Chapter 18, Problem 6PE

(a)

Interpretation Introduction

Interpretation:

Nuclear equation for alpha decay of ${}_{90}^{238}\text{Th}$ 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.

(b)

Interpretation Introduction

Interpretation:

Nuclear equation for alpha decay of ${}_{94}^{239}\text{Pu}$ has to be determined.

Concept Introduction:

Refer to part (a).