Radium- 226 decays by emitting an alpha particle. What is the daughter nucleus? a) \(\mathrm{Rd}\) b) \(\mathrm{Rn}\) c) Bi d) \(\mathrm{Pb}\)

Short Answer

Expert verified
Answer: b) Rn

Step by step solution

01

Identify the isotope undergoing alpha decay

Radium-226 is the isotope undergoing alpha decay. Radium's symbol is Ra, and it has an atomic number of 88 and a mass number of 226.
02

Identify the alpha particle

An alpha particle is equivalent to a helium-4 nucleus (\(\mathrm{He}^{2+}\)) with an atomic number of 2 and a mass number of 4.
03

Write the balanced nuclear equation

In a balanced nuclear equation, the total number of protons and neutrons (mass numbers) on the reactants side must equal the total number of protons and neutrons on the products side. \(\mathrm{^{226}_{88}Ra} \longrightarrow\mathrm{^{4}_{2}He} + \mathrm{^{A}_{Z}X}\) X is the daughter nucleus; we should solve for A and Z.
04

Solve for A and Z

We'll solve for A (mass number) and Z (atomic number) in the daughter nucleus X: Conservation of mass numbers: A + 4 = 226 A = 226 - 4 A = 222 Conservation of atomic numbers: Z + 2 = 88 Z = 88 - 2 Z = 86
05

Identify the daughter nucleus

The daughter nucleus X has an atomic number (Z) of 86 and mass number (A) of 222. Looking up element 86 on the periodic table, we find that it is Radon (Rn). Therefore, the daughter nucleus is \(\mathrm{^{222}_{86}Rn}\). The correct answer is: b) \(\mathrm{Rn}\)

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Key Concepts

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

Daughter Nucleus
In the context of nuclear reactions, particularly alpha decay, a 'daughter nucleus' is the remaining nucleus after an original ('parent') nucleus undergoes a radioactive decay process. During alpha decay, an alpha particle, which is essentially a helium nucleus consisting of two protons and two neutrons, is emitted from the parent nucleus. As a result, the daughter nucleus that is left behind will have a different atomic number and mass number compared to the original nucleus.

For example, when Radium-226 decays by emitting an alpha particle, it transforms into a different element altogether. The change in atomic and mass numbers is due to the loss of the two protons and two neutrons carried away by the alpha particle. Hence, the daughter nucleus in this case has an atomic number decremented by 2 and a mass number decremented by 4. The alpha decay of Radium-226 results in the formation of Radon-222 as the daughter nucleus.
Balanced Nuclear Equation
A balanced nuclear equation is pivotal in nuclear chemistry, as it ensures the law of conservation of mass and energy is upheld within nuclear reactions. In the equation, particles and atomic nuclei are balanced on both sides of the reaction, similar to how atoms are balanced in chemical equations.

To properly balance a nuclear equation for alpha decay, one must account for both the atomic number (representing the number of protons) and the mass number (the total number of protons and neutrons). When Radium-226 decays, it emits an alpha particle (\textstyle{\text{He}}^{2+}) and becomes Radon-222. The balanced equation for this reaction is: \[ \textstyle{^{226}_{88}Ra} \rightarrow \textstyle{^{4}_{2}He} + \textstyle{^{222}_{86}Rn} \]The left side of the arrow shows the parent nucleus and the alpha particle, while the right side displays the daughter nucleus that is being sought.
Conservation of Mass Numbers
The principle of conservation of mass numbers is one of the fundamental principles governing nuclear reactions. It states that the total mass number (the sum of protons and neutrons) in the nuclear system must remain constant before and after the reaction. This means that the total of protons plus neutrons is the same on both the reactant and product sides of the equation.

In simpler terms, if you subtract the mass number of the emitted alpha particle from the original nucleus, you'll obtain the mass number of the daughter nucleus. For instance, in the decay of Radium-226: \[ 226 - 4 = 222 \]Accordingly, after releasing an alpha particle with a mass number of 4, the remaining daughter nucleus will have a mass number of 222. This demonstrates the conservation of mass numbers in the process of alpha decay.
Periodic Table
The periodic table is an essential tool in chemistry and nuclear physics, as it catalogs all known elements based on their atomic number, which is the number of protons found in the nucleus of an atom. Elements in the periodic table are arranged in increasing order of atomic number from left to right and top to bottom. Each element is represented by a specific symbol and positioned in such a manner that it aligns with others in the same group or period possessing similar chemical properties.

To identify a daughter nucleus in an alpha decay reaction, referencing the periodic table is crucial. By determining the atomic number of the daughter nucleus, which is the original atomic number minus two (since two protons are found in the emitted alpha particle), we can find the corresponding element. For Alpha decay involving Radium-226, the atomic number decreases from 88 to 86, leading us to Radon (Rn) in the periodic table, which is the daughter nucleus formed from this specific decay process.

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Most popular questions from this chapter

The binding energy of \({ }_{2}^{3}\) He is lower than that of \({ }_{1}^{3} \mathrm{H} .\) Provide a plausible explanation, considering the Coulomb interaction between two protons in \({ }_{2}^{3}\) He.

A certain radioactive isotope decays to one-eighth its original amount in \(5.00 \mathrm{~h} .\) How long would it take for \(10.0 \%\) of it to decay?

\(^{214} \mathrm{Pb}\) has a half-life of \(26.8 \mathrm{~min}\). How many minutes must elapse for \(90.0 \%\) of a given sample of \({ }^{214} \mathrm{~Pb}\) atoms to decay?

The most common isotope of uranium, \({ }_{92}^{238} \mathrm{U},\) produces radon \({ }_{86}^{222} \mathrm{Rn}\) through the following sequence of decays: $$\begin{array}{c}{ }^{238} \mathrm{U} \rightarrow{ }^{234} \mathrm{Th}+\alpha,{ }^{234} \mathrm{Th} \rightarrow{ }^{234} \mathrm{~Pa}+\beta^{-}+\bar{\nu}_{e}, \\\\{ }_{91}^{234} \mathrm{~Pa} \rightarrow{ }_{92}^{234} \mathrm{U}+\beta+\bar{\nu}_{e},{ }^{234} \mathrm{U} \rightarrow{ }^{230} \mathrm{Th}+\alpha ,\\\\{ }_{91}^{230} \mathrm{Th} \rightarrow{ }_{90}^{226} \mathrm{Ra}+\alpha,{ }_{88}^{226} \mathrm{Ra} \rightarrow{ }_{86}^{222} \mathrm{Rn}+\alpha,\end{array}$$. A sample of \({ }_{92}^{238} \mathrm{U}\) will build up equilibrium concentrations of its daughter nuclei down to \({ }_{88}^{226} \mathrm{Ra} ;\) the concentrations of each are such that each daughter is produced as fast as it decays. The \({ }_{88}^{226} \mathrm{Ra}\) decays to \({ }_{86}^{222} \mathrm{Rn},\) which escapes as a gas. (The \(\alpha\) particles also escape, as helium; this is a source of much of the helium found on Earth.) In high concentrations, the radon is a health hazard in buildings built on soil or foundations containing uranium ores, as it can be inhaled. a) Look up the necessary data, and calculate the rate at which \(1.00 \mathrm{~kg}\) of an equilibrium mixture of \({ }_{92}^{238} \mathrm{U}\) and its first five daughters produces \({ }_{86}^{222} \mathrm{Rn}\) (mass per unit time). b) What activity (in curies per unit time) of radon does this represent?

What is the nuclear configuration of the daughter nucleus associated with the alpha decay of \(\mathrm{Hf}(A=157,\) \(Z=72\) )?

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