Chapter 10, Problem 39P

Interpretation Introduction

(a)

Interpretation:

A balanced nuclear equation for decay of thorium-232 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 particle ( ${}_2^4\text{He}$ ) is termed as alpha emission.

Answer to Problem 39P

The complete nuclear reaction for decay of thorium-232 by a emission is represented as follows:

  ${}_{\text{90}}^{\text{232}}\text{Th}\to {}_{88}^{228}\text{Ra+}{}_2^4\text{He}$

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:

  ${}_{\text{90}}^{\text{232}}\text{Th}\to ?\text{+}{}_2^4\text{He}$

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 $232-4=228$

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:

  ${}_{\text{90}}^{\text{232}}\text{Th}\to {}_{88}^{228}\text{Ra+}{}_2^4\text{He}$

Interpretation Introduction

(b)

Interpretation:

A balanced nuclear equation for decay of sodium-25 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 39P

The complete nuclear reaction for decay of sodium-25 by β emission is represented as follows:

  ${}_{\text{11}}^{\text{25}}\text{Na}\to {}_{12}^{25}\text{Mg+}{}_{-1}^0\text{e}$

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:

  ${}_{\text{11}}^{\text{25}}\text{Na}\to {}_{12}^{25}\text{Mg+}{}_{-1}^0\text{e}$

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 having -1 charge thus atomic number of new nucleus will be obtained by adding 1 to the atomic number of original nucleus $11+1=12$

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

The element having atomic number 12 is magnesium.

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

  ${}_{\text{11}}^{\text{25}}\text{Na}\to {}_{12}^{25}\text{Mg+}{}_{-1}^0\text{e}$

Interpretation Introduction

(c)

Interpretation:

A balanced nuclear equation for decay of xenon-118 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 ( ${}_{+1}^0\text{e}$ ) is termed as positron emission.

Answer to Problem 39P

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

  ${}_{\text{54}}^{\text{118}}\text{Xe}\to {}_{53}^{118}\text{I+}{}_{+1}^0\text{e}$

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:

  ${}_{\text{54}}^{\text{118}}\text{Xe}\to ?\text{+}{}_{+1}^0\text{e}$

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 54. Thus, the atomic number of new nuclei will be $54-1=53$

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

The element having atomic number 53 is Iodine(I).

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

  ${}_{\text{54}}^{\text{118}}\text{Xe}\to {}_{53}^{118}\text{I+}{}_{+1}^0\text{e}$

Interpretation Introduction

(d)

Interpretation:

A balanced nuclear equation for decay of curium-243 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 particle ( ${}_2^4\text{He}$ ) is termed as alpha emission.

Answer to Problem 39P

The complete nuclear reaction for decay of curium through a emission is represented as follows:

  ${}_{96}^{243}\text{Cm}\to {}_{94}^{239}\text{Pu+}{}_2^4\text{He}$

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:

  ${}_{96}^{243}\text{Cm}\to ?\text{+}{}_2^4\text{He}$

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 $96-2=94$

Mass number: The sum of mass number on both side must be equal in a nuclear equation. Since, the particle emitted during decay of curium is having mass number 4. Thus, mass number of new nuclei will be obtained by subtracting 4 from atomic number of curium as $243-4=239$

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

The element having atomic number 94 is Plutonium (Pu).

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

  ${}_{96}^{243}\text{Cm}\to {}_{94}^{239}\text{Pu+}{}_2^4\text{He}$