Chapter 10, Problem 10.42P

Interpretation Introduction

(a)

Interpretation:

A balanced nuclear equation for decay of sulfur-35 through βemission should be predicted.

Concept Introduction:

The reactions involving formation of new nucleus with emission of some radiation along with it from an original nucleus are known as nuclear reactions.

To write a nuclear reaction the chemical equation for the reaction must be balanced that is the sum of atomic number and mass number on both the side should be equal.

A radioactive process in which nucleus of an atom emits a beta particle ( ${}_{-1}^0\text{e}$ ) is termed as beta emission.

Answer to Problem 10.42P

The complete nuclear reaction for decay of sulfur-35 through β emission is represented as follows:

  ${}_{\text{16}}^{\text{35}}\text{S}\to {}_{17}^{35}\text{Cl+}{}_{\text{-1}}^{\text{0}}\text{e}$

Explanation of Solution

To write a balanced chemical reaction following steps should be followed which are given as follows:

1. Write the incomplete reaction showing original nucleus with atomic number and mass number and the particle emitted on the left and right side respectively as follows:

  ${}_{\text{16}}^{\text{35}}\text{S}\to ?\text{+}{}_{\text{-1}}^{\text{0}}\text{e}$

2. Determine atomic number and mass number or particle emitted on right side as follows:

Atomic number: The sum of atomic number on both side must be equal in a chemical reaction. Since, particle emitted during the reaction is having -1 charge thus atomic number of new nucleus will be obtained by adding 1 to the atomic number of original nucleus $16+1=17$

3. The chemical equation is completed using atomic number of new nuclei.

The element having atomic number 17 is Chlorine.

Now, write the element with its atomic number and mass number to complete the chemical equation as follows:

  ${}_{\text{16}}^{\text{35}}\text{S}\to {}_{17}^{35}\text{Cl+}{}_{\text{-1}}^{\text{0}}\text{e}$

Interpretation Introduction

(b)

Interpretation:

A balanced nuclear equation for decay of thorium-225 through a emission should be predicted.

Concept Introduction:

The reactions involving formation of new nucleus with emission of some radiation along with it from an original nucleus are known as nuclear reactions.

To write a nuclear reaction the chemical equation for the reaction must be balanced that is the sum of atomic number and mass number on both the side should be equal.

A radioactive process in which nucleus of an atom emits a ( ${}_2^4\text{He}$ ) particle is termed as alpha emission.

Answer to Problem 10.42P

The complete nuclear reaction for decay of thorium-225 through a emission is represented as follows:

  ${}_{\text{90}}^{\text{225}}\text{Th}\to {}_{88}^{221}\text{Ra+}{}_2^4\text{He}$

Explanation of Solution

To write a balanced chemical reaction following steps should be followed which are given as follows:

1. Write the incomplete reaction showing original nucleus with atomic number and mass number and the particle emitted on the left and right side respectively as follows:

  ${}_{\text{90}}^{\text{225}}\text{Th}\to ?\text{+}{}_2^4\text{He}$

2. Determine atomic number and mass number or particle emitted on right side as follows:

Atomic number: The sum of atomic number on both side must be equal in a chemical reaction. Since, particle emitted during the reaction is having 2 protons thus atomic number of new nucleus will be obtained by subtracting 2 from atomic number of original nucleus $90-2=88$

Mass number: The sum of mass number on both side must be equal in a nuclear equation. Since, the particle emitted during decay of thorium is having mass number 4. Thus, mass number of new nuclei will be obtained by subtracting 4 from atomic number of thorium as $225-4=221$

3. The chemical equation is completed using atomic number of new nuclei.

The element having atomic number 88 is radium.

Now, write the element with its atomic number and mass number to complete the chemical equation as follows:

  ${}_{\text{90}}^{\text{225}}\text{Th}\to {}_{88}^{221}\text{Ra+}{}_2^4\text{He}$

Interpretation Introduction

(c)

Interpretation:

A balanced nuclear equation for decay of rhodium-93 through positron emission should be predicted.

Concept Introduction:

The reactions involving formation of new nucleus with emission of some radiation along with it from an original nucleus are known as nuclear reactions.

To write a nuclear reaction the chemical equation for the reaction must be balanced that is the sum of atomic number and mass number on both the side should be equal.

A radioactive process in which nucleus of an atom emits a positron ${}_{+1}^0\text{e}$ particle is termed as positron emission.

Answer to Problem 10.42P

The complete nuclear reaction for decay of rhodium by positron emission is represented as follows:

  ${}_{\text{45}}^{\text{93}}\text{Rh}\to {}_{44}^{93}\text{Ru+}{}_{+1}^0\text{e}$

Explanation of Solution

To write a balanced chemical reaction following steps should be followed which are given as follows:

1. Write the incomplete reaction showing original nucleus with atomic number and mass number and the particle emitted on the left and right side respectively as follows:

  ${}_{\text{45}}^{\text{93}}\text{Rh}\to ?\text{+}{}_{+1}^0\text{e}$

2. Determine atomic number and mass number or particle emitted on right side as follows:

Atomic number: The sum of atomic number on both side must be equal in a chemical reaction. Since, particle emitted during the reaction is ${}_{\text{+1}}^{\text{0}}\text{e}$ thus atomic number of new nucleus will be obtained by subtracting 1 from the atomic number of original nucleus 45. Thus, the atomic number of new nuclei will be $45-1=44$

3. The chemical equation is completed using atomic number of new nuclei.

The element having atomic number 44 is Ruthenium (Ru).

Now, write the element with its atomic number and mass number to complete the chemical equation as follows:

  ${}_{\text{45}}^{\text{93}}\text{Rh}\to {}_{44}^{93}\text{Ru+}{}_{+1}^0\text{e}$

Interpretation Introduction

(d)

Interpretation:

A balanced nuclear equation for decay of silver-114 through β emission should be predicted.

Concept Introduction:

The reactions involving formation of new nucleus with emission of some radiation along with it from an original nucleus are known as nuclear reactions.

To write a nuclear reaction the chemical equation for the reaction must be balanced that is the sum of atomic number and mass number on both the side should be equal.

A radioactive process in which nucleus of an atom emits a beta particle ${}_{-1}^0\text{e}$ is termed as beta emission.

Answer to Problem 10.42P

The complete nuclear reaction for decay of silver-114 through β emission is represented as follows:

  ${}_{\text{47}}^{\text{114}}\text{Ag}\to {}_{48}^{114}\text{Cd+}{}_{\text{-1}}^{\text{0}}\text{e}$

Explanation of Solution

To write a balanced chemical reaction following steps should be followed which are given as follows:

1. Write the incomplete reaction showing original nucleus with atomic number and mass number and the particle emitted on the left and right side respectively as follows:

  ${}_{\text{47}}^{\text{114}}\text{Ag}\to ?\text{+}{}_{\text{-1}}^{\text{0}}\text{e}$

2. Determine atomic number and mass number or particle emitted on right side as follows:

Atomic number: The sum of atomic number on both side must be equal in a chemical reaction. Since, particle emitted during the reaction is having -1 charge thus atomic number of new nucleus will be obtained by adding 1 to the atomic number of original nucleus $47+1=48$

3. The chemical equation is completed using atomic number of new nuclei.

The element having atomic number 48 is cadmium (Cd).

Now, write the element with its atomic number and mass number to complete the chemical equation as follows:

  ${}_{\text{47}}^{\text{114}}\text{Ag}\to {}_{48}^{114}\text{Cd+}{}_{\text{-1}}^{\text{0}}\text{e}$