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. \({ }_{90}^{232} \mathrm{Th}\) b. \({ }^{220} \mathrm{Rn}\) c. \({ }^{216} \mathrm{Po}\)

Short answer

The balanced nuclear equations for each alpha decay process are: a. \({ }_{90}^{232} \mathrm{Th} \rightarrow { }_{2}^{4} \mathrm{He} + { }_{88}^{228} \mathrm{Ra}\) b. \({ }^{220} \mathrm{Rn} \rightarrow { }_{2}^{4} \mathrm{He} + { }_{84}^{216} \mathrm{Po}\) c. \({ }^{216} \mathrm{Po} \rightarrow { }_{2}^{4} \mathrm{He} + { }_{82}^{212} \mathrm{Pb}\)

Step-by-step solution

a. Balancing the equation for \({ }_{90}^{232} \mathrm{Th}\) alpha decay

To balance the equation for the alpha decay of \({ }_{90}^{232} \mathrm{Th}\), we need to conserve both mass numbers and atomic numbers. We start with the parent nucleus and the emitted alpha particle: \({ }_{90}^{232} \mathrm{Th} \rightarrow { }_{2}^{4} \mathrm{He} + X\) Now, we look for the daughter nucleus (X) with the correct mass and atomic numbers. The mass number of the daughter nucleus would be the initial mass number minus the alpha particle's mass number (232 - 4 = 228). The atomic number of the daughter nucleus would be the initial atomic number minus the alpha particle's atomic number (90 - 2 = 88). Thus, we have: \({ }_{90}^{232} \mathrm{Th} \rightarrow { }_{2}^{4} \mathrm{He} + { }_{88}^{228} \mathrm{Ra}\)

b. Balancing the equation for \({ }^{220} \mathrm{Rn}\) alpha decay

To balance the equation for the alpha decay of \({ }^{220} \mathrm{Rn}\), we follow the same steps as before and start with the parent nucleus and the emitted alpha particle: \({ }^{220} \mathrm{Rn} \rightarrow { }_{2}^{4} \mathrm{He} + X\) The mass number of the daughter nucleus would be (220 - 4) = 216, and the atomic number would be (86 - 2) = 84. So, we have: \({ }^{220} \mathrm{Rn} \rightarrow { }_{2}^{4} \mathrm{He} + { }_{84}^{216} \mathrm{Po}\)

c. Balancing the equation for \({ }^{216} \mathrm{Po}\) alpha decay

To balance the equation for the alpha decay of \({ }^{216} \mathrm{Po}\), we use the same procedure as in the previous examples, taking into account that the atomic number of Polonium is 84. \({ }^{216} \mathrm{Po} \rightarrow { }_{2}^{4} \mathrm{He} + X\) The mass number of the daughter nucleus would be (216 - 4) = 212, and the atomic number would be (84 - 2) = 82. So, we have: \({ }^{216} \mathrm{Po} \rightarrow { }_{2}^{4} \mathrm{He} + { }_{82}^{212} \mathrm{Pb}\)

Key concepts

Radioactive Decay

Radioactive decay is a natural process where unstable atomic nuclei lose energy by emitting radiation. This transformation results in the parent element changing into a different daughter element. For instance, alpha decay is one type of radioactive decay where the nucleus ejects an alpha particle, which consists of two protons and two neutrons. This particle is essentially a helium-4 nucleus ((_2^4)He), and its ejection leads to a decrease in both the mass number and atomic number of the original element.

Alpha decay typically occurs in heavy elements where the strong nuclear force cannot hold the nucleus together due to the large number of protons causing excessive repulsion. The result is a more stable nucleus, albeit of a different element, and the release of part of the parent nucleus as an alpha particle.

Key Aspects of Alpha Decay

• Alpha particles have a charge of +2 and a mass number of 4.
• During alpha decay, the mass number of the parent nucleus decreases by 4, and the atomic number decreases by 2.
• Alpha decay leads to the formation of a new element that is two places lower in the periodic table.

Nuclear Equation Balancing

Balancing nuclear equations is necessary to represent the conservation of mass and atomic numbers during radioactive decay processes. In the case of alpha decay, we ensure that the total mass and atomic numbers before and after the decay are equal, thus obeying the law of conservation of matter and charge.

When balancing a nuclear equation for alpha decay, subtract four from the mass number and two from the atomic number of the parent nucleus to find the daughter nucleus. The alpha particle's mass and atomic numbers are always 4 and 2, respectively. The steps for balancing are simple but crucial for understanding the changes that occur within an atom's nucleus during decay.

Process of Balancing

• Start with the parent nucleus on one side of the equation.
• Add an alpha particle ((_2^4)He) on the product side.
• Find the daughter nucleus by applying the subtraction of mass and atomic numbers from the parent.
• Ensure the sum of the mass and atomic numbers on both sides of the equation are equal.

Nuclear Chemistry

Nuclear chemistry deals with the reactions and processes that occur within the nuclei of atoms. This includes studying radioactive decay, nuclear fission, and nuclear fusion. One of the fundamental principles of nuclear chemistry is the stability of atomic nuclei, which depends on the balance between nuclear forces and electrostatic repulsion among protons.

An unstable nucleus can undergo changes, such as radioactive decay, to reach a more stable configuration. Alpha decay is one example of such a process, often observed in heavier elements, and is an essential concept in the field of nuclear chemistry. Understanding why and how these nuclear reactions occur not only explains natural radioactive phenomena but also has practical applications in medicine, energy production, and other fields.

Applications of Nuclear Chemistry

• Medical imaging and cancer treatment through radiopharmaceuticals.
• Energy generation in nuclear power plants via controlled nuclear reactions.
• Study of the age and composition of materials using radioactive isotopes.