Problem 124 - Introductory Chemistry Atoms First

Question

Given four nuclei: \({ }_{10}^{20} \mathrm{~A},{ }_{9}^{17} \mathrm{~B},{ }_{12}^{28} \mathrm{C}\), and \({ }_{86}^{226} \mathrm{D}\), which is likely to emit a \(\beta^{-}\) particle, and which is likely to emit a \(\beta^{+}\) particle? Explain your choices.

Step-by-Step Solution

Answer

Nucleus C is the most likely to emit a β- particle, as it has the highest neutron-to-proton ratio (N/Z > 1), suggesting more neutrons relative to protons. Either nucleus B or D could potentially emit a β+ particle, as both have N/Z ratios below 1, indicating more protons relative to neutrons. However, it's not possible to determine which one is more likely to emit a β+ particle based solely on N/Z ratios.

Step 1: Calculate the neutron-to-proton ratios of each nucleus

To calculate the neutron-to-proton (N/Z) ratio for each nucleus, divide the number of neutrons by the number of protons: 1. For nucleus A: \( N/Z_A = \frac{20-10}{10} = 1 \) 2. For nucleus B: \( N/Z_B = \frac{17-9}{9} = \frac{8}{9} \) 3. For nucleus C: \( N/Z_C = \frac{28-12}{12} = \frac{4}{3} \) 4. For nucleus D: \( N/Z_D = \frac{226-86}{86} = \frac{35}{43} \)

Step 2: Compare the N/Z ratios

We can compare the N/Z ratios to determine which nuclei are likely to undergo beta decay: 1. Nucleus A has N/Z = 1, which generally indicates stability 2. Nucleus B has N/Z < 1, suggesting it has more protons relative to neutrons, which increases the likelihood of β+ decay 3. Nucleus C has N/Z > 1, suggesting it has more neutrons relative to protons, which increases the likelihood of β- decay 4. Nucleus D has N/Z < 1, suggesting it has more protons relative to neutrons, so it's also a candidate for β+ decay

Step 3: Determine the most likely candidates for beta decay

Based on the N/Z ratios and the rules for beta decay, we identify the most likely candidates for beta-minus and beta-plus decay: - Nucleus C has the highest N/Z ratio, suggesting it's the most likely candidate to undergo β- decay - Nucleus B and D both have N/Z ratios below 1 so could be candidates for β+ decay, although it's not possible to select which one based solely on N/Z ratios as other factors come into play In conclusion, nucleus C is the most likely to emit a β- particle, and either nucleus B or D could potentially emit a β+ particle. However, we cannot determine which of these two is more likely to emit a β+ particle based solely on the neutron-to-proton ratios.

Key Concepts

Beta-minus decay

Beta-minus decay occurs when a neutron in the nucleus of an atom transforms into a proton, emitting an electron (called a beta-minus particle) and an antineutrino in the process. This transformation increases the proton count by one while reducing the neutron count by one.

During beta-minus decay, you may notice the following:

• Neutron to proton conversion: This is the hallmark of beta-minus decay, where a neutron becomes a proton.
• Emission of a beta-minus particle: An electron is emitted, which carries away energy.
• Antineutrino is also emitted, but it is often not detected because it interacts weakly with matter.

In our exercise, nucleus C, with its neutron-to-proton ratio greater than 1 (\( N/Z_C = \frac{4}{3} \)), has an excess of neutrons. This makes it a candidate for beta-minus decay as it strives to achieve a more stable balance of neutrons and protons.

Beta-plus decay

Beta-plus decay is a type of radioactive decay wherein a proton transforms into a neutron while emitting a positron (a beta-plus particle) and a neutrino. This process decreases the proton number while increasing the neutron number, altering the neutron-to-proton ratio towards stability.

Characteristics of beta-plus decay include:

• Proton to neutron conversion: Key to beta-plus decay is the conversion of a proton into a neutron.
• Emission of a positron: The positively charged counterpart of an electron, known as a positron, is emitted.
• Neutrino emission: A neutrino is also released, facilitating energy conservation.

In our exercise, both nuclei B and D have neutron-to-proton ratios less than 1, indicating more protons than neutrons. This imbalance suggests they might undergo beta-plus decay to increase nuclear stability.

Neutron-to-proton ratio

The neutron-to-proton (N/Z) ratio is a crucial factor in determining the nuclear stability of an atom. This ratio helps predict the likelihood of a nucleus undergoing radioactive decay processes like beta decay.

Here's how it works:

• If N/Z is close to 1, the nucleus is likely stable, with sufficient protons and neutrons to maintain balance.
• A high N/Z ratio indicates a larger number of neutrons, implying the potential for beta-minus decay as the nucleus seeks to convert some of its neutrons to protons.
• A low N/Z ratio means there are more protons, suggesting beta-plus decay could occur to convert protons into neutrons for better stability.

For our four nuclei in the exercise, calculating the N/Z ratio was crucial in identifying potential decay pathways. Nucleus A showed stability with an N/Z ratio of 1, while nucleus C had a higher ratio, leaning towards beta-minus decay. Meanwhile, nuclei B and D had lower ratios, indicating beta-plus decay potentials.

Nuclear stability

Nuclear stability refers to the ability of a nucleus to resist changes or decay. Stable nuclei do not spontaneously change form, and decay is their way of seeking stability through processes like alpha, beta-minus, or beta-plus decay.

Several factors influence nuclear stability:

• Neutron-to-proton ratio: This ratio affects stability, with around 1 being preferable for lighter elements. Heavier elements may require higher ratios for stability.
• Nuclear forces: The strong nuclear force holds protons and neutrons together, contributing to stability. As the number of nucleons increases, this force must balance repulsive electrostatic forces between protons.
• Energy states: Nuclei tend to favor lower energy states. Radioactive decay often results from a nucleus seeking to lower its energy state and become more stable.

In this exercise, we examined the nuclear stability of different nuclei through their N/Z ratios. Nucleus A was identified as stable, while the others, B, C, and D, displayed tendencies towards decay by virtue of their N/Z imbalances.