Question

Complete each of the following nuclear equations by supplying the missing particle. a. \(\quad_{84}^{210} \mathrm{Po} \rightarrow_{82}^{206} \mathrm{Pb}+?\) b. \(_{57}^{140} \mathrm{La} \rightarrow ?+2_{0}^{1} \mathrm{n}\) c. \(\frac{235}{92} \mathrm{U}+? \rightarrow_{54}^{143} \mathrm{Xe}+_{38}^{90} \mathrm{Sr}+3_{0}^{1} \mathrm{n}\)

Short answer

The short answers to the nuclear equations are: a) \(_{84}^{210}\mathrm{Po} \rightarrow _{82}^{206}\mathrm{Pb} + _2^4\mathrm{He}\) b) \(_{57}^{140}\mathrm{La} \rightarrow _{57}^{138}\mathrm{Ba} + 2_{0}^{1}\mathrm{n}\) c) \(\frac{235}{92}\mathrm{U} + _0^1\mathrm{n} \rightarrow _{54}^{143}\mathrm{Xe} + _{38}^{90}\mathrm{Sr} + 3_{0}^{1}\mathrm{n}\)

Step-by-step solution

Identify the given particles

We are given the following equation: \(_{84}^{210} \mathrm{Po} \rightarrow _{82}^{206} \mathrm{Pb} + ?\) We want to find the missing particle in the equation.

Conservation of Atomic Numbers and Mass Numbers

Apply the conservation of atomic numbers and mass numbers in the equation: \(_{84}^{210}\mathrm{Po} \rightarrow _{82}^{206}\mathrm{Pb} + _{Z_3}^{A_3}? \) where Z_3 refers to the atomic number (number of protons) and A_3 refers to the mass number (total nucleons) of the missing particle. Applying the conservation rules, we get: 84 = 82 + Z_3, which implies Z_3 = 2 210 = 206 + A_3, which implies A_3 = 4

Identify the Missing Particle

The missing particle has an atomic number of 2 and a mass number of 4, which corresponds to an alpha particle, symbolized as \(_2^4\mathrm{He}\). The completed equation is: \(_{84}^{210}\mathrm{Po} \rightarrow _{82}^{206}\mathrm{Pb} + _2^4\mathrm{He}\) b)

Identify the given particles

We are given the following equation: \(_{57}^{140} \mathrm{La} \rightarrow ? + 2_{0}^{1} \mathrm{n}\) We want to find the missing particle in the equation.

Conservation of Atomic Numbers and Mass Numbers

Apply the conservation of atomic numbers and mass numbers in the equation: \(_{57}^{140}\mathrm{La} \rightarrow _{Z_3}^{A_3}? + 2_{0}^{1}\mathrm{n}\) Applying the conservation rules, we get: 57 = Z_3 + 0, which implies Z_3 = 57 140 = A_3 + 2, which implies A_3 = 138

Identify the Missing Particle

The missing particle has an atomic number of 57 and a mass number of 138, which corresponds to Barium, symbolized as \(_{57}^{138}\mathrm{Ba}\). The completed equation is: \(_{57}^{140}\mathrm{La} \rightarrow _{57}^{138}\mathrm{Ba} + 2_{0}^{1}\mathrm{n}\) c)

Identify the given particles

We are given the following equation: \(\frac{235}{92}\mathrm{U} + ? \rightarrow _{54}^{143}\mathrm{Xe} + _{38}^{90}\mathrm{Sr} + 3_{0}^{1}\mathrm{n} \) We want to find the missing particle in the equation.

Conservation of Atomic Numbers and Mass Numbers

Apply the conservation of atomic numbers and mass numbers in the equation: \(\frac{235}{92} \mathrm{U} +_‌{Z_3}^{A_3}? \rightarrow_{54}^{143}\mathrm{Xe}+_{38}^{90} \mathrm{Sr}+3_{0}^{1}\mathrm{n}\) Applying the conservation rules, we get: 92 + Z_3 = 54 + 38 + 0, which implies Z_3 = 0 235 + A_3 = 143 + 90 + 3, which implies A_3 = 1

Identify the Missing Particle

The missing particle has an atomic number of 0 and a mass number of 1, which corresponds to a neutron, symbolized as \(_0^1\mathrm{n}\). The completed equation is: \(\frac{235}{92}\mathrm{U} + _0^1\mathrm{n} \rightarrow _{54}^{143}\mathrm{Xe} + _{38}^{90}\mathrm{Sr} + 3_{0}^{1}\mathrm{n}\)

Key concepts

Nuclear Chemistry

Nuclear chemistry is a fascinating field that explores the reactions and processes involving the nuclei of atoms. It involves understanding how elements change from one form to another at the nuclear level, which is very distinct from chemical reactions that involve the electrons surrounding the nucleus.

One of the key types of reactions studied in nuclear chemistry is radioactive decay, where an unstable nucleus emits radiation to become more stable. This can include alpha decay, beta decay, and gamma emission, among others. Investigating these kinds of reactions not only provides insights into the forces and particles within an atom but also has practical applications, such as in nuclear energy production, medical treatments, and tracing chemical processes in the environment using isotopes.

Alpha Particle

An alpha particle is a type of ionizing radiation ejected from the nuclei of certain radioactive isotopes during alpha decay, one of the common modes of radioactive decay. An alpha particle is equivalent to a helium-4 nucleus, which means it contains 2 protons and 2 neutrons. It is denoted symbolically as \( _2^4\mathrm{He} \).

Alpha particles are relatively large and carry a +2 charge. Because of their size and charge, alpha particles have a limited ability to penetrate materials and can be stopped by a sheet of paper or even the outer layers of human skin. However, if alpha-emitting substances are ingested or inhaled, they can cause significant biological damage due to their high ionization power.

Conservation of Mass and Atomic Numbers

The concept of conservation of mass and atomic numbers is pivotal in balancing nuclear equations. It's based on the law of conservation of mass, which states that mass cannot be created or destroyed, and the principle of conservation of charge, which implies the number of protons (atomic number) must remain constant during a reaction.

In the context of the exercise, this means that the sum of the mass numbers (protons plus neutrons) and the sum of the atomic numbers (protons only) must be the same on both sides of a nuclear equation. This rule allows us to determine the identity of unknown products or reactants in nuclear reactions. For instance, if a nucleus emits an alpha particle, the mass number will decrease by 4 and the atomic number will decrease by 2, leading to the formation of a new element that is lighter and has a lower atomic number.