Chapter 21, Problem 21.78QP

Interpretation Introduction

Interpretation:

Nuclear equation for each step in the decay of thorium isotope has to be given.

Concept Introduction:

Nuclear reaction is a physical process in which there is a change in identity of an atomic nucleus. Natural radioactive decays, artificial radioactive decays... are considered as nuclear reactions because these processes make changes in the identity of an atomic nucleus.

Common particles in radioactive decay and nuclear transformations are mentioned below,

$\begin{array}{cc}\text{Particle}& \text{Symbol}\\ \text{Proton}& {}_{\text{1}}^{\text{1}}\text{H}\text{or}{}_{\text{1}}^{\text{1}}\text{P}\\ \text{Neutron}& {}_{\text{0}}^{\text{1}}\text{n}\\ \text{Electron}& {}_{\text{-1}}^{\text{0}}\text{e}\\ \text{Alpha}\text{particle}& {}_{\text{2}}^{\text{4}}\text{He}\text{or}{}_{\text{2}}^{\text{4}}\text{α}\\ \text{Beta}\text{particle}& {}_{\text{-1}}^{\text{0}}\text{e}\text{or}{}_{\text{-1}}^{\text{0}}\text{β}\\ \text{Positron}& {}_{\text{1}}^{\text{0}}\text{e}\end{array}$

There are various types of nuclear processes. The changes in atomic number and mass number accompanying radioactive decay are mentioned below,

$\begin{array}{cccccc}\text{Type}\text{of}\text{Decay}& \text{Symbol}& \text{charge}& \text{Mass}& \begin{array}{l}\text{Change}\\ \text{in}\text{atomic}\text{number}\end{array}& \begin{array}{l}\text{Change}\\ \text{in}\text{mass}\text{number}\end{array}\\ \text{Beta}& {}_{\text{-1}}^{\text{0}}\text{e}\text{or}{}_{\text{-1}}^{\text{0}}\text{β}& \text{-1}& \text{0}& \text{+1}& \text{none}\\ \text{Positron}& {}_{\text{+1}}^{\text{0}}\text{e}& \text{+1}& \text{0}& \text{-1}& \text{none}\\ \text{Alpha}& {}_{\text{2}}^{\text{4}}\text{He}\text{or}{}_{\text{2}}^{\text{4}}\text{α}& \text{+2}& \text{4}& \text{-2}& \text{-4}\\ \text{Gamma}& {}_{\text{0}}^{\text{0}}\text{γ}& \text{0}& \text{0}& \text{none}& \text{none}\end{array}$

An isotope can be represented using ${}_Z^{A}X$ symbol

$\begin{array}{l}\text{Z}\to \text{Atomic}\text{number}=\text{Number}\text{of}\text{proton}\\ \text{A}\to \text{Mass}\text{number}=\text{Number}\text{of}\text{protons}\text{+}\text{Number}\text{of}\text{neutrons}\end{array}$