Balanced nuclear equations for decay series of thorium-232 has to be written from the given data. Concept Introduction: Nuclear reaction: A nuclear reaction in which a lighter nucleus fuses together into new stable nuclei or a heavier nucleus split into stable daughter nuclei with the release of large amount of energy. Balancing nuclear reaction equation: The balanced nuclear reaction should conserve both mass number and atomic number. The sum of the mass numbers of the reactants should be equal to the sum of mass numbers of the products in the reaction. The sum of atomic numbers (or the atomic charge) of the reactants should be equal to the sum of atomic numbers (or the atomic charge) of the products in the reaction.

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Answer 1

Question

Chapter 24, Problem 24.132P

Interpretation Introduction

Interpretation:

Balanced nuclear equations for decay series of thorium-232 has to be written from the given data.

Concept Introduction:

Nuclear reaction: A nuclear reaction in which a lighter nucleus fuses together into new stable nuclei or a heavier nucleus split into stable daughter nuclei with the release of large amount of energy.

Balancing nuclear reaction equation: The balanced nuclear reaction should conserve both mass number and atomic number.

The sum of the mass numbers of the reactants should be equal to the sum of mass numbers of the products in the reaction.

The sum of atomic numbers (or the atomic charge) of the reactants should be equal to the sum of atomic numbers (or the atomic charge) of the products in the reaction.

Expert Solution & Answer

Explanation of Solution

In the decay series of $232Th$ , the sequence of particles emitted in each step follows as,

$α, β−, β−, α, α, α, α, β−, β−, α$

According to the law of conservation of atomic mass and number, the unknown element can be predicted.

The elemental symbol of an element: $ZAX = Atomic numberMass numberX$

Step (1): Emission of $α$ particle from $232Th$

An alpha particle is emitted and an alpha particle is $24α.$ Atomic number of Thorium is 90.

The given unbalanced nuclear equation is,

$90232Th→ ? + 24α$

Product formed by $α$ decaying of $232Th$ is determined as follows,

$Sum of Atomic number of Reactant = Product90(Th)= x + 2(alpha)x = 88Element with Z = 88 ⇒ Radium(Ra)Sum of Mass number of Reactant = Product90232Th = 82xRa + 4(alpha)x = 228$

The elemental symbol of the unknown element is $88228Ra$ . Therefore, the balanced nuclear equation is,

$90232Th→ 88228Ra + 24α_$

Step (2): Emission of $β−$ particle from $88228Ra$

A $β−$ particle is emitted and a $β−$ particle is $−10β$ .

The given unbalanced nuclear equation is,

$88228Ra→ −10β + ?$

Product formed by $β−$ decaying of $88228Ra$ is determined as follows,

$Sum of Atomic number of Reactant = Product88(Ra) = x+ −1(β−) x = 89Element with Z = 89 ⇒ Actinium(Ac)Sum of Atomic number of Reactant = Product88228Ra = 89xAc + 0(β−)x = 228$

The elemental symbol of the unknown element is $89228Ac$ . Therefore, the balanced nuclear equation is,

$88228Ra→ −10β + 89228Ac_$

Step (3): Emission of $β−$ particle from $89228Ac$

The given unbalanced nuclear equation is,

$89228Ac → −10β + ?$

Product formed by $β−$ decaying of $89228Ac$ is determined as follows,

$Sum of Atomic number of Reactant = Product89(Ac) = x+ −1(β−) x = 90Element with Z = 90 ⇒ Thorium(Th)Sum of Atomic number of Reactant = Product89228Ac = 90xTh + 0(β−)x = 228$

The elemental symbol of the unknown element is $90228Th$ . Therefore, the balanced nuclear equation is,

$89228Ac → −10β + 90228Th_$

Step (4): Emission of $α$ particle from $90228Th$

The given unbalanced nuclear equation is,

$90228Th→ ? + 24α$

Product formed by $α$ decaying of $90228Th$ is determined as follows,

$Sum of Atomic number of Reactant = Product90(Th)= x + 2(alpha)x = 88Element with Z = 88 ⇒ Radium(Ra)Sum of Mass number of Reactant = Product90228Th = 88xRa + 4(alpha)x = 224$

