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Write balanced equations for (a) \({ }_{92}^{238} \mathrm{U}(\alpha, \mathrm{n})_{{ }_{94}^{24}}^{24} \mathrm{Pu}\), (b) \({ }_{7}^{14} \mathrm{~N}(\alpha, \mathrm{p})_{8}^{17} \mathrm{O},(\mathrm{c}){ }_{26}^{56} \mathrm{Fe}\left(\alpha, \beta^{-}\right)_{29}^{60} \mathrm{Cu}\).

Short Answer

Expert verified
The balanced equations for the given nuclear reactions are: (a) \({ }_{92}^{238} \mathrm{U} + { }_{2}^{4}\alpha \rightarrow { }_{94}^{24} \mathrm{Pu} + { }_{0}^{1}\mathrm{n}\) (b) \({ }_{7}^{14}\mathrm{~N} + { }_{2}^{4}\alpha \rightarrow { }_{8}^{17}\mathrm{O} + { }_{1}^{1}\mathrm{p}\) (c) \({ }_{26}^{56}\mathrm{Fe} + { }_{2}^{4}\alpha \rightarrow { }_{29}^{60}\mathrm{Cu} + \beta^{-}\)

Step by step solution

01

(a) Balanced Equation for U-alpha-n-Pu reaction

For the given reaction: \({ }_{92}^{238} \mathrm{U}(\alpha, \mathrm{n}){ }_{94}^{24} \mathrm{Pu}\), let's balance the equation by adding the alpha particle and neutron. \({ }_{92}^{238} \mathrm{U} + { }_{2}^{4}\alpha \rightarrow { }_{94}^{24} \mathrm{Pu} + { }_{0}^{1}\mathrm{n}\) Now, let's verify the conservation of protons and neutrons. On both sides of the equation, protons and neutrons should be equal: Left side: Protons (Z) = 92 + 2 = 94 Neutrons (N) = (238 - 92) + (4 - 2) = 146 + 2 = 148 Right side: Protons (Z) = 94 Neutrons (N) = 24 - 94 = 150 - 1 = 148 Since the protons and neutrons are equal on both sides, the equation is balanced.
02

(b) Balanced Equation for N-alpha-p-O reaction

For the given reaction: \({ }_{7}^{14}\mathrm{~N}(\alpha, \mathrm{p})_{8}^{17}\mathrm{O}\), let's balance the equation by adding the alpha particle and proton. \({ }_{7}^{14}\mathrm{~N} + { }_{2}^{4}\alpha \rightarrow { }_{8}^{17}\mathrm{O} + { }_{1}^{1}\mathrm{p}\) Now, let's verify the conservation of protons and neutrons: Left side: Protons (Z) = 7 + 2 = 9 Neutrons (N) = (14 - 7) + (4 - 2) = 7 + 2 = 9 Right side: Protons (Z) = 8 + 1 = 9 Neutrons (N) = (17 - 8) = 9 Since the protons and neutrons are equal on both sides, the equation is balanced.
03

(c) Balanced Equation for Fe-alpha-beta-Cu reaction

For the given reaction: \({ }_{26}^{56} \mathrm{Fe}\left(\alpha,\beta^{-}\right)_{29}^{60} \mathrm{Cu}\), let's balance the equation by adding the alpha particle and beta particle. \({ }_{26}^{56}\mathrm{Fe} + { }_{2}^{4}\alpha \rightarrow { }_{29}^{60}\mathrm{Cu} + \beta^{-}\) Now, let's verify the conservation of protons and neutrons: Left side: Protons (Z) = 26 + 2 = 28 Neutrons (N) = (56 - 26) + (4 - 2) = 30 + 2 = 32 Right side: Protons (Z) = 29 - 1 = 28 Neutrons (N) = (60 - 29) = 31 + 1 = 32 Since the protons and neutrons are equal on both sides, the equation is balanced.

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Key Concepts

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

Nuclear Reactions
Understanding nuclear reactions is fundamental to grasping how elements transform into different elements, which is at the heart of nuclear chemistry. A nuclear reaction involves the change in an atom's nucleus and can lead to the transmutation of one element into another. These reactions are governed by several principles, including the conservation of mass and charge, and feature various types of particles, including alpha and beta particles, protons, and neutrons.

Key types of nuclear reactions include fission, where a nucleus splits into smaller parts, and fusion, where lighter nuclei combine to form a heavier nucleus. Both reactions result in the release or absorption of energy and can be represented by balanced nuclear equations that illustrate the principle of conservation by showing that the number of protons and neutrons remains constant before and after the reaction.
Alpha Particles
Alpha particles are a type of ionizing radiation ejected from the nuclei of certain radioactive substances. An alpha particle is essentially a helium-4 nucleus, consisting of two protons and two neutrons, symbolized as \({ }_{2}^{4}\text{He}\) or simply \(\text{α}\). Owing to their composition, alpha particles are relatively massive and carry a +2 charge. In nuclear equations, the emission of an alpha particle from a nucleus results in a decrease in both atomic number and mass number of the original atom, due to the loss of the two protons and two neutrons.

