Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 18: Problem 61

Actinium-217 decays by releasing an alpha particle. Write an equation for this decay process, and determine what element is formed.

Short Answer

Expert verified
In the alpha decay of Actinium-217, an alpha particle is released, resulting in the formation of the new element, Francium-213. The nuclear decay equation representing this process is: \[_{89}^{217}Ac \rightarrow _{2}^{4}He + _{87}^{213}Fr.\]

Step by step solution

01

Identifying The Initial Nucleus

We are dealing with Actinium-217. In nuclear notation, this is represented as: \(_{89}^{217}Ac\), in which 89 is the atomic number (number of protons) and 217 is the mass number (number of protons + neutrons).
02

Writing The Alpha Decay Equation

The symbol for an alpha particle is \(_2^4He\). Knowing that the atomic number decreases by 2 and the mass number by 4 during alpha decay, the equation for the decay of Actinium-217 should look like this: \[_{89}^{217}Ac \rightarrow _{2}^{4}He + _{(89-2)}^{(217-4)}X\].
03

Identifying The Element Formed

After isotopic notation, we find that \(_{89}^{217}Ac \rightarrow _{2}^{4}He + _{87}^{213}X\). Here, \(X\) represents the new element. We can find this element in the periodic table by looking for the atomic number 87, which corresponds to Francium (Fr). So, the equation should now look like \[_{89}^{217}Ac \rightarrow _{2}^{4}He + _{87}^{213}Fr.\] Thus, the element formed is Francium-213
04

Checking The Equation

We need to ensure that the total atomic numbers and mass numbers on both the reactant side and product side are equal, satisfying the law of conservation of mass and energy. When we add up the atomic numbers on both sides, we get 89, and when we add up the mass number, we get 217. Thus, the alpha decay equation of Actinium-217 checks out perfectly.

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!

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

What is a quark?

Why is the constant rate of decay of radioac- tive nuclei so important in radioactive dating?

Complete and balance the following nuclear equations: \begin{equation}a. \quad_{75}^{187} \mathrm{Re}+? \longrightarrow_{75}^{188} \mathrm{Re}+_{1}^{1} \mathrm{H}\end{equation} \begin{equation}b.\quad_{4}^{9} \mathrm{Be}+_{2}^{4} \mathrm{He} \longrightarrow ?+_{0}^{1} n \end{equation} \begin{equation}c.\quad_{11}^{22} \mathrm{Na}+? \longrightarrow_{10}^{22} \mathrm{Ne}\end{equation}

How many milligrams remain of a 15.0 \(\mathrm{mg}\) sample of radium-226 after 6396 \(\mathrm{y} ?\) The half-life of this isotope is 1599 \(\mathrm{y}\) .

What is the relationship between an alpha particle and a helium nucleus?
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Chemical Analysis

Read Explanation

Chemical Reactions

Read Explanation

Making Measurements

Read Explanation

The Earths Atmosphere

Read Explanation

Inorganic Chemistry

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