Question

One type of commercial smoke detector contains a minute amount of radioactive americium-241 ( \(\left.^{(241} \mathrm{Am}\right),\) which decays by \(\alpha\) -particle production. The \(\alpha\) particles ionize molecules in the air, allowing it to conduct an electric current. When smoke particles enter, the conductivity of the air is changed and the alarm buzzes. a. Write the equation for the decay of \(^{241}_{95} \mathrm{Am}\) by \(\alpha\) -particle production. b. The complete decay of \(^{241} \mathrm{Am}\) involves successively \(\alpha, \alpha\) \(\boldsymbol{\beta}, \alpha, \alpha, \boldsymbol{\beta}, \alpha, \alpha, \alpha, \boldsymbol{\beta}, \alpha,\) and \(\boldsymbol{\beta}\) production. What is the final stable nucleus produced in this decay series? c. Identify the 11 intermediate nuclides.

Short answer

The decay equation for Am-241 by alpha-particle production is: \[ ^{241}_{95}\mathrm{Am}\rightarrow ^{237}_{93}\mathrm{Np} + ^{4}_{2}\mathrm{He} \] The final stable nucleus after the given decay series is \(\left.^{209}_{81}\mathrm{Tl}\right.\). The 11 intermediate nuclides in the decay series are: 1. \(\left.^{237}_{93}\mathrm{Np}\right.\) 2. \(\left.^{233}_{91}\mathrm{Pa}\right.\) 3. \(\left.^{233}_{92}\mathrm{U}\right.\) 4. \(\left.^{229}_{90}\mathrm{Th}\right.\) 5. \(\left.^{225}_{88}\mathrm{Ra}\right.\) 6. \(\left.^{225}_{89}\mathrm{Ac}\right.\) 7. \(\left.^{221}_{87}\mathrm{Fr}\right.\) 8. \(\left.^{217}_{85}\mathrm{At}\right.\) 9. \(\left.^{213}_{83}\mathrm{Bi}\right.\) 10. \(\left.^{213}_{84}\mathrm{Po}\right.\) 11. \(\left.^{209}_{82}\mathrm{Pb}\right.\).

Step-by-step solution

Write the equation for the decay of Am-241 by alpha-particle production.

To write the equation for the decay of \(\left.^{241}_{95} \mathrm{Am}\right.\) by alpha-particle production, we need to know that an alpha particle is a helium nucleus, represented as \(\left.^{4}_{2} \mathrm{He}\right.\). The decay process is represented as: \[ ^{241}_{95}\mathrm{Am}\rightarrow ^{A}_{Z}\mathrm{X} + ^{4}_{2}\mathrm{He} \] Where A and Z are the atomic mass and atomic number of the resulting nucleus (X), respectively. To find the values for A and Z, use the conservation of nucleons (protons and neutrons): A = 241 - 4 Z = 95 - 2 #(Step 2) Final Stable Nucleus in Decay Series#

Determine the final stable nucleus after the given decay process.

The decay series given is: \(\alpha, \alpha, \beta, \alpha, \alpha, \beta, \alpha, \alpha, \alpha, \beta, \alpha, \beta\). Using this decay series, we can calculate the change in atomic mass and atomic number after each step: 1. After the first two alpha decays, Am loses 2 alpha particles, so the change in atomic mass number is -8 and the change in the atomic number is -4. 2. Following the beta decay, there is no change in the atomic mass but an increase in the atomic number by 1. 3. As we proceed through the decay series, the total change in atomic mass and atomic number after all these decays will be -32 and -14, respectively. Now, to find the final stable nucleus: Final Atomic Mass = 241 - 32 = 209 Final Atomic Number = 95 - 14 = 81 The final stable nucleus is \(\left.^{209}_{81}\mathrm{Tl}\right.\). #(Step 3) Identify Intermediate Nuclides#

Determine the 11 intermediate nuclides in the decay series.

