Chapter 18, Problem 16RQ

Interpretation Introduction

Interpretation:

Nuclide formed after emission of seven $\text{α}$ particles and four $\text{β}$ particles 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.