Question

Cobalt has a stable isotope, \({ }_{27}^{59} \mathrm{Co},\) and 22 radioactive isotopes. The most stable radioactive isotope is \({ }_{27}^{60} \mathrm{Co} .\) What is the dominant decay mode of this isotope? a) \(\beta^{+}\) b) \(\beta^{-}\) c) proton emission d) neutron emission

Short answer

Answer: The dominant decay mode for the radioactive isotope ${}_{27}^{60} \mathrm{Co}$ is \(\beta^{-}\) decay.

Step-by-step solution

Identify the atomic number and mass number of the stable isotope

For the stable isotope \({ }_{27}^{59} \mathrm{Co}\), the atomic number (Z) is 27 and the atomic mass number (A) is 59.

Identify the atomic number and mass number of the radioactive isotope

For the radioactive isotope \({ }_{27}^{60} \mathrm{Co}\), the atomic number (Z) is still 27, but the atomic mass number (A) is now 60.

Compare the radioactive isotope to the stable isotope

Comparing both isotopes, we can see that the atomic number (Z) remains constant, while the atomic mass number (A) increases by 1. Since the atomic number hasn't changed, it means the number of protons remains the same in both isotopes. This excludes the options of proton and neutron emission, as they would change the number of protons or neutrons respectively.

Determine the possible radioactive decay mode

Between \(\beta^{+}\) and \(\beta^{-}\) decay, we need to determine which one will result in the change in the mass number while keeping the number of protons constant. The \(\beta^{+}\) decay releases positrons, which convert a proton into a neutron in the nucleus. This process does not change the mass number and hence doesn't match what we observe in the radioactive isotope. Therefore, the dominant decay mode must be \(\beta^{-}\) decay. In \(\beta^{-}\) decay, a neutron gets converted into a proton, releasing an electron and an antineutrino. This process doesn't change the mass number but increases the atomic number (Z) by 1. However, since the radioactive isotope precedes the decay, it has a higher atomic number than the product (not the stable isotope). Therefore, it will return to its stable form through \(\beta^-\) decay. So, the dominant decay mode for \({ }_{27}^{60} \mathrm{Co}\) is: b) \(\beta^{-}\) decay

Key concepts

Isotopes

Isotopes are versions of a chemical element that have the same number of protons but different numbers of neutrons in their nuclei. As a result, while they share the same atomic number (Z), which defines the element, they have different atomic mass numbers (A).

For example, cobalt, the element in our exercise, has various isotopes, including the stable \text{\({ }_{27}^{59} \text{Co}\)} and the radioactive \text{\({ }_{27}^{60} \text{Co}\)}. These numbers, 59 and 60, represent the total count of protons and neutrons in each nucleus. Since the atomic number is the same for both (27, indicating 27 protons), the difference lies in the number of neutrons. The neutron number impacts the stability of the nucleus, with certain neutron-proton ratios leading to radioactive decay as the means to achieve a more stable state.

Beta Decay

Beta decay is a type of radioactive decay in which a beta particle, either an electron or a positron, is emitted from an atomic nucleus. There are two main types of beta decay: beta-minus (\text{\(\beta^-\)}) and beta-plus (\text{\(\beta^+\)}).

In \text{\(\beta^-\)} decay, a neutron in the nucleus is transformed into a proton, and an electron and an antineutrino are emitted. This increases the atomic number by 1, as seen in the decay of \text{\({ }_{27}^{60} \text{Co}\)} in our exercise, but leaves the mass number unchanged because the total count of protons and neutrons remains the same.

\text{\(\beta^+\)} decay, on the other hand, involves the conversion of a proton into a neutron with the emission of a positron and a neutrino, decreasing the atomic number by 1. Notably, both decay processes are significant in medical applications, including radiotherapy and radiotracing. The ability to predict and understand the behavior of isotopes through beta decay is crucial in nuclear physics and various practical technologies.

Nuclear Physics

Nuclear physics is the branch of physics that studies the constituents and interactions of atomic nuclei. This field is fundamental in understanding the forces and particles that comprise the nucleus, such as protons and neutrons, and the processes that can alter the composition of the nucleus, including radioactive decay.

Nuclear physics explains why certain isotopes are stable and why others, like \text{\({ }_{27}^{60} \text{Co}\)}, are radioactive and decay over time. It plays a critical role in many aspects of modern life, from the generation of electricity in nuclear power plants to the development of medical imaging techniques and cancer treatments. Additionally, the principles of nuclear physics underpin our understanding of cosmological phenomena and the evolution of the universe.