Write a nuclear equation for the beta decay of each nuclide. (a) Pb-211 (b) Tl-207 (c) Th-234 (d) Pa-234

Short Answer

Expert verified
Beta decay equations are: (a) \( _{82}^{211}Pb \rightarrow _{83}^{211}Bi + e^{-} + \text{antineutrino} \), (b) \( _{81}^{207}Tl \rightarrow _{82}^{207}Pb + e^{-} + \text{antineutrino} \), (c) \( _{90}^{234}Th \rightarrow _{91}^{234}Pa + e^{-} + \text{antineutrino} \), (d) \( _{91}^{234}Pa \rightarrow _{92}^{234}U + e^{-} + \text{antineutrino} \).

Step by step solution

01

Understanding Beta Decay

Beta decay is a nuclear process where a beta particle (which is an electron or a positron) is emitted. For beta-minus decay, a neutron is converted into a proton with the emission of an electron (beta particle) and an antineutrino. In beta-plus decay, a proton is converted into a neutron with the emission of a positron (beta particle) and a neutrino. The atomic number changes by one, but the mass number remains the same.
02

Writing the Equation for Pb-211

Lead-211 (Pb-211) undergoes beta-minus decay. This means a neutron in its nucleus is converted into a proton, and a beta particle, which is an electron, is emitted along with an antineutrino. The atomic number increases by one.Pb-211 -> Bi-211 + e- + antineutrino\[ \text{{_^{211}_{82}Pb }} \rightarrow \text{{_^{211}_{83}Bi}} + \text{e}^{-} + \text{antineutrino} \]
03

Writing the Equation for Tl-207

Thallium-207 (Tl-207) undergoes beta-minus decay. This results in the emission of an electron and an antineutrino, increasing the atomic number by one.tl-207 -> Pb-207 + e- + antineutrino\[ \text{{_^{207}_{81}Tl }} \rightarrow \text{{_^{207}_{82}Pb}} + \text{e}^{-} + \text{antineutrino} \]
04

Writing the Equation for Th-234

Thorium-234 (Th-234) undergoes beta-minus decay, emitting an electron and an antineutrino, which increases the atomic number by one while the mass number remains unchanged.tH-234 -> Pa-234 + e- + antineutrino\[ \text{{_^{234}_{90}Th }} \rightarrow \text{{_^{234}_{91}Pa}} + \text{e}^{-} + \text{antineutrino} \]
05

Writing the Equation for Pa-234

Protactinium-234 (Pa-234) undergoes beta-minus decay. During the decay, it emits an electron and an antineutrino, causing the atomic number to increase by one.Pa-234 -> U-234 + e- + antineutrino\[ \text{{_^{234}_{91}Pa }} \rightarrow \text{{_^{234}_{92}U}} + \text{e}^{-} + \text{antineutrino} \]

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Key Concepts

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

Nuclear Chemistry
Nuclear chemistry is the branch of chemistry that deals with changes in the nucleus of atoms. The focus is on reactions that involve changes in atomic number, mass number, or energy states within the nucleus. These nuclear reactions are not to be confused with chemical reactions, which involve changes in the electron configuration surrounding an atom's nucleus but do not affect the nucleus itself.

In nuclear chemistry, isotopes play a crucial role as they can often be unstable and subject to radioactive decay. The decay processes transform one element into another and are accompanied by the emission of various types of radiation. These include alpha particles, beta particles, and gamma rays. By understanding nuclear equations and the types of emissions, we can predict the behavior of radioactive substances and harness nuclear reactions for beneficial uses such as medical treatments, energy production, and dating ancient materials.
Radioactive Decay
Radioactive decay is a spontaneous process by which an unstable atomic nucleus loses energy by emitting radiation. There are several different types of radioactive decay, including alpha decay, beta decay, and gamma decay. Each type involves different particles and changes in the nucleus.

During radioactive decay, the unstable nucleus spontaneously transforms into a more stable configuration. This process changes the number of protons and/or neutrons in the nucleus, leading to the formation of a different element or a different isotope of the same element. Radioactive decay is a natural process that occurs in all isotopes that have an unstable nucleus, and it cannot be altered or sped up by chemical means. The rate of decay is characterized by the half-life, which is the time it takes for half of a sample of the radioactive substance to decay.
Beta Particle Emission
Beta particle emission is one type of radioactive decay. In beta-minus decay, a neutron within the nucleus is transformed into a proton, and a beta particle—which is essentially an electron—is released. Alternately, in beta-plus decay, a proton is transformed into a neutron, and a positron (the antiparticle of an electron) is emitted.

When a beta particle is emitted, the atomic number of the atom changes by one unit, reflecting the change in the number of protons. However, the mass number remains constant because a proton and a neutron have approximately the same mass. Therefore, the nucleus maintains its total nucleon count (protons plus neutrons).

By writing down the nuclear equation for beta decay, such as those seen in the steps for nuclides like Pb-211 and Tl-207, we can keep track of these changes. The equations confirm the conservation of charge and nucleon number, which are fundamental principles in nuclear reactions.
Neutrinos in Beta Decay
In the process of beta decay, alongside the beta particle, another very small and elusive particle known as a neutrino (in beta-plus decay) or its antiparticle, the antineutrino (in beta-minus decay), is emitted. Neutrinos are neutral particles with very little mass and are incredibly difficult to detect because they interact with matter only very weakly.

For every beta particle emitted, a neutrino or antineutrino is also released to conserve lepton number, an important conservation law in particle physics. These particles carry away energy and momentum, ensuring that energy is conserved in the decay process. They play a crucial role in several phenomena in the universe, including the nuclear reactions that occur in the sun and other stars. In nuclear equations, neutrinos are often represented by the Greek letter nu (ν), and their presence is critical for the balance of the equation.

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