Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Fill in the blanks in these radioactive decay series: (a) \(^{232} \mathrm{Th} \stackrel{\alpha}{\longrightarrow}\) _______ \(\stackrel{\beta}{\longrightarrow}\) ________ \(\stackrel{\beta}{\longrightarrow}{ }^{228} \mathrm{Th}\) (b) \({ }^{235} \mathrm{U} \stackrel{\alpha}{\longrightarrow}\) ________ \(\stackrel{\beta}{\longrightarrow}\) _________ \(\stackrel{\alpha}{\longrightarrow}^{227} \mathrm{Ac}\) (c) _______ \(\stackrel{\alpha}{\longrightarrow}{ }^{233} \mathrm{~Pa} \stackrel{\beta}{\longrightarrow}\) ___________ \(\stackrel{\alpha}{\longrightarrow}\) ________.

Short Answer

Expert verified
The filled decay series are: (a) \(^{232} \mathrm{Th} \stackrel{\alpha}{\longrightarrow}{ }^{228} \mathrm{Ra}\)\(\stackrel{\beta}{\longrightarrow}{ }^{228} \mathrm{Fr}\) \(\stackrel{\beta}{\longrightarrow}{ }^{228} \mathrm{Th}\), (b) \({ }^{235} \mathrm{U} \stackrel{\alpha}{\longrightarrow}{ }^{231} \mathrm{Th}\) \(\stackrel{\beta}{\longrightarrow}{ }^{231} \mathrm{Pa}\) \(\stackrel{\alpha}{\longrightarrow}^{227} \mathrm{Ac}\), (c) \({ }^{237} \mathrm{Th} \stackrel{\alpha}{\longrightarrow}{ }^{233} \mathrm{~Pa} \stackrel{\beta}{\longrightarrow}{ }^{233} \mathrm{Th}\) \(\stackrel{\alpha}{\longrightarrow}{ }^{229} \mathrm{Fr}\)

Step by step solution

01

Solve for (a)

In alpha decay, atomic number decreases by 2 and atomic mass decreases by 4. Thus, after the alpha decay of Thorium 232, we get Radon 228.After the beta decay, the atomic number increases by one and we get Francium 228.After the second beta decay, we again increase the atomic number by one so we finally get Thorium 228. Hence, the series is: \(^{232} \mathrm{Th} \stackrel{\alpha}{\longrightarrow}{ }^{228} \mathrm{Ra}\)\(\stackrel{\beta}{\longrightarrow}{ }^{228} \mathrm{Fr}\) \(\stackrel{\beta}{\longrightarrow}{ }^{228} \mathrm{Th}\)
02

Solve for (b)

Using the same logic we infer that after alpha decay from Uranium 235, we get Thorium 231.After the beta decay, the atomic number increases by one so, we get Protactinium 231.After another alpha decay, we get Actinium 227. Hence, the series is: \({ }^{235} \mathrm{U} \stackrel{\alpha}{\longrightarrow}{ }^{231} \mathrm{Th}\) \(\stackrel{\beta}{\longrightarrow}{ }^{231} \mathrm{Pa}\) \(\stackrel{\alpha}{\longrightarrow}^{227} \mathrm{Ac}\)
03

Solve for (c)

The series starts with an alpha decay. So we increase the atomic number by 2 and the mass by 4 to the Pa-233, obtaining Thorium 237.The Thorium 237 then undergoes a beta decay, transforming into Protactinium 237. Lastly, after another alpha decay, the result is Francium 233. Hence, the series is: \({ }^{237} \mathrm{Th} \stackrel{\alpha}{\longrightarrow}{ }^{233} \mathrm{~Pa} \stackrel{\beta}{\longrightarrow}{ }^{233} \mathrm{Th}\) \(\stackrel{\alpha}{\longrightarrow}{ }^{229} \mathrm{Fr}\).

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.