The elemental symbol of the unknown element is $88224Ra$ . Therefore, the balanced nuclear equation is,

$90228Th→ 88224Ra+ 24α_$

Step (5): Emission of $α$ particle from $88224Ra$

The given unbalanced nuclear equation is,

$88224Ra→ ? + 24α$

Product formed by $α$ decaying of $88224Ra$ is determined as follows,

$Sum of Atomic number of Reactant = Product88(Ra)= x + 2(alpha)x = 86Element with Z = 86 ⇒ Radon(Rn)Sum of Mass number of Reactant = Product88224Ra = 86xRn + 4(alpha)x = 220$

The elemental symbol of the unknown element is $86220Rn$ . Therefore, the balanced nuclear equation is,

$88224Ra→ 86220Rn + 24α_$

Step (6): Emission of $α$ particle from $86220Rn$

The given unbalanced nuclear equation is,

$86220Rn→ ? + 24α$

Product formed by $α$ decaying of $86220Rn$ is determined as follows,

$Sum of Atomic number of Reactant = Product86(Rn)= x + 2(alpha)x = 84Element with Z = 84 ⇒ Polonium(Po)Sum of Mass number of Reactant = Product86220Rn = 84xPo + 4(alpha)x = 216$

The elemental symbol of the unknown element is $84216Po$ . Therefore, the balanced nuclear equation is,

$86220Rn→ 84216Po + 24α_$

Step (7): Emission of $α$ particle from $84216Po$

The given unbalanced nuclear equation is,

$84216Po→ ? + 24α$

Product formed by $α$ decaying of $84216Po$ is determined as follows,

$Sum of Atomic number of Reactant = Product84(Po)= x + 2(alpha)x = 82Element with Z = 82 ⇒ Lead(Pb)Sum of Mass number of Reactant = Product84216Po = 82xPb + 4(alpha)x = 212$

The elemental symbol of the unknown element is $82212Pb$ . Therefore, the balanced nuclear equation is,

$84216Po→ 82212Pb + 24α_$

Step (8): Emission of $β−$ particle from $82212Pb$

The given unbalanced nuclear equation is,

$82212Pb→ −10β + ?$

Product formed by $β−$ decaying of $82212Pb$ is determined as follows,

$Sum of Atomic number of Reactant = Product82(Pb) = x+ −1(β−) x = 83Element with Z = 83 ⇒ Bismuth(Bi)Sum of Atomic number of Reactant = Product82212Pb = 83xBi + 0(β−)x = 212$

The elemental symbol of the unknown element is $83212Bi$ . Therefore, the balanced nuclear equation is,

$82209Pb → −10β + 83212Bi_$

Step (9): Emission of $β−$ particle from $83212Bi$

The given unbalanced nuclear equation is,

$83212Bi → −10β + ?$

Product formed by $β−$ decaying of $83212Bi$ is determined as follows,

$Sum of Atomic number of Reactant = Product83(Bi) = x+ −1(β−) x = 84Element with Z = 84 ⇒ Polonium(Po)Sum of Atomic number of Reactant = Product83212Bi = 84xPo + 0(β−)x = 212$

The elemental symbol of the unknown element is $84212Po$ . Therefore, the balanced nuclear equation is,

$83212Bi → −10β + 84212Po_$

Step (10): Emission of $α$ particle from $84212Po$

The given unbalanced nuclear equation is,

$84212Po→ ? + 24α$

Product formed by $α$ decaying of $84212Po$ is determined as follows,

$Sum of Atomic number of Reactant = Product84(Po)= x + 2(alpha)x = 82Element with Z = 82 ⇒ Lead(Pb)Sum of Mass number of Reactant = Product84212Po = 82xPb + 4(alpha)x = 208$

The elemental symbol of the unknown element is $82208Pb$ . Therefore, the balanced nuclear equation is,

$84212Po→ 82208Pb + 24α_$

Therefore, balanced nuclear equations for decay series of thorium-232 follows as,

$90232Th→ 88228Ra + 24α88228Ra→ −10β + 89228Ac89228Ac → −10β + 90228Th90228Th→ 88224Ra+ 24α88224Ra→ 86220Rn + 24α86220Rn→ 84216Po + 24α84216Po→ 82212Pb + 24α82209Pb → −10β + 83212Bi83212Bi → −10β + 84212Po84212Po→ 82208Pb + 24α$

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Answer 2

Question

Chapter 24, Problem 24.132P

Interpretation Introduction

Interpretation:

Balanced nuclear equations for decay series of thorium-232 has to be written from the given data.

