Question

Each of the following nuclides is known to undergo radioactive decay by production of a beta particle, \({ }_{-1}^{0} \mathrm{e}\). Write a balanced nuclear equation for each process. a. \({ }_{6}^{14} \mathrm{C}\) b. \({ }_{55}^{140} \mathrm{Cs}\) c. \({ }^{234} \mathrm{Th}\)

Short answer

The balanced nuclear equations for each process are: a. \({}_{6}^{14}\mathrm{C} \rightarrow {}_{7}^{14}\mathrm{N} + {}_{-1}^{0}\mathrm{e}\) b. \({}_{55}^{140}\mathrm{Cs} \rightarrow {}_{56}^{140}\mathrm{Ba} + {}_{-1}^{0}\mathrm{e}\) c. \({}_{90}^{234}\mathrm{Th} \rightarrow {}_{91}^{234}\mathrm{Pa} + {}_{-1}^{0}\mathrm{e}\)

Step-by-step solution

Identify the Parent Nuclide and the Beta Particle

In this case, the parent nuclide is \({}_{6}^{14}\mathrm{C}\). A beta particle is represented as \({ }_{-1}^{0} \mathrm{e}\).

Write the Balanced Nuclear Equation

As the atomic number of the parent nuclide increases by 1, the daughter nucleus will have an atomic number of 7 (6+1). The mass number remains unchanged (14). Thus, the daughter nucleus is \({}_{7}^{14}\mathrm{N}\). The balanced nuclear equation is: \({}_{6}^{14}\mathrm{C} \rightarrow {}_{7}^{14}\mathrm{N} + {}_{-1}^{0}\mathrm{e}\) b. Cesium-140 undergoes beta decay

Identify the Parent Nuclide and the Beta Particle

Here, the parent nuclide is \({}_{55}^{140}\mathrm{Cs}\). The beta particle is represented as \({ }_{-1}^{0} \mathrm{e}\).

Write the Balanced Nuclear Equation

As the atomic number of the parent nuclide increases by 1, the daughter nucleus will have an atomic number of 56 (55+1). The mass number remains unchanged (140). Thus, the daughter nucleus is \({}_{56}^{140}\mathrm{Ba}\). The balanced nuclear equation is: \({}_{55}^{140}\mathrm{Cs} \rightarrow {}_{56}^{140}\mathrm{Ba} + {}_{-1}^{0}\mathrm{e}\) c. Thorium-234 undergoes beta decay

Identify the Parent Nuclide and the Beta Particle

In this case, the parent nuclide is \({}^{234}\mathrm{Th}\). The notation does not include the atomic number, but we know that the atomic number of thorium is 90. So, the parent nuclide can also be written as \({}_{90}^{234}\mathrm{Th}\). The beta particle is represented as \({ }_{-1}^{0} \mathrm{e}\).

Write the Balanced Nuclear Equation

As the atomic number of the parent nuclide increases by 1, the daughter nucleus will have an atomic number of 91 (90+1). The mass number remains unchanged (234). Thus, the daughter nucleus is \({}_{91}^{234}\mathrm{Pa}\). The balanced nuclear equation is: \({}_{90}^{234}\mathrm{Th} \rightarrow {}_{91}^{234}\mathrm{Pa} + {}_{-1}^{0}\mathrm{e}\)

Key concepts

Radioactive Decay

Radioactive decay is a natural process by which an unstable atomic nucleus loses energy by emitting particles or radiation. This transformation leads to the creation of a different element or a different isotope of the same element. There are several types of radioactive decay, including alpha decay, beta decay, gamma decay, and others.

Beta decay, one of the most common forms, involves the transformation of a neutron into a proton within the nucleus, releasing a beta particle, which is an electron, in the process. The release of a beta particle is a way for the nucleus to achieve a more stable configuration by decreasing its neutron-to-proton ratio. Educationally, it's essential to visualize beta decay as a 'rebalancing act' where the nucleus endeavours to find stability by altering its very core constituents.

Balanced Nuclear Equation

A balanced nuclear equation details the process of radioactive decay and is vital for understanding which particles are involved. In order to be balanced, both sides of the equation must have the same total atomic mass and the same total atomic number. These conservation laws ensure that mass and charge are preserved during the decay process.

For instance, during beta decay, the mass number stays the same because the transformation of a neutron to a proton does not change the total number of nucleons (protons and neutrons combined). However, the atomic number increases by 1 because the new proton contributes to the element's atomic number. The emitted beta particle—an electron—has an atomic number of -1 as it carries a negative charge, but it has no impact on the mass number because its mass is negligible compared to that of nucleons. The inclusion of the beta particle on the product side of the equation completes the balance of both mass and atomic number.

Beta Particle Production

Beta particle production is a signature event in beta decay, resulting in the emission of an electron from the nucleus. It's a curious phenomenon because the nucleus of an atom typically contains protons and neutrons, not electrons. It is mediated by the weak nuclear force, one of the four fundamental forces in physics.

In an intuitive sense, a neutron 'transforms' into a proton and an emitted electron, along with a neutrino, which is another subatomic particle that is not usually represented in the nuclear equation. This process not only changes the identity of the element (by changing the atomic number) but also releases energy that was previously part of the binding energy holding the nucleus together. Understanding beta particle production reveals the interconnected nature of matter and energy at the nuclear level and the incredible forces that operate within the atom to shape the material universe.