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 the world of nuclear physics, alpha decay is a type of radioactive decay where an alpha particle is emitted from a heavy nucleus. An alpha particle is made up of 2 protons and 2 neutrons, essentially forming a helium nucleus. During alpha decay, the parent nucleus loses two protons and two neutrons, resulting in a decrease in atomic number by 2 and atomic mass by 4. This transforms the original atom into a new element entirely. For example, in the decay of thorium-232: - Thorium-232 undergoes alpha decay. - It transforms into radium-228. It's essential to understand this process because it explains how elements change over time naturally. Alpha decay often occurs among heavier elements like uranium and thorium because they are unstable and seek lower energy states by releasing alpha particles.
Beta Decay
Beta decay is another form of radioactive decay in which a beta particle—either an electron or a positron—is emitted from an atomic nucleus. This process changes a neutron to a proton or a proton to a neutron, leading to an increase or decrease in the atomic number by one, respectively. There are two types of beta decay: - **Beta-minus decay**: A neutron turns into a proton, emitting an electron and an antineutrino. The atomic number increases by 1 since a new proton is formed. - **Beta-plus decay**: A proton turns into a neutron, emitting a positron and a neutrino, decreasing the atomic number by 1. In the decay series exercise given, during a beta decay: - The atomic number of the nucleus rises by one. - For example, Radon-228 changes to Francium-228. Understanding beta decay is crucial because it acts as a counter mechanism to alpha decay, continually altering the nuclei of elements, which helps explain the variety of elements we see around us.
Thorium Decay Series
The thorium decay series is a series of radioactive decay reactions beginning with thorium-232 and ending in a stable isotope. The decay paths involve a combination of alpha and beta decays which gradually lead to the formation of stable lead isotopes. Here's how the series progresses: - **Begins with Thorium-232**: Undergoes successive alpha decay. - **Transformations**: Each transformation results in a different element with its own properties. - **Continues through various elements**: It includes elements like radium, radon, and eventually ends with lead. This series is significant in the study of radioactive substances and nuclear chemistry because it represents natural transmutations that occur over long periods. Thorium series is also immensely important in geological dating and tracing studies, helping scientists understand Earth's formation and the age of different rocks and minerals.

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 radioactive isotope \({ }^{238} \mathrm{Pu},\) used in pacemakers, decays by emitting an alpha particle with a half-life of 86 yr. (a) Write an equation for the decay process. (b) The energy of the emitted alpha particle is \(9.0 \times 10^{-13} \mathrm{~J}\), which is the energy per decay. Assume that all the alpha particle energy is used to run the pacemaker, calculate the power output at \(t=0\) and \(t=10 \mathrm{yr}\). Initially \(1.0 \mathrm{mg}\) of \({ }^{238} \mathrm{Pu}\) was present in the pacemaker (Hint: After \(10 \mathrm{yr}\), the activity of the isotope decreases by 8.0 percent. Power is measured in watts or \(\mathrm{J} / \mathrm{s}\).).

Consider the decay series \(\mathrm{A} \longrightarrow \mathrm{B} \longrightarrow \mathrm{C} \longrightarrow \mathrm{D}\) where \(A, B,\) and \(C\) are radioactive isotopes with halflives of \(4.50 \mathrm{~s}, 15.0\) days, and \(1.00 \mathrm{~s},\) respectively, and \(\mathrm{D}\) is nonradioactive. Starting with 1.00 mole of A, and none of \(\mathrm{B}, \mathrm{C},\) or \(\mathrm{D},\) calculate the number of moles of \(\mathrm{A}, \mathrm{B}, \mathrm{C},\) and \(\mathrm{D}\) left after 30 days.

Consider this redox reaction: $$ \begin{array}{r} \mathrm{IO}_{4}^{-}(a q)+2 \mathrm{I}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \\ \mathrm{I}_{2}(s)+\mathrm{IO}_{3}^{-}(a q)+2 \mathrm{OH}^{-}(a q) \end{array} $$ When \(\mathrm{KIO}_{4}\) is added to a solution containing iodide ions labeled with radioactive iodine- \(128,\) all the radioactivity appears in \(\mathrm{I}_{2}\) and none in the \(\mathrm{IO}_{3}^{-}\) ion. What can you deduce about the mechanism for the redox process?

In 1997 a scientist at a nuclear research center in Russia placed a thin shell of copper on a sphere of highly enriched uranium-235. Suddenly, there was a huge burst of radiation, which turned the air blue. Three days later, the scientist died of radiation damage. Explain what caused the accident. (Hint: Copper is an effective metal for reflecting neutrons.)

For each pair of elements listed, predict which one has more stable isotopes: (a) Co or \(\mathrm{Ni},\) (b) \(\mathrm{F}\) or \(\mathrm{Se}\) (c) Ag or Cd.

See all solutions

Recommended explanations on Chemistry Textbooks

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