Alpha particles play a crucial role in nuclear reactions such as radioactive decay, where a heavier nucleus releases an alpha particle to become a lighter element. This is significantly relevant in studying the decay series of heavy elements like uranium or thorium.
Conservation of Mass and Charge
The principle of conservation of mass and charge is vital in balancing nuclear equations. It states that during a nuclear reaction, the total mass and the total charge must be conserved – they should remain unchanged from the reactants to the products.

When balancing nuclear equations, one must ensure that the sum of atomic numbers (protons) and mass numbers (total of protons and neutrons) on the left side of the equation equals that on the right side. This principle explains why in an alpha decay process, the parent nucleus loses two protons and two neutrons – reflected in a decrease of four in mass number and two in atomic number. The conservation of charge is similarly critical; the charge of the nucleus changes based on the particles it emits or captures, and these changes must be accounted for to balance the equation.
Neutron Balance
Neutron balance refers to the conservation of neutrons before and after a nuclear reaction. Neutrons, along with protons, make up the majority of an atom's mass and are fundamental in stabilizing the nucleus. Unlike protons, neutrons do not carry an electrical charge, but they still play a significant role in nuclear reactions.

To achieve neutron balance in nuclear equations, one must count the number of neutrons in both the reactants and products and ensure they are equal. This involves calculating the difference between the mass number and the atomic number for each nuclide involved in the reaction. If neutrons are neither created nor destroyed during a reaction, such as in alpha or proton emissions, this balance is straightforward. However, in reactions involving beta decay, a neutron is transformed into a proton (or vice versa), and an electron or positron is emitted, requiring additional consideration to achieve neutron balance.

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Most popular questions from this chapter

(a) Which of the following statements about the uranium used in nuclear reactors is or are true? (i) Natural uranium has too little \({ }^{295} \mathrm{U}\) to be used as a fuel. (ii) \({ }^{24} \mathrm{U}\) cannot be used as a fucl because it forms a supereritical mass too casily. (iii) To be used as fuel, uranium must be enriched so that it is more than \(50 \%{ }^{2.35} \mathrm{U}\) in composition. (iv) The neutron-induced fission of \({ }^{235} \mathrm{U}\) releases more neutrons per nucleus than fission of \({ }^{2.85} \mathrm{U}\). (b) Which of the following statements about the plutonium shown in the chapter-opening photograph explains why it cannot be used for nuclear power plants or nuclear weapons? (i) None of the isotopes of Pu possess the characteristics needed to support nuclear fission chain reactions. (ii) The orange glow indicates that the only radioactive decay products are heat and visible light. (iii) The particular isotope of plutonium used for RTGs is incapable of sustaining a chain reaction. (iv) Plutonium can be used as a fuel, but only atter it decays to uranium.

Methyl acetate \(\left(\mathrm{CH}_{3} \mathrm{COOCH}_{3}\right)\) is formed by the reaction of acetic acid with methyl alcohol. If the methyl alcohol is labcled with oxygen-18, the oxygen-18 ends up in the methyl acetate: CC(=O)CCCCCC(=O)O (a) Do the \(\mathrm{C}-\mathrm{OH}\) bond of the acid and the \(\mathrm{O}-\mathrm{H}\) bond of the alcohol break in the reaction, or do the \(\mathrm{O}-\mathrm{H}\) bond of the acid and the \(\mathrm{C}-\mathrm{OH}\) bond of the alcohol break? (b) Imagine a similar experiment using the radioisotope \({ }^{3} \mathrm{H}\), which is called tritium and is usually denoted \(\mathrm{T}\). Would the reaction between \(\mathrm{CH}_{3} \mathrm{COOH}\) and \(\mathrm{TOCH}_{3}\) provide the same information about which bond is broken as does the above experiment with \(\mathrm{H}^{18} \mathrm{OCH}_{3}\) ?

Potassium-40 decays to argon-40 with a half-life of \(1.27 \times 10^{9} \mathrm{yr}\), What is the age of a rock in which the mass ratio of \({ }^{40} \mathrm{Ar}\) to \({ }^{40} \mathrm{~K}\) is \(4.2\) ?

Iodine-131 is a convenient radioisotope to monitor thyroid activity in humans. It is a beta emitter with a half-life of \(8.02\) days. The thyroid is the only gland in the body that uses iodine. A person undergoing a test of thyroid activity drinks a solution of Nal, in which only a small fraction of the iodide is radioactive. (a) Why is NaI a good choice for the source of iodine? (b) If a Geiger counter is placed near the person's thyroid (which is near the neck) right after the sodium iodide solution is taken, what will the data look like as a function of time? (c) A normal thyroid will take up about \(12 \%\) of the ingested iodide in a few hours. How long will it take for the radioactive iodide taken up and held by the thyroid to decay to \(0.01 \%\) of the original amount?

According to current regulations, the maximum permissible dose of strontium- 90 in the body of an adult is \(1 \mu \mathrm{Ci}\left(1 \times 10^{-6} \mathrm{Ci}\right)\). Using the relationship rate \(=k N\), calculate the number of atoms of strontium- 90 to which this dose corresponds. To what mass of strontium- 90 does this correspond? The half-life for strontium- 90 is \(28.8 \mathrm{yr}\).

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