Question

Each of the following nuclides is known to undergo radioactive decay by production of an alpha particle, \({ }_{2}^{4} \mathrm{He}\). Write a balanced nuclear equation for each process. a. \({ }_{92}^{234} \mathrm{U}\) b. \({ }^{222} \mathrm{Rn}\) c. \({ }^{162} \mathrm{Re}\)

Short answer

The balanced nuclear equations for the given nuclides undergoing alpha decay are as follows: a. \({ }_{92}^{234} \mathrm{U} \rightarrow { }_{2}^{4} \mathrm{He} + { }_{90}^{230} \mathrm{Th}\) b. \({ }_{86}^{222} \mathrm{Rn} \rightarrow { }_{2}^{4} \mathrm{He} + { }_{84}^{218} \mathrm{Po}\) c. \({ }^{162} \mathrm{Re} \rightarrow { }_{2}^{4} \mathrm{He} + { }_{73}^{158} \mathrm{Ta}\)

Step-by-step solution

Nuclide U

For the nuclide U, we have \({ }_{92}^{234} \mathrm{U}\), and an alpha particle \({ }_{2}^{4} \mathrm{He}\)

Nuclide Rn

For the nuclide Rn, we have \({ }^{222} \mathrm{Rn}\) (note that the atomic number is not given; radon is a noble gas, and its atomic number is 86), and an alpha particle \({ }_{2}^{4} \mathrm{He}\)

Nuclide Re

For the nuclide Re, we have \({ }^{162} \mathrm{Re}\) (note that the atomic number is not given; rhenium is a transition metal, and its atomic number is 75), and an alpha particle \({ }_{2}^{4} \mathrm{He}\) #Step 2: Balance each nuclear equation#

Balancing the nuclear equation for nuclide U

In order to balance the equation for nuclide U, we need to find the atomic number and mass number of the resulting decay product. The atomic number decreases by 2 (from 92 to 90) and the mass number decreases by 4 (from 234 to 230). The resulting product is \({ }_{90}^{230} \mathrm{Th}\). So the balanced equation is: \({ }_{92}^{234} \mathrm{U} \rightarrow { }_{2}^{4} \mathrm{He} + { }_{90}^{230} \mathrm{Th}\)

Balancing the nuclear equation for nuclide Rn

In order to balance the equation for nuclide Rn, we need to find the atomic number and mass number of the resulting decay product. The atomic number decreases by 2 (from 86 to 84) and the mass number decreases by 4 (from 222 to 218). The resulting product is \({ }_{84}^{218} \mathrm{Po}\). So the balanced equation is: \({ }_{86}^{222} \mathrm{Rn} \rightarrow { }_{2}^{4} \mathrm{He} + { }_{84}^{218} \mathrm{Po}\)

Balancing the nuclear equation for nuclide Re

In order to balance the equation for nuclide Re, we need to find the atomic number and mass number of the resulting decay product. The atomic number decreases by 2 (from 75 to 73) and the mass number decreases by 4 (from 162 to 158). The resulting product is \({ }_{73}^{158} \mathrm{Ta}\). So the balanced equation is: \({ }^{162} \mathrm{Re} \rightarrow { }_{2}^{4} \mathrm{He} + { }_{73}^{158} \mathrm{Ta}\)

Key concepts

Nuclide

A nuclide is a term used in nuclear physics to represent a distinct kind of atom or nucleus characterized by a specific number of protons and neutrons. It refers to a species of atom as distinguished by its atomic number (protons) and atomic mass (protons plus neutrons).
Throughout the process of radioactive decay, nuclides transform into different elements or isotopes.
This change happens due to the emission of particles such as alpha particles, which alter the composition of the nucleus. Identifying nuclides involves examining:

• Atomic number (number of protons)
• Mass number (total number of protons and neutrons)
• Symbol of the element (for example, U for Uranium)

A balanced nuclear equation must accurately reflect the initial and final nuclide states post-decay.

Alpha Particle

An alpha particle is a type of ionizing radiation ejected by some radioactive substances like Uranium and Radon during the process of decay. It is identical to a helium nucleus, comprising two protons and two neutrons, and hence it's represented by \(_{2}^{4} \mathrm{He}\).These particles have several distinctive characteristics:

• Mass and Charge: With a mass number of 4 and a charge of +2, they are relatively heavy and positively charged compared to other forms of radiation like beta and gamma rays.
• Penetration Ability: Alpha particles have low penetration ability, usually stopped by a sheet of paper or even the outer layer of human skin.
• Impact on Parent Nuclide: The emission of an alpha particle from a nuclide decreases its atomic number by 2 and its mass number by 4, leading to the formation of a new nuclide.

In the balanced nuclear equations provided, each initial nuclide produces an alpha particle as part of the decay process.

Balanced Nuclear Equation

A balanced nuclear equation represents the transformation of one nuclide into another as a result of radioactive decay. It's an essential tool in understanding and calculating changes in atomic and mass numbers during decay.To balance a nuclear equation, one must ensure:

• Conservation of Atomic Number: The total count of protons (atomic number) on both sides of the equation must be equal.
• Conservation of Mass Number: The total mass numbers, which include both protons and neutrons, must also be equal on both sides.

For instance:

• \[_{92}^{234} \mathrm{U} \rightarrow { }_{2}^{4} \mathrm{He} + { }_{90}^{230} \mathrm{Th}\]Ensures balance by decreasing the atomic number from 92 to 90 and the mass number from 234 to 230.

By writing and balancing nuclear equations, we can understand the progression and outcome of radioactive decays, as observed in various nuclides such as Uranium, Radon, and Rhenium.