Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 18: Problem 22

The Th-232 disintegration series starts with \(\frac{232}{90} \mathrm{Th}\) and emits the following rays successively: \(\alpha, \beta, \beta, \alpha, \alpha, \alpha, \beta, \alpha, \beta, \alpha\). The series ends with the stable \({ }_{82}^{208} \mathrm{~Pb}\). Write the formula for each nuclide in the series.

Short Answer

Expert verified
The nuclides in order are: Th-232, Ra-228, Ac-228, Th-228, Ra-224, Rn-220, Po-216, At-216, Bi-212, Po-212, Pb-208.

Step by step solution

01

- Understand the decay types

Recall that an alpha \( \alpha \) decay decreases the atomic number by 2 and the mass number by 4. A beta \( \beta \) decay increases the atomic number by 1 while keeping the mass number unchanged.
02

- Apply the first alpha decay

Starting with \( \frac{232}{90} \mathrm{Th} \), applying alpha decay: \[ \frac{232}{90} \mathrm{Th} \rightarrow \frac{228}{88} \mathrm{Ra} + \alpha \]
03

- Apply the first beta decay

From \( \frac{228}{88} \mathrm{Ra} \), applying beta decay: \[ \frac{228}{88} \mathrm{Ra} \rightarrow \frac{228}{89} \mathrm{Ac} + \beta \]
04

- Apply the second beta decay

From \( \frac{228}{89} \mathrm{Ac} \), applying another beta decay: \[ \frac{228}{89} \mathrm{Ac} \rightarrow \frac{228}{90} \mathrm{Th} + \beta \]
05

- Apply the second alpha decay

From \( \frac{228}{90} \mathrm{Th} \), applying alpha decay: \[ \frac{228}{90} \mathrm{Th} \rightarrow \frac{224}{88} \mathrm{Ra} + \alpha \]
06

- Apply the third alpha decay

From \( \frac{224}{88} \mathrm{Ra} \), applying alpha decay: \[ \frac{224}{88} \mathrm{Ra} \rightarrow \frac{220}{86} \mathrm{Rn} + \alpha \]
07

- Apply the fourth alpha decay

From \( \frac{220}{86} \mathrm{Rn} \), applying alpha decay: \[ \frac{220}{86} \mathrm{Rn} \rightarrow \frac{216}{84} \mathrm{Po} + \alpha \]
08

- Apply the third beta decay

From \( \frac{216}{84} \mathrm{Po} \), applying beta decay: \[ \frac{216}{84} \mathrm{Po} \rightarrow \frac{216}{85} \mathrm{At} + \beta \]
09

- Apply the fifth alpha decay

From \( \frac{216}{85} \mathrm{At} \), applying alpha decay: \[ \frac{216}{85} \mathrm{At} \rightarrow \frac{212}{83} \mathrm{Bi} + \alpha \]
10

- Apply the fourth beta decay

From \( \frac{212}{83} \mathrm{Bi} \), applying beta decay: \[ \frac{212}{83} \mathrm{Bi} \rightarrow \frac{212}{84} \mathrm{Po} + \beta \]
11

- Apply the sixth alpha decay

From \( \frac{212}{84} \mathrm{Po} \), applying alpha decay: \[ \frac{212}{84} \mathrm{Po} \rightarrow \frac{208}{82} \mathrm{Pb} + \alpha \]

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.

alpha decay
In alpha decay, the nucleus of an atom loses an alpha particle. An alpha particle consists of 2 protons and 2 neutrons. When an atom undergoes alpha decay, its atomic number decreases by 2 and its mass number decreases by 4. For example, during the first step of the Th-232 disintegration series: \( \frac{232}{90} \text{Th} \rightarrow \frac{228}{88} \text{Ra} + \text{alpha particle} \). This means the thorium-232 converts into radium-228, losing an alpha particle in the process.
beta decay
Beta decay occurs when a neutron in an atom’s nucleus converts into a proton, emitting a beta particle (an electron). Unlike alpha decay, beta decay increases the atomic number by one while keeping the mass number unchanged. As shown in step 3 of the exercise: \( \frac{228}{88} \text{Ra} \rightarrow \frac{228}{89} \text{Ac} + \text{beta particle} \). Here, radium-228 becomes actinium-228 after a beta decay.
radioactive decay series
A radioactive decay series is a sequence of radioactive decays that certain radioactive elements go through until they reach a stable nuclide. Each decay produces a new element, which may itself be radioactive. The Th-232 disintegration series, for example, starts with thorium-232 and goes through multiple decays (both alpha and beta) to finally become lead-208, a stable end product. Understanding each step in this sequence helps us see how heavy elements break down gradually into stable forms.
nuclide notation
Nuclide notation is a way of representing different isotopes and their properties. It shows the mass number (total protons and neutrons) at the top left and the atomic number (total protons) at the bottom left of the element symbol. For instance, \( \frac{232}{90} \text{Th} \) represents thorium-232: 232 is the mass number, and 90 is the atomic number. Throughout the Th-232 disintegration series, nuclide notation helps us track the changes in atomic and mass numbers through each decay step.

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

The nuclide \(\frac{223}{8}\) Fr emits three particles losing 8 mass units and 3 atomic number units. Propose a radioactive decay series and write the symbol for the resulting nuclide.

When the nuclide \({ }_{98}^{249} \mathrm{Cf}\) was bombarded with \({ }_{7}^{15} \mathrm{~N}\), four neutrons and a new transuranium element were formed. Write the nuclear equation for this transmutation.

Strontium-90 has a half-life of 28 years. If a sample was tested in 1980 and found to be emitting 240 counts/min, in what year would the same sample be found to be emitting 30 counts/min? How much of the original Sr-90 would be left?

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.

The \({ }_{6}^{14} \mathrm{C}\) content of an ancient piece of wood was found to be one-sixteenth of that in living trees. How many years old is this piece of wood if the half-life of carbon-14 is 5730 years?
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Chemical Analysis

Read Explanation

Physical Chemistry

Read Explanation

Chemical Reactions

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