Question

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

Short answer

The radioactive nuclides in each pair are: (a) ${}_{20}^{45}\mathrm{Ca}$, due to the excess of neutrons; (b) ${}^{14}\mathrm{C}$, due to the excess of neutrons; (c) _thorium-230_, due to having an N/Z ratio farther from the ideal ratio for heavy elements.

Step-by-step solution

Counting the number of neutrons and protons in each nuclide

First, let's find the number of protons (Z) and neutrons (N) in each nuclide. We can find the number of protons from the atomic number, and the number of neutrons can be determined by subtracting the atomic number from the mass number (A - Z). $$\{\begin{array}{l}(a)\\ {}_{20}^{40}\mathrm{Ca}:Z=20,N=40-20=20\\ {}_{20}^{45}\mathrm{Ca}:Z=20,N=45-20=25\end{array}$$ $$\{\begin{array}{l}(b)\\ {}^{12}\mathrm{C}:Z=6,N=12-6=6\\ {}^{14}\mathrm{C}:Z=6,N=14-6=8\end{array}$$ $$\{\begin{array}{l}(c)\\ \mathrm{Pb}-206:Z=82,N=206-82=124\\ \mathrm{Th}-230:Z=90,N=230-90=140\end{array}$$

Analyzing the Nuclides' Stability

Now let's analyze the stability of each nuclide. (a) For ${}_{20}^{40}\mathrm{Ca}$ and ${}_{20}^{45}\mathrm{Ca}$, both are Isotopes of calcium and have the same number of protons (20). Since the number of neutrons and protons are equal in ${}_{20}^{40}\mathrm{Ca}$, it is a stable nuclide, while ${}_{20}^{45}\mathrm{Ca}$ has an excess of neutrons, making it more likely to be unstable and radioactive. (b) For ${}^{12}\mathrm{C}$ and ${}^{14}\mathrm{C}$, these carbon isotopes have the same number of protons (6), but different numbers of neutrons. In the case of ${}^{12}\mathrm{C}$, the number of neutrons and protons are equal, which makes it a stable nuclide. ${}^{14}\mathrm{C}$ has 8 neutrons, and due to the neutron excess, it is more likely to be unstable and thus radioactive. (c) For lead-206 and thorium-230, we can analyze their stability based on the ratio of neutrons to protons (N/Z). $$\frac{N}{Z}(\mathrm{Pb}-206)=\frac{124}{82}\approx 1.51$$ $$\frac{N}{Z}(\mathrm{Th}-230)=\frac{140}{90}\approx 1.56$$ The ideal N/Z ratio for heavy elements is around 1.5. Since the ratio for lead-206 is closer to 1.5, it is more likely to be stable, while thorium-230 is less stable and likely to be radioactive.

Predicting and Explaining the Radioactive Nuclide in Each Pair

Based on our analysis, we can predict the radioactive nuclide in each pair: (a)${}_{20}^{40}\mathrm{Ca}$ is _stable_, and ${}_{20}^{45}\mathrm{Ca}$ is _radioactive_ due to the excess of neutrons. (b)${}^{12}\mathrm{C}$ is _stable_, and ${}^{14}\mathrm{C}$ is _radioactive_ due to the excess of neutrons. (c) _Lead-206_ is _stable_, and _thorium-230_ is _radioactive_ due to having an N/Z ratio farther from the ideal ratio for heavy elements.

Key concepts

Nuclide Stability

Understanding nuclide stability is essential for predicting whether a nuclide will be stable or radioactive. A nuclide is a specific type of nucleus defined by its number of protons and neutrons. The stability of a nuclide depends largely on the forces between these particles in the nucleus.
The key force that holds the nucleus together is the nuclear strong force, which acts over a very short range and is significantly stronger than the electromagnetic repulsion between protons. A nuclide is considered stable if the strong nuclear force can overcome this repulsive force, effectively bonding the nuclear particles together.
An instability arises when these forces are no longer balanced, often because of an imbalance in the number of neutrons relative to protons. When a nucleus is unstable, it tends to undergo radioactive decay to reach a more stable form. This decay can occur through various processes, such as beta or alpha decay, depending on the specific nuclide's configuration.

Neutron-to-Proton Ratio

The neutron-to-proton (N/Z) ratio is a critical factor in determining the stability of a nuclide. Generally, lighter elements have a N/Z ratio close to 1:1, meaning the number of neutrons is roughly equal to the number of protons.
For heavier elements, however, more neutrons are needed to stabilize the nucleus due to the increased electrostatic repulsion between the many protons. That's why their typical stable N/Z ratios are greater than 1.
A nuclide's stability heavily depends on this ratio. It provides an intuitive way to predict which isotopes are stable and which might undergo radioactive decay.

• For lighter elements, such as carbon, a nearly equal number of neutrons and protons (like in ${}^{12}\text{C}$) indicates stability.
• For heavier isotopes, N/Z ratios closer to 1.5 are ideal. As deviations from this ratio increase, so does the likelihood of decay.

Carbon Isotopes

Carbon has two well-known isotopes: ${}^{12}\text{C}$ and ${}^{14}\text{C}$. These isotopes are abundant in various natural processes and are crucial in radiocarbon dating.
${}^{12}\text{C}$ is remarkably stable because its neutron-to-proton ratio is 1:1, perfectly balancing the forces in the nucleus. It constitutes the majority of carbon found on Earth.

• Stable because the even N/Z ratio stabilizes the internal nuclear forces.

${}^{14}\text{C}$, on the other hand, is radioactive. It has two extra neutrons, leading to an N/Z ratio of 1.33.
While it is unstable, ${}^{14}\text{C}$ plays an important role in dating historical biological artifacts, as it decays into nitrogen-14 with a known half-life.

• Useful for dating ancient objects.

Calcium Isotopes

Calcium isotopes vary, but important examples include ${}^{40}\text{Ca}$ and ${}^{45}\text{Ca}$. Both have the same number of protons, as all calcium atoms do, but they differ in neutron counts.
${}^{40}\text{Ca}$ has a neutron-to-proton ratio of 1:1, making it very stable. It is one of the most abundant isotopes of calcium and is found in many biological structures such as bones and shells because of its stability.
In contrast, ${}^{45}\text{Ca}$ is radioactive. With an additional five neutrons, it strays from the ideal N/Z ratio, leading to instability.

• This nuclide tends to decay to reach a more stable state.

Knowing the stability of various calcium isotopes helps scientists understand geological and biological processes.

Heavy Elements Stability

The stability of heavy elements is a fascinating topic in nuclear chemistry. As elements get heavier, the challenge of stability increases due to the higher number of protons and the greater electromagnetic repulsion within the nucleus.
For heavy elements like lead and thorium, maintaining stability typically requires a higher neutron-to-proton ratio. This is because more neutrons can help to dilute the repulsive forces among the numerous protons.
For example, lead-206 has an N/Z ratio close to 1.5, placing it within a stable range for such a heavy element.

• Lead-206 is stable because its N/Z ratio is near the ideal for heavy elements.

Thorium-230, however, is less stable. It has an N/Z ratio greater than 1.5, implying it is on the verge of instability. Such isotopes often seek stability through radioactive decay, emitting particles to adjust their internal ratios closer to stability.

• Instability leads to radioactive decay.