Chapter 10, Problem 10.35P

Interpretation Introduction

(a)

Interpretation:

The following nuclear equation should be completed:

  ${}_{\text{26}}^{\text{59}}\text{Fe}\to \text{?}\text{+}{}_{\text{-1}}^{\text{0}}\text{e}$

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.

Answer to Problem 10.35P

The complete nuclear equation should be represented as follows:

  ${}_{\text{26}}^{\text{59}}\text{Fe}\to {}_{\text{27}}^{\text{59}}\text{Co}\text{+}{}_{\text{-1}}^{\text{0}}\text{e}$

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 on the left and right side respectively as follows:

  ${}_{\text{26}}^{\text{59}}\text{Fe}\to \text{?}\text{+}{}_{\text{-1}}^{\text{0}}\text{e}$

2. Determine atomic number and mass number of particle on left 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 charge -1 and original nuclei has atomic number 26. Thus, atomic number of substance on right side will be obtained by adding 1 to the atomic number of element on left side that is $26+1=27$

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

The element having atomic number 27 is cobalt.

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

  ${}_{\text{26}}^{\text{59}}\text{Fe}\to {}_{\text{27}}^{\text{59}}\text{Co}\text{+}{}_{\text{-1}}^{\text{0}}\text{e}$

Interpretation Introduction

(b)

Interpretation:

The following nuclear equation should be completed:

  ${}_{\text{78}}^{\text{190}}\text{Pt}\to \text{?}\text{+}{}_2^{\text{4}}\text{He}$

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.

Answer to Problem 10.35P

The complete nuclear equation should be represented as follows:

  ${}_{\text{78}}^{\text{190}}\text{Pt}\to {}_{\text{76}}^{\text{186}}\text{Os}\text{+}{}_{\text{2}}^{\text{4}}\text{He}$

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:

  ${}_{\text{78}}^{\text{190}}\text{Pt}\to \text{?}\text{+}{}_2^{\text{4}}\text{He}$

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 platinum that is original nucleus as $78-2=76$

Mass number of new nucleus will be calculated by subtracting 4 from mass number of original nucleus as follows:

  $190-4=186$

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

The element having atomic number 76 is osmium (Os).

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

  ${}_{\text{78}}^{\text{190}}\text{Pt}\to {}_{\text{76}}^{\text{186}}\text{Os}\text{+}{}_{\text{2}}^{\text{4}}\text{He}$

Interpretation Introduction

(c)

Interpretation:

The following nuclear equation should be completed:

  ${}_{\text{80}}^{\text{178}}\text{Hg}\to \text{?}\text{+}{}_{\text{+1}}^{\text{0}}\text{e}$

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.

Answer to Problem 10.35P

The complete nuclear equation should be represented as follows:

  ${}_{\text{80}}^{\text{178}}\text{Hg}\to {}_{\text{79}}^{\text{178}}\text{Au}\text{+}{}_{\text{+1}}^{\text{0}}\text{e}$

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:

  ${}_{\text{80}}^{\text{178}}\text{Hg}\to \text{?}\text{+}{}_{\text{+1}}^{\text{0}}\text{e}$

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 charge +1 and original nuclei has atomic number 80. Thus, atomic number of substance on right side will be obtained by subtracting 1 from the atomic number of element on left side that is $80-1=79$

3. The chemical equation is completed using atomic number of new nuclei.The element having atomic number 79 is gold (Au).

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

  ${}_{\text{80}}^{\text{178}}\text{Hg}\to {}_{\text{79}}^{\text{178}}\text{Au}\text{+}{}_{\text{+1}}^{\text{0}}\text{e}$