Question

One nuclide in each of these pairs is radioactive. Predict which is radioactive and which is stable: (a) \({ }_{20}^{40} \mathrm{Ca}\) and \({ }_{20}^{45} \mathrm{Ca}\), (b) \({ }^{12} \mathrm{C}\) and \({ }^{14} \mathrm{C}\), (c) lead-206 and thorium-230. Fxplain your choice in each case.

Short answer

(a) \({ }_{20}^{40} \mathrm{Ca}\) is stable and \({ }_{20}^{45} \mathrm{Ca}\) is radioactive due to the odd number of neutrons in \({ }^{45} \mathrm{Ca}\). (b) \({ }^{12} \mathrm{C}\) is stable and \({ }^{14} \mathrm{C}\) is radioactive because the additional neutrons in \({ }^{14} \mathrm{C}\) make it more unstable. (c) lead-206 is stable and thorium-230 is radioactive as elements with atomic numbers greater than 82 typically undergo radioactive decay.

Step-by-step solution

Determine the number of protons and neutrons in each nuclide

First, let's determine the number of protons and neutrons in each nuclide. This will help us in predicting if the nuclide is stable or radioactive. (a) \({ }_{20}^{40} \mathrm{Ca}\) and \({ }_{20}^{45} \mathrm{Ca}\) Both have 20 protons. Number of neutrons: \({ }^{40} \mathrm{Ca}\) has 20 neutrons and \({ }^{45} \mathrm{Ca}\) has 25 neutrons. (b) \({ }^{12} \mathrm{C}\) and \({ }^{14} \mathrm{C}\) Carbon has 6 protons. Number of neutrons: \({ }^{12} \mathrm{C}\) has 6 neutrons and \({ }^{14} \mathrm{C}\) has 8 neutrons. (c) lead-206 and thorium-230 For lead-206, it has 82 protons and 124 neutrons. For thorium-230, it has 90 protons and 140 neutrons.

Evaluate stability using the number of protons and neutrons

Next, let's use the information about protons and neutrons to predict which nuclide in each pair is likely to be more stable. (a) \({ }_{20}^{40} \mathrm{Ca}\) and \({ }_{20}^{45} \mathrm{Ca}\) \({ }^{40} \mathrm{Ca}\) has even numbers of both protons and neutrons (20 of each), which indicates greater stability. On the other hand, \({ }^{45} \mathrm{Ca}\) has an odd number of neutrons (25), making it more likely to be unstable and therefore radioactive. (b) \({ }^{12} \mathrm{C}\) and \({ }^{14} \mathrm{C}\) Both nuclides have an even number of protons (6). However, \({ }^{12} \mathrm{C}\) has an even number of neutrons (6), while \({ }^{14} \mathrm{C}\) has an even number of neutrons (8), which makes it more likely to be unstable and therefore radioactive. (c) lead-206 and thorium-230 Lead-206 has an even number of both protons (82) and neutrons (124), indicating greater stability. Thorium-230 has an even number of protons (90) but an even number of neutrons (140) as well. However, it is important to note that elements with atomic numbers greater than 82 are generally unstable and undergo radioactive decay. Thus, among these two nuclides, thorium-230 is more likely to be radioactive.

Final predictions and explanations

Based on the above analysis, we can make the following predictions: (a) \({ }_{20}^{40} \mathrm{Ca}\) is stable and \({ }_{20}^{45} \mathrm{Ca}\) is radioactive. The odd number of neutrons in \({ }^{45} \mathrm{Ca}\) makes it more unstable. (b) \({ }^{12} \mathrm{C}\) is stable and \({ }^{14} \mathrm{C}\) is radioactive. The additional neutrons in \({ }^{14} \mathrm{C}\) could make it more unstable. (c) lead-206 is stable and thorium-230 is radioactive. Although both have even numbers of protons and neutrons, thorium-230 is more likely to be radioactive due to the general rule that elements with atomic numbers greater than 82 undergo radioactive decay.

Key concepts

Nuclide Stability

Understanding why certain nuclides are stable while others are not is key in the field of nuclear chemistry. The stability of a nuclide is influenced by the balance between the forces at play within an atomic nucleus. Protons, which are positively charged, repel each other due to their like charges. Neutrons, with no electrical charge, act as a buffer, reducing the repulsive forces between protons.

Nuclides with a balanced ratio of neutrons to protons tend to be more stable. This is because the additional neutrons help to offset the electrostatic repulsion between the positively charged protons. On the other hand, nuclides with a disproportionate number of neutrons or protons are less stable and more likely to undergo nuclear decay to reach a more stable state.

Moreover, nuclides with both even proton and neutron numbers are generally more stable than those with odd numbers. This is based on observed patterns known as the 'even-odd rule', which suggests that nuclides with even numbers of both particles are less likely to be radioactive.

Nuclear Decay

As nuclides seek stability, those with unstable configurations undergo nuclear decay. Nuclear decay is the 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, and gamma rays, or the nuclide may simply rearrange into another state or a different element altogether.

There are various types of decay, including alpha decay (where the nucleus emits an alpha particle), beta decay (emission of an electron or a positron), and gamma decay (emission of high-energy photons). Each type of decay changes the nuclide in a different way, potentially altering the number of protons and neutrons or simply changing the energy state of the nucleus.

The rate at which a nuclide decays is characterized by its half-life—the time it takes for half of the atoms in a sample to decay. This property is fundamental in radiometric dating techniques, such as those used to estimate the age of carbon-based artifacts through carbon-14 decay.

Neutron-to-Proton Ratio

The neutron-to-proton (n/p) ratio is crucial in determining the stability of a nuclide. The stability pattern across the periodic table shows that nuclides with an n/p ratio close to 1 are generally more stable when the atomic number is low. However, as the number of protons increases, the repulsion between them becomes stronger, and a greater number of neutrons are needed to stabilize the nucleus.

A stable n/p ratio is typically around 1 for light elements (lighter than calcium), but this ratio grows for heavier elements. For example, stable lead isotopes have n/p ratios higher than 1.5. An imbalance in this ratio leads to instability, prompting nuclear decay as the nuclide attempts to reach a more stable configuration. This concept is crucial when predicting whether a given nuclide will be stable or radioactive.

A simple method to predict the stability of a nuclide incorporating this ratio follows the Belt of Stability, a graphical representation that shows stable nuclides lying within certain bounds on a chart plotting the number of neutrons versus the number of protons. Nuclides that fall outside this belt are prone to decay towards it. As such, knowing the neutron and proton count of nuclides brings vital insight into their potential stability.