To find the intermediate nuclides, we will follow the same decay series and find the resulting nuclei after each decay. Nuclide 1: After the first alpha decay, Atomic Mass = 241 - 4 = 237 Atomic Number = 95 - 2 = 93 The first intermediate nuclide is \(\left.^{237}_{93}\mathrm{Np}\right.\). Nuclide 2: After the second alpha decay, Atomic Mass = 237 - 4 = 233 Atomic Number = 93 - 2 = 91 The second intermediate nuclide is \(\left.^{233}_{91}\mathrm{Pa}\right.\). Nuclide 3: After the beta decay, Atomic Mass = 233 Atomic Number = 91 + 1 = 92 The third intermediate nuclide is \(\left.^{233}_{92}\mathrm{U}\right.\). Follow this process for the rest of the decay series to find all 11 intermediate nuclides. The complete list will be: 1. \(\left.^{237}_{93}\mathrm{Np}\right.\) 2. \(\left.^{233}_{91}\mathrm{Pa}\right.\) 3. \(\left.^{233}_{92}\mathrm{U}\right.\) 4. \(\left.^{229}_{90}\mathrm{Th}\right.\) 5. \(\left.^{225}_{88}\mathrm{Ra}\right.\) 6. \(\left.^{225}_{89}\mathrm{Ac}\right.\) 7. \(\left.^{221}_{87}\mathrm{Fr}\right.\) 8. \(\left.^{217}_{85}\mathrm{At}\right.\) 9. \(\left.^{213}_{83}\mathrm{Bi}\right.\) 10. \(\left.^{213}_{84}\mathrm{Po}\right.\) 11. \(\left.^{209}_{82}\mathrm{Pb}\right.\).

Key concepts

Beta Decay

In the realm of nuclear physics, beta decay is a fascinating process where an unstable nucleus transforms by emitting a beta particle. A beta particle typically involves either an electron or a positron. This process occurs because a neutron in the nucleus is converted into a proton, or vice versa. The conversion results in the emission of a beta particle.
- **Beta Minus Decay**: Here, a neutron in the nucleus is transformed into a proton, releasing an electron and an antineutrino. It can be represented by the formula\[ ^{A}_{Z} ext{X} \rightarrow ^{A}_{Z+1} ext{Y} + e^- + \bar{u}_e \]- **Beta Plus Decay**: In this scenario, a proton is converted into a neutron, emitting a positron and a neutrino. It is not as common because it requires energy input or an external source to occur.
Beta decay increases the atomic number by one without changing the mass number, as no nucleons are lost. This change is significant because it affects the chemical element identity, turning one element into another!

Radioactive Decay Series

Radioactive decay series is a sequence of successive radioactive decays that certain isotopes undergo until they reach a stable state. This series is characterized by the type of decay, whether it be alpha or beta.
Alpha decay reduces the mass number by four and the atomic number by two, while beta decay increases the atomic number by one without altering the mass number. - **Example of a Decay Series**: Starting with a heavy unstable nucleus like Americium-241, the series can involve multiple steps through different intermediate nuclides until reaching a stable isotope. - **Completion of Series**: The endpoint of this decay chain is the conversion into a stable nucleus, commonly a lead isotope, which marks the conclusion of the series.
Understanding radioactive decay series is crucial in fields like geology and archaeology for dating samples and tracing nuclear material pathways.

Intermediate Nuclides

During radioactive decay, there are several transitional forms known as intermediate nuclides. These nuclides are formed at each step of the decay process prior to reaching a stable atom.
**Characteristics of Intermediate Nuclides**:

• They stem from either alpha or beta decay steps.
• Each intermediate nuclide features different atomic and mass numbers as it moves through the decay series.
• They are crucial transitional forms helping in understanding the complete decay path.

To identify intermediate nuclides, tracking the decay steps and calculating the resultant atomic mass and atomic number accurately is key. They are important for understanding the overall change in nuclear properties throughout the decay process.

Stable Nucleus

A stable nucleus is the endpoint of a radioactive decay series. It is the final product after a series of transformations, no longer subject to radioactive decay.
**Features of a Stable Nucleus**:

• It has a balanced ratio of protons to neutrons, making it stable against both radioactivity and further decay.
• Stable nuclei do not emit radiation since their nucleons are in a lower energy state.
• In decay series, heavy and unstable isotopes eventually transmute into a stable form, often ending with lead isotopes in the case of many natural decay series.

The journey to a stable nucleus involves a fascinating transition through various radioactive transformations, reshaping atoms until stability is achieved, marking the end of radioactive decay.