Chapter 18, Problem 22AE

Interpretation Introduction

Interpretation:

The formula for all nuclides in $\text{Th-232}$ disintegration series has to be written.

Concept Introduction:

Radioactivity is a nuclear process. It is defined as the spontaneous disintegration of a nucleus which is done by the emission of radiation. The elements which show this property is said to be radioactive elements. Radioactive elements undergo decay by emission of alpha particles and beta particles.

An alpha particle emission causes a decrease of atomic number by 2 and mass number by 4 as the alpha particle is a helium nuclide with two protons and two neutrons whereas emission of beta particle results in an increase of atomic number by 1.

Beta particle is represented as $\text{β}$. Loss of beta particle results in no change in mass number and increase of atomic number by 1.

Explanation of Solution

In the 1^st step, $\text{T}\text{h}$ emits $\text{α}$ particle and reaction is as follows:

    $\text{T}\text{h}\stackrel{\text{α}}{\to }\text{X}$

Here, $\text{X}$ is radium. So, nuclide symbol is $\text{R}\text{a}$.

In the 2^nd step, $\text{R}\text{a}$ emits $\text{β}$ particle and reaction is as follows:

    $\text{R}\text{a}\stackrel{\text{β}}{\to }\text{X}$

Here, $\text{X}$ is actinium. So, nuclide symbol will be $\text{A}\text{c}$.

In the 3^rd step, $\text{A}\text{c}$ emits $\text{β}$ particle and reaction is as follows:

    $\text{A}\text{c}\stackrel{\beta }{\to }\text{X}$

Here, $\text{X}$ is thorium. So, nuclide symbol will be $\text{T}\text{h}$.

In the 4^th step, $\text{T}\text{h}$ emits $\text{α}$ particle and reaction is as follows:

    $\text{T}\text{h}\stackrel{\alpha }{\to }\text{X}$

Here, $\text{X}$ is radium. So, nuclide symbol will be $\text{R}\text{a}$.

In the 5^th step, $\text{R}\text{a}$ emits $\text{α}$ particle and reaction is as follows:

    $\text{R}\text{a}\stackrel{\alpha }{\to }\text{X}$

Here, $\text{X}$ is radon. So, nuclide symbol will be $\text{R}\text{n}$.

In the 6^th step, $\text{R}\text{n}$ emits $\text{α}$ particle and reaction is as follows:

    $\text{R}\text{n}\stackrel{\alpha }{\to }\text{X}$

Here, $\text{X}$ is polonium. So, nuclide symbol will be $\text{P}\text{o}$.

In the 7^th step $\text{P}\text{o}$ emits $\text{β}$ particle and reaction is as follows:

    $\text{P}\text{o}\stackrel{\beta }{\to }\text{X}$

Here, $\text{X}$ is astatine. So, nuclide symbol will be $\text{A}\text{t}$.

In the 8^th step $\text{A}\text{t}$ emits $\text{α}$ particle and reaction is as follows:

    $\text{A}\text{t}\stackrel{\alpha }{\to }\text{X}$

Here, $\text{X}$ is bismuth. So, nuclide symbol will be $\text{B}\text{i}$.

In the 9^th step $\text{B}\text{i}$ emits $\text{β}$ particle and reaction is as follows:

    $\text{B}\text{i}\stackrel{\beta }{\to }\text{X}$

Here, $\text{X}$ is polonium. So, nuclide symbol will be $\text{P}\text{o}$.

In the 10^th step $\text{P}\text{o}$ $\text{α}$ particle and reaction is as follows:

    $\text{P}\text{o}\stackrel{\alpha }{\to }\text{X}$

Here, $\text{X}$ is lead. So, nuclide symbol will be $\text{P}\text{b}$.

So, final nuclide in the disintegration series will be $\text{P}\text{b}$.