Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 18: Problem 36

Many home smoke detectors use americunium-241 as a radiation source. In the smoke detector, \({ }^{241}\) Am emits alpha particles. These alpha particles are detected by the detector unless smoke particles block their passage. Thus, as long as there are no smoke particles, the detector gets a continuous stream of alpha particles hitting it; smoke causes a disruption in this stream and activates the alarm. Write out the nuclear equation for the decay of \({ }^{241}\) Am by alpha emission.

Short Answer

Expert verified
\({}^{241}_{95}\)Am \rightarrow \({}^{237}_{93}\)Np + \({}^{4}_{2}\)He.

Step by step solution

01

- Identify the parent nucleus

The parent nucleus is Americium-\({}^{241}\), denoted as \({}^{241}_{95}\)Am.
02

- Understand alpha emission

Alpha emission involves the release of an alpha particle, which is a helium nucleus represented as \({}^{4}_{2}\)He. This particle has 2 protons and 2 neutrons.
03

- Write down the alpha emission equation

When \({}^{241}_{95}\)Am undergoes alpha emission, it releases a \({}^{4}_{2}\)He particle, leaving behind a new nucleus. The original mass number decreases by 4 (241 - 4 = 237), and the atomic number decreases by 2 (95 - 2 = 93).
04

- Identify the daughter nucleus

The daughter nucleus will have a mass number of 237 and an atomic number of 93. This corresponds to Neptunium-\({}^{237}\), denoted as \({}^{237}_{93}\)Np.
05

- Write the complete nuclear equation

Combine all components to write the nuclear equation: \({}^{241}_{95}\)Am \rightarrow \({}^{237}_{93}\)Np + \({}^{4}_{2}\)He.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

nuclear reactions
Nuclear reactions involve changes in an atom's nucleus, often resulting in the transformation of elements. One type of nuclear reaction is alpha decay, where an unstable nucleus releases an alpha particle.
In our example, Americium-241 undergoes alpha decay. This process changes the nucleus by reducing its mass and atomic numbers. Americium-241 (^{241}_{95}Am) emits an alpha particle (^{4}_{2}He), converting into Neptunium-237 (^{237}_{93}Np).
Nuclear reactions are vital in various applications, including energy production and medical treatments. They demonstrate the powerful energy stored within an atom's nucleus.
radioactive decay
Radioactive decay is how unstable atomic nuclei lose energy. This happens when the nucleus releases particles, transforming into a new element. There are different types of radioactive decay: alpha, beta, and gamma decay.
In alpha decay, the nucleus emits an alpha particle, which consists of 2 protons and 2 neutrons. This particle is highly ionizing but has low penetration power. In our exercise, Americium-241 undergoes alpha decay, producing Neptunium-237 and an alpha particle.
Radioactive decay is a naturally occurring process crucial for dating archaeological finds, medical imaging, and even household smoke detectors, like the one in our example.
chemistry education
Understanding nuclear reactions and radioactive decay is critical in chemistry education. These concepts explain the behavior of atoms and kinds of radiation they emit.
By learning about alpha decay, students appreciate the transformation of elements and the forces within nuclei. Solving problems like the Americium-241 decay equation enhances critical thinking and comprehension of atomic structures.
Effective chemistry education combines theoretical knowledge with practical examples to illustrate the relevance of nuclear science in everyday life, like smoke detectors, radiotherapy in cancer treatment, and power generation in nuclear reactors.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Rubidium-87, a beta emitter, is the product of positron emission. Identify (a) the product of rubidium- 87 decay (b) the precursor of rubidium- 87

Cobalt-60 has a half-life of \(5.26\) years. If \(1.00 \mathrm{~g}\) of \({ }^{60} \mathrm{Co}\) were allowed to decay, how many grams would be left after (a) one half-life? (b) two half-lives? (c) four half-lives? (d) ten half-lives?

Clearly distinguish between fission and fusion. Give an example of each.

Consider the fission reaction $$ { }_{92}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{35}^{94} \mathrm{Sr}+{ }_{54}^{139} \mathrm{Xe}+3{ }_{0}^{1} \mathrm{n}+\text { energy } $$ Calculate the following using these mass data \((1.0 \mathrm{~g}\) is equivalent to \(9.0 \times 10^{13} \mathrm{~J}\) ): $$ \begin{array}{rlr} \mathrm{U}-235=235.0439 \mathrm{amu} & \text { Sr-94 }=93.9154 \mathrm{amu} \\\ \mathrm{Xe}-139=138.9179 \mathrm{amu} & \mathrm{n}=1.0087 \mathrm{amu} \end{array} $$ (a) the energy released in joules for a single event (one uranium atom splitting) (b) the energy released in joules per mole of uranium splitting (c) the percentage of mass lost in the reaction

Write a nuclear equation for the conversion of \({ }_{13}^{26} \mathrm{Al}\) to \({ }_{12}^{26} \mathrm{Mg}\).
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Kinetics

Read Explanation

Chemistry Branches

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free