Question

Which isotope in each pair is more likely to decay by electron capture? (a) \({ }^{13} \mathrm{~B}\) or \({ }^{8} \mathrm{~B}\) (b) \({ }^{209} \mathrm{Bi}\) or \({ }^{194} \mathrm{Bi}\)

Short answer

The isotopes more likely to decay by electron capture are \({ }^{8}\mathrm{B}\) (with an n/p ratio of 0.6) and \({ }^{194}\mathrm{Bi}\) (with an n/p ratio of 1.34).

Step-by-step solution

Determine the number of protons and neutrons

To find the n/p ratios, we first need to determine the number of protons (Z) and neutrons (N) in each isotope. We can do this using the mass number (A) of each isotope: A = Z + N. (a) \({ }^{13}\mathrm{B}\) isotope: A = 13, since B is Boron, which has an atomic number Z = 5 (5 protons) N = A - Z = 13 - 5 = 8 (8 neutrons) \({ }^{8}\mathrm{B}\) isotope: A = 8, Z = 5 (Boron) N = A - Z = 8 - 5 = 3 (3 neutrons) (b) \({ }^{209}\mathrm{Bi}\) isotope: A = 209, since Bi is Bismuth, which has an atomic number Z = 83 (83 protons) N = A - Z = 209 - 83 = 126 (126 neutrons) \({ }^{194}\mathrm{Bi}\) isotope: A = 194, Z = 83 (Bismuth) N = A - Z = 194 - 83 = 111 (111 neutrons)

Calculate n/p ratios for each isotope

Next, we'll calculate the neutron-to-proton ratio for each of the isotopes. \({ }^{13}\mathrm{B}\): n/p = 8/5 = 1.6 \({ }^{8}\mathrm{B}\): n/p = 3/5 = 0.6 \({ }^{209}\mathrm{Bi}\): n/p = 126/83 ≈ 1.52 \({ }^{194}\mathrm{Bi}\): n/p = 111/83 ≈ 1.34

Compare the n/p ratios and determine which isotope is more likely to decay by electron capture

Now that we have the n/p ratios for each isotope, we can determine which one in each pair is more likely to decay by electron capture. (a) Since \({ }^{8}\mathrm{B}\) has a lower n/p ratio (0.6) compared to \({ }^{13}\mathrm{B}\) (1.6), \({ }^{8}\mathrm{B}\) is more likely to decay by electron capture. (b) In this pair, \({ }^{194}\mathrm{Bi}\) has a lower n/p ratio (1.34) compared to \({ }^{209}\mathrm{Bi}\) (1.52), making \({ }^{194}\mathrm{Bi}\) more likely to decay by electron capture. So the isotopes more likely to decay by electron capture are \({ }^{8}\mathrm{B}\) and \({ }^{194}\mathrm{Bi}\).

Key concepts

Isotopes

When we talk about atoms of the same element having different numbers of neutrons, we refer to these variations as isotopes. Despite having the same number of protons—and therefore being the same element—isotopes can have very different properties, particularly when it comes to nuclear stability.

For example, carbon-12 (C^{12}), with 6 protons and 6 neutrons, is a stable isotope of carbon. However, carbon-14 (C^{14}), with 6 protons and 8 neutrons, is radioactive and will decay over time. This illustrates how the neutron count can affect an atom's behavior and stability, which is precisely what makes studying isotopes fascinating and significant.

Neutron-to-Proton Ratio

The neutron-to-proton ratio, often abbreviated as n/p ratio, is a key concept in nuclear physics and helps predict the stability of an isotope. Atoms with a balanced n/p ratio tend to be more stable. As atoms get larger, they need a higher number of neutrons compared to protons to remain stable.

However, when the n/p ratio becomes too low or too high, the nucleus can become unstable, leading to what we call radioactive decay. Isotopes with a particularly low n/p ratio are candidates for a type of decay known as electron capture, where an electron from an inner orbit is captured by the nucleus and combines with a proton to form a neutron.

Radioactive Decay

Radioactive decay is a spontaneous process where an unstable atomic nucleus loses energy by emitting radiation. There are several types of decay, including alpha, beta, and gamma decay, as well as electron capture.

In the step-by-step solutions for isotopes prone to electron capture, we focused on comparing their n/p ratios to predict which isotope is more likely to undergo this specific type of decay. Electron capture is a unique type of decay in that instead of emitting particles, the nucleus draws in an electron to achieve a more stable n/p ratio.

Nuclear Stability

Nuclear stability is governed by various forces and factors, such as the strong nuclear force that holds the nucleus together and the balance between the number of neutrons and protons (the n/p ratio). The stability of a nucleus is also influenced by its energy levels and the arrangement of nucleons within it.

Isotopes with extremely low or high n/p ratios often undergo radioactive decay to reach a more stable state. It's interesting to note that the Zone of Stability, a region on a graph plotting number of neutrons versus number of protons, shows us where stable nuclei tend to lie. It's a band where the n/p ratio is balanced, and isotope pairs falling outside of this band, like the ones in the exercise, help us predict decay paths like electron capture.