Chapter 18, Problem 9PE

(a)

Interpretation Introduction

Interpretation:

Type of emission occurs in transition from ${}_{88}^{226}\text{Ra}$ to ${}_{86}^{222}\text{Rn}$ 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.

Gamma rays $\left(\text{γ}\right)$ are energy photons with more energy than X-rays. Loss of gamma ray causes no change in mass number as well as atomic number.

Answer to Problem 9PE

Type of emission that occurs in transition from ${}_{88}^{226}\text{Ra}$ to ${}_{86}^{222}\text{Rn}$ is alpha emission.

Explanation of Solution

Skeleton equation for transition from ${}_{88}^{226}\text{Ra}$ to ${}_{86}^{222}\text{Rn}$ is as follows:

  ${}_{88}^{226}\text{Ra}{\to }_{86}^{222}\text{Rn}{+}_Z^A\text{X}$

Loss of alpha particle results in decrease of mass number by 4 and atomic number by 2.

Since transition causes loss in mass number by 4 and in atomic number by 2 hence 1 alpha particle is lost in this transition. Therefore alpha emission is taken place in this transition.

Hence resultant nuclear equation for alpha emission of ${}_{88}^{226}\text{Ra}$ is as follows:

  ${}_{88}^{226}\text{Ra}{\to }_{86}^{222}\text{Rn}{+}_2^4\text{He}$

(b)

Interpretation Introduction

Interpretation:

Type of emissions occurs in transition from ${}_{88}^{222}\text{Rn}$ to ${}_{87}^{222}\text{Fr}$ and then ${}_{87}^{222}\text{Fr}$ to ${}_{87}^{222}\text{Fr}$ have to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 9PE

Type of emissions occurs in transition from ${}_{88}^{222}\text{Rn}$ to ${}_{87}^{222}\text{Fr}$ and then ${}_{87}^{222}\text{Fr}$ to ${}_{87}^{222}\text{Fr}$ are beta and gamma emission respectively.

Explanation of Solution

Skeleton equation for transition from ${}_{88}^{222}\text{Rn}$ to ${}_{87}^{222}\text{Fr}$  is as follows:

  ${}_{88}^{222}\text{Rn}{\to }_{87}^{222}\text{Fr}{+}_Z^A\text{X}$

Since this transition causes atomic number to increase by 1 thus this transition is called beta emission. Resultant nuclear equation for beta emission of ${}_{88}^{226}\text{Ra}$ is as follows:

  ${}_{88}^{222}\text{Rn}{\to }_{87}^{222}\text{Fr}{+}_{-1}^0\text{e}$

Skeleton equation for transition from ${}_{87}^{222}\text{Fr}$ to ${}_{87}^{222}\text{Fr}$  is as follows:

  ${}_{87}^{222}\text{Fr}{\to }_{87}^{222}\text{Fr}{+}_Z^A\text{X}$

Since this transition cause no change in atomic number as well as in mass number thus this transition is gramma emission. Resultant nuclear equation for gamma emission of ${}_{87}^{222}\text{Fr}$ is as follows:

  ${}_{87}^{222}\text{Fr}{\to }_{87}^{222}\text{Fr}+\text{energy}$

(c)

Interpretation Introduction

Interpretation:

Type of emission occurs in transition from ${}_{92}^{238}\text{U}$ to ${}_{93}^{238}\text{Np}$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 9PE

Type of emission that occurs in transition from ${}_{92}^{238}\text{U}$ to ${}_{93}^{238}\text{Np}$ is beta emission.

Explanation of Solution

Skeleton equation for transition from ${}_{92}^{238}\text{U}$ to ${}_{93}^{238}\text{Np}$  is as follows:

  ${}_{92}^{238}\text{U}{\to }_{93}^{238}\text{Np}{+}_Z^A\text{X}$

Since this transition causes atomic number to increase by 1 thus this transition is called beta emission. Resultant nuclear equation for beta emission of ${}_{92}^{238}\text{U}$ is as follows:

  ${}_{92}^{238}\text{U}{\to }_{93}^{238}\text{Np}{+}_{-1}^0\text{e}$