Problem 22 The Th-232 disintegration series... [FREE SOLUTION]

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

- 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.

- Apply the first alpha decay

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

- Apply the first beta decay

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

- Apply the second beta decay

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

- Apply the second alpha decay

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

- Apply the third alpha decay

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

- Apply the fourth alpha decay

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

- Apply the third beta decay

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

- Apply the fifth alpha decay

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

- Apply the fourth beta decay

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

- Apply the sixth alpha decay

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

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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鈥檚 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.

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