Question

Which of the following nuclides would you expect to be radioactive: \(_{26}^{58} \mathrm{Fe}, _{27}^{60} \mathrm{Co},\) \(_{41}^{92} \mathrm{Nb},\) mercury-202, radium-226? Justify your choices.

Short answer

The radioactive nuclides among the given list are \(_{27}^{60} \mathrm{Co}\) and \(_{41}^{92} \mathrm{Nb}\), as they have an excess of neutrons and their atomic and mass numbers fall within the criteria for possible instability. The other nuclides (\(_{26}^{58} \mathrm{Fe}\), mercury-202, and radium-226) are likely to be stable.

Step-by-step solution

Determine Neutron Numbers for Each Nuclide

\(_{26}^{58} \mathrm{Fe}\) has 26 protons and \(58-26 = 32\) neutrons. \(_{27}^{60} \mathrm{Co}\) has 27 protons and \(60-27 = 33\) neutrons. \(_{41}^{92} \mathrm{Nb}\) has 41 protons and \(92-41 = 51\) neutrons. Mercury-202 (Hg, atomic number 80) has 80 protons and \(202-80 = 122\) neutrons. Radium-226 (Ra, atomic number 88) has 88 protons and \(226-88 = 138\) neutrons.

Detect Stability Based on Atomic and Mass Numbers

\(_{26}^{58} \mathrm{Fe}\) has even atomic and mass numbers, so it is generally stable. \(_{27}^{60} \mathrm{Co}\) has an odd atomic number and even mass number; it might be stable. \(_{41}^{92} \mathrm{Nb}\) has an odd atomic number and an even mass number; it might be stable. Mercury-202 has even atomic and mass numbers, so it is generally stable. Radium-226 has even atomic and mass numbers, so it is generally stable.

Consider Second Isotopes

Fe and Hg both have stable isotopes aside from the ones given. The stable isotopes are \(_{26}^{56} \mathrm{Fe}\) and \(_{80}^{200} \mathrm{Hg}\). Since \(_{27}^{60} \mathrm{Co}\) has an excess of neutrons and Co has no other stable isotopes, we can expect it to be radioactive. \(_{41}^{92} \mathrm{Nb}\) also has an excess of neutrons, making it likely to be radioactive.

Conclusion

The radioactive nuclides among the given list are \(_{27}^{60} \mathrm{Co}\) and \(_{41}^{92} \mathrm{Nb}\). Their atomic and mass numbers fall within the criteria for possible instability, and they have an excess of neutrons. Other nuclides on the list (\(_{26}^{58} \mathrm{Fe}\), mercury-202, and radium-226) are likely to be stable.

Key concepts

Neutron Number Determination

Determining the neutron number in a nucleus is crucial to understanding the structure of an atom. Neutrons play a vital role in the nucleus by contributing to the mass of an atom and providing the necessary force to keep the positively charged protons within the nucleus from repelling each other.

The neutron number can be calculated by subtracting the atomic number, which represents the number of protons, from the mass number of the nuclide. For instance, in the exercise, iron-58 (_{26}^{58}Fe) has an atomic number of 26 (protons) and a mass number of 58, leading to a calculation of 58-26 = 32 neutrons. Knowing the number of neutrons helps us understand the isotope of the element and plays a significant role in assessing an atom's stability.

Isotope Stability

The stability of isotopes is influenced by the ratio of neutrons to protons in the nucleus. Isotopes with a balanced ratio tend to be more stable, while those with too many or too few neutrons can be unstable and thereby radioactive. Radioactive isotopes tend to undergo decay to reach a more stable state.

In the exercise, cobalt-60 (_{27}^{60}Co) and niobium-92 (_{41}^{92}Nb) were identified as likely radioactive. This is because their neutron to proton ratio is higher compared to their stable counterparts, making their nuclei unstable. As a general rule, nuclides with an even number of protons and neutrons have greater stability. This rule can be applied in the determination process, as seen in the step-by-step solution. It's also important to note that elements with many known isotopes, like iron (Fe) and mercury (Hg), often have more stable isotopes to consider.

Atomic and Mass Numbers

The atomic number and the mass number are fundamental characteristics of an element's atom. The atomic number (Z) represents the number of protons in the nucleus and defines the element itself. The mass number (A), on the other hand, is the total number of protons and neutrons in the nucleus.

In the provided exercise, this information helps us predict the radioactivity of certain nuclides. Generally, a nuclide with an even atomic number and an even mass number has a higher chance of being stable, such as mercury-202 (_{80}^{202}Hg) and radium-226 (_{88}^{226}Ra). This is due to the pairing of particles within the nucleus, which tends to lead to a more energetically favorable configuration. It's also integral to compare with known stable isotopes of the elements in question, as seen in steps 3 and 4, to determine if the nuclide would likely be stable or radioactive.