Chapter 18, Problem 15RQ

Interpretation Introduction

Interpretation:

Formation of $\text{P}\text{b}$ from $\text{T}\text{h}$ by successive $\text{α}$, $\text{β}$, $\text{β}$, $\text{α}$, $\text{α}$, $\text{α}$, $\text{α}$, $\text{β}$, $\text{β}$, $\text{α}$ particle emissions has to be determined.

Concept Introduction:

Alpha particle $\left(\text{α}\right)$ represents nucleus of helium atom. It contains two protons and two neutrons. Emission of this alpha particle is called alpha decay. Symbol of alpha particle is ${}_2^4\text{He}$.

General equation for alpha decay is as follows:

  ${}_Z^A\text{X}{\to }_{Z-2}^{A-4}\text{Y}{+}_2^4\text{He}$

Here,

$A$ is the mass or nucleon number.

$Z$ is the atomic number.

$\text{X}$ is the symbol of the element.

Beta emission $\left({\text{β}}^{-}\right)$ process represents conversion of neutron into 1 proton and 1 electron. This proton stays within nucleus and electron is emitted out of atom. This electron is called the beta particle $\left({}_{-1}^0\text{e}\right)$.

The generic equation of the beta decay is as follows:

  ${}_Z^A\text{X}{\to }_{Z+1}^A\text{Y}{+}_{-1}^0\text{e}$

Here,

$A$ is the mass or nucleon number.

$Z$ is the atomic number.

$\text{X}$ and Y is the symbol of the element.

Explanation of Solution

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

    $\text{T}\text{h}\to \text{X}{+}_2^4\text{He}$

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:

    $\text{R}\text{a}\to \text{X}{+}_{-1}^0\text{e}$

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:

    $\text{A}\text{c}\to \text{X}{+}_{-1}^0\text{e}$

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:

    $\text{T}\text{h}\to \text{X}{+}_2^4\text{He}$

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:

    $\text{R}\text{a}\to \text{X}{+}_2^4\text{He}$

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:

    $\text{R}\text{n}\to \text{X}{+}_2^4\text{He}$

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

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

    $\text{P}\text{o}\to \text{X}{+}_2^4\text{He}$

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

In the 8^th step $\text{P}\text{b}$ emits $\text{β}$ particle and reaction is as follows:

    $\text{P}\text{b}\to \text{X}{+}_{-1}^0\text{e}$

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:

    $\text{B}\text{i}\to \text{X}{+}_{-1}^0\text{e}$

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

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

    $\text{P}\text{o}\to \text{X}{+}_2^4\text{He}$

Here, $\text{X}$ is lead. So, nuclide symbol will be $\text{P}\text{b}$. So, final nuclide in the disintegration series will be $\text{P}\text{b}$.