Question

The thorium isotope \({ }_{90}^{232}\) Th is radioactive and decays by ejecting two protons and two neutrons from its nucleus. (a) Where is this atom likely to lie in the band of stability? (b) Write a nuclear reaction for this decay process. (c) Which type of decay is this?

Short answer

The thorium isotope \({}_{90}^{232}\mathrm{Th}\) lies below the band of stability due to its neutron-rich nature with a \(Z/N\) ratio of approximately 0.634. The nuclear reaction for this decay process is \(\ce{_{90}^{232}Th ->[\alpha] _{88}^{230}Ra + _{2}^{4}He}\), where it undergoes alpha decay, emitting an alpha particle to become more stable.

Step-by-step solution

(a) Determine where this atom is likely to lie in the band of stability.

Since the thorium isotope is radioactive, it is unstable and therefore lies outside the band of stability. The band of stability is determined by the ratio of protons to neutrons (\(Z/N\)) of an atom. For heavy nuclei like thorium, the stable nuclei tend to have more neutrons than protons, so the ratio \(Z/N\) is typically greater than 1. In the case of \({}_{90}^{232}\mathrm{Th}\), the ratio \(Z/N\) is given by: \[\frac{Z}{N} = \frac{90}{(232-90)} = \frac{90}{142} \approx 0.634\] Since this ratio is less than 1, the thorium isotope is likely to be below the band of stability. It means that this atom is neutron-rich and needs to reduce the number of neutrons to become more stable.

(b) Write a nuclear reaction for this decay process.

In this decay process, the thorium isotope ejects two protons and two neutrons from its nucleus. These four particles together form an alpha particle, which is a helium nucleus with 2 protons and 2 neutrons represented by \({}^4\mathrm{He }\) or \(\alpha\). The nuclear reaction for this decay process is: \[\ce{_{90}^{232}Th ->[\alpha] _{88}^{230}Ra + _{2}^{4}He}\] The product of this reaction is radium, which has an atomic number 2 less than thorium, and an atomic mass 4 less than thorium.

(c) Identify the type of decay.

Since this decay process involves the ejection of an alpha particle, which is a helium nucleus containing 2 protons and 2 neutrons, this is an example of alpha decay. Alpha decay is a type of radioactive decay in which an unstable atom ejects an alpha particle to become more stable.

Key concepts

Radioactive Decay

Radioactive decay is a fundamental nuclear process by which an unstable atomic nucleus loses energy by emitting radiation. This spontaneous transformation can result in the emission of particles such as alpha particles, beta particles, or gamma rays. For instance, heavy elements like uranium or thorium undergo decay to reach a more stable state. It's like a rock rolling down a hill as the nucleus releases energy to attain stability, much the way objects move to lower energy levels when possible. Each type of decay leads to the formation of a different element and changes the number of protons or the overall mass of the nucleus.

In our exercise example, thorium-232 emits an alpha particle. This process not only changes the identity of the element (from thorium to radium) but also releases a significant amount of energy. Therefore, understanding radioactive decay is crucial for fields ranging from nuclear energy production to medical imaging and treatments.

Band of Stability

The band of stability is a concept used to describe the balance between protons and neutrons in a stable nucleus. Think of it like a 'safe zone' on a chart where stable isotopes live. Nuclei that have the right neutron-to-proton ratio ({N/Z}) tend to be stable and resist radioactive decay. As you move higher up or lower down in atomic number, this ratio changes, and stability becomes harder to maintain.

In the exercise, thorium-232 lies outside this band, indicating it's not stable and prone to decay. Typically, lighter elements require a 1:1 ratio for stability, but heavier elements like thorium need a greater number of neutrons to counterbalance the repulsive forces between protons. That's why thorium, with a {N/Z} ratio less than 1, requires a transformation to get into the 'safe zone' of the band of stability.

Nuclear Reaction

A nuclear reaction occurs when a nucleus, with a distinctive number of protons and neutrons, undergoes a change due to natural processes or induced events. This can result in a change in the element's identity, as atomic numbers (the number of protons) and atomic masses alter. In our thorium decay example, the nuclear reaction succinctly captures its transformation to radium with a byproduct of an alpha particle.

The balanced equation provided in the nuclear reaction is crucial for understanding the conservation of mass and charge. Just like balancing a checkbook, nuclear reactions keep the 'account' of protons and neutrons in check. This allows scientists and students alike to predict the outcome of these powerful but tiny changes occurring within the nucleus.

Isotopes

Isotopes are like different flavors of the same element—each has the same number of protons but varies in neutron count. Picture lemons with different numbers of seeds, all still lemons but slightly differing on the inside. Isotopes can be stable or unstable (radioactive), where stable isotopes don't change over time, and radioactive isotopes decay, altering their structure. Our example, thorium-232, is a radioactive isotope. Its instability provokes the need to shed particles, leading to decay.

Understanding isotopes is crucial since their properties affect everything from the dating of archaeological artifacts (like with carbon-14) to the power generation in nuclear reactors and even cancer treatments. They truly show the diversity within the elements of the periodic table.

Neutron-to-Proton Ratio

The neutron-to-proton ratio ({N/Z}) is the critical number that helps us understand whether a nucleus is likely to be stable or not. This ratio is akin to a recipe balance—just as the right ingredients in precise amounts create the perfect cake, the right {N/Z} ratio results in a stable nucleus. Elements up to iron tend to prefer a 1:1 ratio, but for heavier elements, such as thorium, the ratio must be higher to maintain nuclear stability due to the increased repulsion between the greater number of protons.

The exercise showcases how thorium-232 falls short of this stability marker with an {N/Z} ratio of approximately 0.634, prompting it to undergo alpha decay to approach a more favorable ratio. It's vital for students to grasp that the further a nucleus is from the ideal ratio, the more unstable it is and the more likely it will undergo decay to seek equilibrium.