Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 18: Problem 70

It takes about \(10^{6}\) y for just half the samar- ium-149 in nature to decay by alpha-particle emission. Write the decay equation, and find the isotope that is produced by the reaction.

Short Answer

Expert verified
The decay equation is \(^{149}_{62}Sm → ^{4}_{2}He + ^{145}_{60}Nd\). The isotope produced in the decay reaction is Neodymium-145.

Step by step solution

01

Writing the Decay Equation

Samarium-149 undergoes alpha decay. An alpha particle is essentially a helium nucleus and therefore can be represented as \(^{4}_{2}He\). The decay reaction can be written as follows: \(^{149}_{62}Sm → ^{4}_{2}He + X\). Here, \(X\) represents the nucleus produced in the decay process.
02

Determine the Produced Isotope

Nuclear reactions obey conservation of mass and charge. So, the sum of the mass numbers (atomic masses) on the right side should equal to the mass number on the left, and the sum of atomic numbers (charges) on the right should equal to the atomic number on the left. Therefore, \(149 = 4 + A\) and \(62 = 2 + Z\), where \(A\) is the mass number and \(Z\) is the atomic number of the new nucleus. Solving these equations gives \(A = 145\) and \(Z = 60\). The isotope produced by the reaction is therefore \(^{145}_{60}Nd\), which is Neodymium-145.

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 the relationship between mass defect and binding energy?

Is the decay of an unstable isotope into a stable isotope always a one-step process? Explain.

Explain why charged particles do not pene- trate matter deeply.

Medium-mass nuclei have larger binding energies per nucleon than heavier nuclei do. What can you conclude from this fact?

Complete the following nuclear reactions. \begin{equation}a. _{5}^{12} \mathrm{B} \rightarrow_{6}^{12} \mathrm{C}+?\end{equation} \begin{equation}b. _{89}^{225} \mathrm{Ac} \longrightarrow_{87}^{221} \mathrm{Fr}+?\end{equation} \begin{equation}\mathbf{c.}_{28}^{63} \mathrm{Ni} \longrightarrow ?+_{-1}^{0} e\end{equation} \begin{equation}\mathbf{d} \cdot_{83}^{212} \mathrm{Bi} \rightarrow ?+_{2}^{4} \mathrm{He}\end{equation}
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

Organic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Nuclear 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