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. Nuclear Transmutations (Section 21.3)

Short answer

Isoptopes \( _{27}^{60} \mathrm{Co}\) and \( _{41}^{92} \mathrm{Nb}\) are more likely to be radioactive due to their odd nucleons and relatively higher N/P ratios. Mercury-202 and Radium-226 have higher N/P ratios, indicating an increased likelihood of radioactivity, but both have even protons and neutrons, causing uncertainty in their stability.

Step-by-step solution

Calculate Neutron-to-Proton Ratios for each nuclide

For each of the given nuclides, calculate the neutron-to-proton ratio (N/P) using the atomic number (Z) and the mass number (A): N/P = (A-Z)/Z 1. \( _{26}^{58} \mathrm{Fe}\): N/P = (58-26)/26 = 32/26 ≈ 1.23 2. \( _{27}^{60} \mathrm{Co}\): N/P = (60-27)/27 ≈ 33/27 ≈ 1.22 3. \( _{41}^{92} \mathrm{Nb}\): N/P = (92-41)/41 = 51/41 ≈ 1.24 4. Mercury-202 (Hg): Atomic number Z = 80, N/P = (202-80)/80 = 122/80 ≈ 1.53 5. Radium-226 (Ra): Atomic number Z = 88, N/P = (226-88)/88 = 138/88 ≈ 1.57

Determine if even or odd number of neutron and protons

For each nuclide, determine if the number of neutrons and protons are even or odd. 1. \( _{26}^{58} \mathrm{Fe}\): even number of protons (26) and even number of neutrons (32) 2. \( _{27}^{60} \mathrm{Co}\): odd number of protons (27) and odd number of neutrons (33) 3. \( _{41}^{92} \mathrm{Nb}\): odd number of protons (41) and even number of neutrons (51) 4. Mercury-202: even number of protons (80) and even number of neutrons (122) 5. Radium-226: even number of protons (88) and even number of neutrons (138)

Determine radioactivity based on N/P ratio and even/odd nucleons

Using the information from Steps 1 and 2, predict which nuclides are likely to be radioactive: 1. \( _{26}^{58} \mathrm{Fe}\): N/P ≈ 1.23, both even nucleons; likely to be stable. 2. \( _{27}^{60} \mathrm{Co}\): N/P ≈ 1.22, odd nucleons; more likely to be radioactive. 3. \( _{41}^{92} \mathrm{Nb}\): N/P ≈ 1.24, odd nucleons; more likely to be radioactive. 4. Mercury-202: N/P ≈ 1.53, both even nucleons; radioactivity uncertain. 5. Radium-226: N/P ≈ 1.57, both even nucleons; radioactivity uncertain. Based on the above information, isotope \( _{27}^{60} \mathrm{Co}\) and \( _{41}^{92} \mathrm{Nb}\) are more likely to be radioactive due to their odd nucleons and relatively higher N/P ratios. Mercury-202 and Radium-226 have higher N/P ratios, indicating an increased likelihood of radioactivity, however, they both have even protons and neutrons, causing uncertainty in their stability.

Key concepts

Neutron-to-Proton Ratio

Understanding the neutron-to-proton ratio (N/P) is critical when determining the stability of a particular nuclide. This ratio helps us predict whether a nucleus will be radioactive.

Atoms crave balance. In the nucleus, this balance is achieved when there is a certain proportion of neutrons to protons. As a rule of thumb, if the N/P ratio is too high or too low, the nuclide is more likely to be unstable and thus radioactive. The ideal range varies, but for lighter elements, a 1:1 ratio is closer to stability, while heavier elements can have more neutrons than protons. The exercise showcased nuclides with varying N/P ratios, hinting at their likelihood to be unstable and therefore radioactive. For instance, nuclides such as Mercury-202 and Radium-226, with higher ratios, signal potential instability.

By assessing this ratio, students and scientists alike can use it as a telltale sign for predicting radioactivity without performing actual physical experiments, which underscores its educational and practical significance.

Radioactive Isotopes

Radioactive isotopes, or radioisotopes, are variants of elements that have unstable atomic nuclei and emit radiation as a result. This radioactivity occurs due to an imbalance in the neutron-to-proton ratio, which drives the nucleus to decay into a more stable state. It's this principle that helps students predict radioactivity in exercises.

Activities such as the decay of Cobalt-60 into Nickel-60 involve the emission of particles which are essential for processes like medical imaging and cancer treatments. Cobalt-60, as identified in the exercise, has an odd number of protons and neutrons, contributing to its instability and resulting in its classification as radioactive. When discussing radioisotopes, it is important not only to recognize their significance in science and medicine but to understand the underlying cause of their instability.

Nuclear Stability

The quest for nuclear stability is the innateness of nuclides to remain intact and not undergo radioactive decay. It is the 'holy grail' for atomic nuclei. Several factors determine this stability, including the neutron-to-proton ratio and the presence of odd or even numbers of nucleons.

Stable nuclides tend to have even numbers of both neutrons and protons, as demonstrated by Iron-58 in the exercise. When we encounter nuclides with an imbalance of neutrons or protons, or if they have odd numbers of nucleons like in Niobium-92, we expect them to seek stability through radioactive decay. Nuclear stability is not just a theoretical concept; it provides practical understanding in fields ranging from energy generation to the synthesis of new elements.

Odd and Even Nucleons

The odd or even nature of the number of nucleons (protons and neutrons) within a nucleus has a profound influence on its stability. Odd numbers of either protons, neutrons, or both typically confer a level of instability to the nuclide, making it more likely to be radioactive.

Nuclei that house even numbers of both protons and neutrons are generally more stable, a phenomenon known as the even-even effect. The exercise revealed that nuclides like Mercury-202, while showing higher N/P ratios, presented an 'even-even' scenario, which tends to associate with greater stability and sometimes creates uncertainty about their radioactivity. On the other hand, radioactivity is more confidently predicted in 'odd-odd' nuclides, such as Cobalt-60. Understanding this odd-even concept helps to simplify and clarify predictions about stability and radioactivity of the elements.