Chapter 18, Problem 10PE

(a)

Interpretation Introduction

Interpretation:

Type of emission occurs in transition from ${}_{82}^{210}\text{Pb}$ to ${}_{82}^{210}\text{Pb}$ 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:

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

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 10PE

Type of emission that occurs in transition from ${}_{82}^{210}\text{Pb}$ to ${}_{82}^{210}\text{Pb}$ is gamma emission.

Explanation of Solution

Skeleton equation for transition from ${}_{82}^{210}\text{Pb}$ to ${}_{82}^{210}\text{Pb}$  is as follows:

  ${}_{82}^{210}\text{Pb}{\to }_{82}^{210}\text{Pb}{+}_Z^A\text{X}$

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 ${}_{82}^{210}\text{Pb}$ is as follows:

  ${}_{82}^{210}\text{Pb}{\to }_{82}^{210}\text{Pb}+\text{energy}$

(b)

Interpretation Introduction

Interpretation:

Type of emissions occurs in transition from ${}_{91}^{234}\text{Pa}$ to ${}_{89}^{230}\text{Ac}$ and then ${}_{89}^{230}\text{Ac}$ to ${}_{90}^{230}\text{Th}$ have to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 10PE

Type of emissions occurs in transition from ${}_{91}^{234}\text{Pa}$ to ${}_{89}^{230}\text{Ac}$ and then ${}_{89}^{230}\text{Ac}$ to ${}_{90}^{230}\text{Th}$ are alpha and beta emission respectively.

Explanation of Solution

Skeleton equation for transition from ${}_{91}^{234}\text{Pa}$ to ${}_{89}^{230}\text{Ac}$ is as follows:

  ${}_{91}^{234}\text{Pa}{\to }_{89}^{230}\text{Ac}{+}_Z^A\text{X}$

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 ${}_{91}^{234}\text{Pa}$ is as follows:

  ${}_{91}^{234}\text{Pa}{\to }_{89}^{230}\text{Ac}{+}_2^4\text{He}$

Skeleton equation for transition from ${}_{89}^{230}\text{Ac}$ to ${}_{90}^{230}\text{Th}$ is as follows:

  ${}_{89}^{230}\text{Ac}{\to }_{90}^{230}\text{Th}{+}_Z^A\text{X}$

Since this transition causes atomic number to increase by 1 thus this transition is called beta emission. Resultant nuclear equation for beta emission of ${}_{89}^{230}\text{Ac}$ is as follows:

  ${}_{89}^{230}\text{Ac}{\to }_{90}^{230}\text{Th}{+}_{-1}^0\text{e}$

(c)

Interpretation Introduction

Interpretation:

Type of emissions occurs in transition from ${}_{90}^{234}\text{Th}$ to ${}_{88}^{230}\text{Ra}$ and then ${}_{88}^{230}\text{Ra}$ to ${}_{88}^{230}\text{Ra}$ have to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 10PE

Type of emissions occurs in transition from  ${}_{90}^{234}\text{Th}$ to ${}_{88}^{230}\text{Ra}$ and then ${}_{88}^{230}\text{Ra}$ to ${}_{88}^{230}\text{Ra}$ are alpha and gamma emission respectively.

Explanation of Solution

Skeleton equation for transition from ${}_{90}^{234}\text{Th}$ to ${}_{88}^{230}\text{Ra}$ is as follows:

  ${}_{90}^{234}\text{Th}{\to }_{88}^{230}\text{Ra}{+}_Z^A\text{X}$

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 ${}_{90}^{234}\text{Th}$ is as follows:

  ${}_{90}^{234}\text{Th}{\to }_{88}^{230}\text{Ra}{+}_2^4\text{He}$

Skeleton equation for transition from ${}_{88}^{230}\text{Ra}$ to ${}_{88}^{230}\text{Ra}$  is as follows:

  ${}_{88}^{230}\text{Ra}{\to }_{88}^{230}\text{Ra}{+}_Z^A\text{X}$

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 ${}_{88}^{230}\text{Ra}$ is as follows:

  ${}_{88}^{230}\text{Ra}{\to }_{88}^{230}\text{Ra}+\text{energy}$