation: 4 + A = 26, resulting in A = 22
  • Final Charge Equation: 2 + Z = 13, resulting in Z = 11
Thus, the resulting nucleus is sodium-22.
Particle Emission
Particle emission in nuclear reactions is a fascinating process through which unstable nuclei release energy to achieve stability. This can occur through the release of particles such as alpha ( ^4_2He), beta (β), or gamma (γ) rays. Each type of emission affects the nucleus differently.

In the specific context of the exercise:
  • Alpha particles are typically emitted in reactions involving heavy nuclei, reducing both mass and atomic numbers by 4 and 2, respectively.
  • In beta decay, a neutron transforms into a proton, increasing the atomic number by 1 but leaving the mass number unchanged, due to the emission of a beta particle (an electron or positron).
  • Gamma emission often accompanies other types of decay, causing no change in mass or charge but releasing energy as the nucleus transitions to a lower energy state.
Thus, understanding the type of particle emitted in a reaction helps in predicting the resulting isotopes and stabilizing effects on the nucleus.
Isotopes
Isotopes are variations of a particular element that have the same atomic number but different mass numbers. The difference in mass arises because isotopes have differing numbers of neutrons while retaining the same number of protons.

This concept is crucial in nuclear reactions like those in the exercise. When a nuclear reaction involves a specific isotope:
  • The proton number, and therefore the chemical element itself, remains consistent.
  • Reactions altering neutron numbers transform one isotope into another.
Consider magnesium-24 in the reactions. A magnesium isotope can absorb a deuteron or release an alpha particle, thereby forming a new isotope, preserving the magnesium identity but with a different total of nucleons.

An understanding of isotopes allows us to explore and predict reaction products, knowing that different isotopes of the same element can display varied physical and chemical properties. This variation is pivotal for applications across nuclear chemistry and physics.

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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 halflives?

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 uraniumbearing 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 \({ }^{238} \mathrm{U}\) 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.

The isotope potassium-40 is a \(\beta\) -emitter. Write out the parentdaughter equation for the decay.

Using the exponential function. The curve is an exponential decay curve, and it is expressed by the equation $$ \frac{N}{N_{0}}=e^{-\lambda t} $$ where \(\lambda\) is the decay constant, and \(N / N_{0}\) is the fraction of the original \(N_{0}\) particles that remain undecayed after a time \(t\). Inasmuch as \(\lambda t_{1 / 2}=0.693, \lambda=0.693 / t_{1 / 2}=0.132 /\) year and \(N / N_{0}=\) 0.333. Thus, $$ 0.333=e^{-0.132 t / y e a r} $$ Take the natural logarithm of each side to find $$ \ln (0.333)=-0.132 t / \text { year } $$ from which \(t=8.3\) years.

A 5.6-MeV alpha particle is shot directly at a uranium atom \((Z=\) 92). About how close will it get to the center of the uranium nucleus? At such high energies the alpha particle will easily penetrate the electron cloud and the effects of the atomic electrons can be ignored. We also assume the uranium atom to be so massive that it does not move appreciably. Then the original KE of the alpha particle will be changed into electrostatic potential energy. This energy, for a charge \(q^{\prime}\) at a distance \(r\) from a point charge \(q\), $$ \text { Potential energy }=q^{\prime} V=k_{0} \frac{q q^{\prime}}{r} $$ Equating the KE of the alpha particle to this potential energy, $$ \left(5.6 \times 10^{6} \mathrm{eV}\right)\left(1.60 \times 10^{-19} \mathrm{~J} / \mathrm{eV}\right)=\left(8.99 \times 10^{9}\right) \frac{(2 e)(92 e)}{r} $$ where \(e=1.60 \times 10^{-19} \mathrm{C}\). We find from this that \(r=4.7 \times 10^{-14}\) \(\mathrm{m} .\)

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