Question

Complete each of the following nuclear equations by supplying the missing particle. a. \({ }_{88}^{226} \mathrm{Ra} \rightarrow{ }_{86}^{222} \mathrm{Rn}+?\) b. \({ }_{86}^{222} \mathrm{Rn} \rightarrow{ }_{84}^{218} \mathrm{Po}+?\) c. \({ }_{1}^{2} \mathrm{H}+{ }_{1}^{3} \mathrm{H} \rightarrow{ }_{2}^{4} \mathrm{He}+?\)

Short answer

The missing particles for each nuclear equation are: a. \({ }_{88}^{226} \mathrm{Ra} \rightarrow{ }_{86}^{222} \mathrm{Rn} + { }_{2}^{4} \mathrm{He}\) b. \({ }_{86}^{222} \mathrm{Rn} \rightarrow{ }_{84}^{218} \mathrm{Po} + { }_{2}^{4} \mathrm{He}\) c. \({ }_{1}^{2} \mathrm{H}+{ }_{1}^{3} \mathrm{H} \rightarrow{ }_{2}^{4} \mathrm{He} + { }_{0}^{1} \mathrm{n}\)

Step-by-step solution

a. \({ }_{88}^{226} \mathrm{Ra} \rightarrow{ }_{86}^{222} \mathrm{Rn}+?\)

We need to balance the atomic mass and atomic number on both sides of the equation. Let the missing particle's atomic mass be A, and its atomic number be Z. Then we have: Atomic mass balance: \(226 = 222 + A\) Atomic number balance: \(88 = 86 + Z\) Now, we can calculate the missing particle's atomic mass and atomic number: A = 226 - 222 = 4 Z = 88 - 86 = 2 So, the missing particle is \({ }_{2}^{4}\mathrm{He}\). The complete nuclear equation is: \({ }_{88}^{226} \mathrm{Ra} \rightarrow{ }_{86}^{222} \mathrm{Rn} + { }_{2}^{4} \mathrm{He}\)

b. \({ }_{86}^{222} \mathrm{Rn} \rightarrow{ }_{84}^{218} \mathrm{Po}+?\)

Again, we need to balance the atomic mass and atomic number on both sides of the equation. Let the missing particle's atomic mass be A, and its atomic number be Z. Then we have: Atomic mass balance: \(222 = 218 + A\) Atomic number balance: \(86 = 84 + Z\) Now, we can calculate the missing particle's atomic mass and atomic number: A = 222 - 218 = 4 Z = 86 - 84 = 2 So, the missing particle is \({ }_{2}^{4}\mathrm{He}\). The complete nuclear equation is: \({ }_{86}^{222} \mathrm{Rn} \rightarrow{ }_{84}^{218} \mathrm{Po} + { }_{2}^{4} \mathrm{He}\)

c. \({ }_{1}^{2} \mathrm{H}+{ }_{1}^{3} \mathrm{H} \rightarrow{ }_{2}^{4} \mathrm{He}+?\)

Once more, we need to balance the atomic mass and atomic number on both sides of the equation. Let the missing particle's atomic mass be A, and its atomic number be Z. Then we have: Atomic mass balance: \(2 + 3 = 4 + A\) Atomic number balance: \(1 + 1 = 2 + Z\) Now, we can calculate the missing particle's atomic mass and atomic number: A = 2 + 3 - 4 = 1 Z = 1 + 1 - 2 = 0 So, the missing particle is \({ }_{0}^{1}\mathrm{n}\), which is a neutron. The complete nuclear equation is: \({ }_{1}^{2} \mathrm{H}+{ }_{1}^{3} \mathrm{H} \rightarrow{ }_{2}^{4} \mathrm{He} + { }_{0}^{1} \mathrm{n}\)

Key concepts

Nuclear Chemistry

Nuclear chemistry is the study of the chemical and physical properties of elements as influenced by changes in the structure of the atomic nucleus. Nuclear reactions involve changes in an atom's nucleus and thus can result in the transformation of one element into another. Unlike chemical reactions, which involve the electrons surrounding the nucleus and do not change one element into another, nuclear reactions can alter the number of protons within a nucleus and are thus tied to the identity of an element.

Different types of nuclear reactions include fusion, fission, and radioactive decay. Each of these reactions is governed by certain principles of conservation, such as the conservation of mass-energy, the conservation of atomic number, and the conservation of charge.

Balancing Nuclear Reactions

Balancing a nuclear reaction is analogous to balancing a chemical equation, but with a focus on the particles in the nucleus. The key principles involved are the conservation of mass number and atomic number. The mass number represents the total number of protons and neutrons in a nucleus, while the atomic number is the number of protons.

In balancing equations, we ensure that both the mass number and the atomic number are conserved across the reaction. For instance, if a particle with a certain mass number and atomic number is lost from the nucleus during the reaction, another particle with the same totals must be gained to maintain balance. Understanding these principles allows us to identify unknown particles in nuclear reactions, as was done in the provided exercise.

Alpha Decay

Alpha decay is a type of radioactive decay in which an atomic nucleus ejects an alpha particle, thereby reducing its mass number by four and its atomic number by two. An alpha particle is essentially a helium nucleus, which means it consists of two protons and two neutrons (symbolized as \( { }_{2}^{4}\mathrm{He} \)).

This process leads to the formation of a new element that is two places to the left on the periodic table. For example, radium-226 undergoing alpha decay to produce radon-222, as described in the given exercise, is showing how an alpha particle is emitted. This reaction showcases the application of balancing nuclear reactions, one of the key aspects of nuclear chemistry.

Neutron Emission

Neutron emission is a type of radioactive decay where a neutron is ejected from an atom's nucleus. This process can occur in various unstable isotopes and is one of the less common forms of radioactive decay. The emission of a neutron from a nucleus decreases its mass number by one, but the atomic number remains unchanged because the identity of the element is not altered by the number of neutrons.

In the exercise provided, the fusion of deuterium and tritium produces a helium nucleus and a free neutron, formalized as \( { }_{0}^{1}\mathrm{n} \). Neutron emission plays a crucial role in various applications, including nuclear reactors and nuclear weapons, where it contributes to the chain reactions that sustain the fission process.