Concept Introduction:

Nuclear reaction: A nuclear reaction in which a lighter nucleus fuses together into new stable nuclei or a heavier nucleus split into stable daughter nuclei with the release of large amount of energy.

Balancing nuclear reaction equation: The balanced nuclear reaction should conserve both mass number and atomic number.

The sum of the mass numbers of the reactants should be equal to the sum of mass numbers of the products in the reaction.

The sum of atomic numbers (or the atomic charge) of the reactants should be equal to the sum of atomic numbers (or the atomic charge) of the products in the reaction.

Expert Solution & Answer

Explanation of Solution

In the decay series of $232Th$ , the sequence of particles emitted in each step follows as,

$α, β−, β−, α, α, α, α, β−, β−, α$

According to the law of conservation of atomic mass and number, the unknown element can be predicted.

The elemental symbol of an element: $ZAX = Atomic numberMass numberX$

Step (1): Emission of $α$ particle from $232Th$

An alpha particle is emitted and an alpha particle is $24α.$ Atomic number of Thorium is 90.

The given unbalanced nuclear equation is,

$90232Th→ ? + 24α$

Product formed by $α$ decaying of $232Th$ is determined as follows,

$Sum of Atomic number of Reactant = Product90(Th)= x + 2(alpha)x = 88Element with Z = 88 ⇒ Radium(Ra)Sum of Mass number of Reactant = Product90232Th = 82xRa + 4(alpha)x = 228$

The elemental symbol of the unknown element is $88228Ra$ . Therefore, the balanced nuclear equation is,

$90232Th→ 88228Ra + 24α_$

Step (2): Emission of $β−$ particle from $88228Ra$

A $β−$ particle is emitted and a $β−$ particle is $−10β$ .

The given unbalanced nuclear equation is,

$88228Ra→ −10β + ?$

Product formed by $β−$ decaying of $88228Ra$ is determined as follows,

$Sum of Atomic number of Reactant = Product88(Ra) = x+ −1(β−) x = 89Element with Z = 89 ⇒ Actinium(Ac)Sum of Atomic number of Reactant = Product88228Ra = 89xAc + 0(β−)x = 228$

The elemental symbol of the unknown element is $89228Ac$ . Therefore, the balanced nuclear equation is,

$88228Ra→ −10β + 89228Ac_$

Step (3): Emission of $β−$ particle from $89228Ac$

The given unbalanced nuclear equation is,

$89228Ac → −10β + ?$

Product formed by $β−$ decaying of $89228Ac$ is determined as follows,

$Sum of Atomic number of Reactant = Product89(Ac) = x+ −1(β−) x = 90Element with Z = 90 ⇒ Thorium(Th)Sum of Atomic number of Reactant = Product89228Ac = 90xTh + 0(β−)x = 228$

The elemental symbol of the unknown element is $90228Th$ . Therefore, the balanced nuclear equation is,

$89228Ac → −10β + 90228Th_$

Step (4): Emission of $α$ particle from $90228Th$

The given unbalanced nuclear equation is,

$90228Th→ ? + 24α$

Product formed by $α$ decaying of $90228Th$ is determined as follows,

$Sum of Atomic number of Reactant = Product90(Th)= x + 2(alpha)x = 88Element with Z = 88 ⇒ Radium(Ra)Sum of Mass number of Reactant = Product90228Th = 88xRa + 4(alpha)x = 224$

The elemental symbol of the unknown element is $88224Ra$ . Therefore, the balanced nuclear equation is,

$90228Th→ 88224Ra+ 24α_$

Step (5): Emission of $α$ particle from $88224Ra$

The given unbalanced nuclear equation is,

$88224Ra→ ? + 24α$

Product formed by $α$ decaying of $88224Ra$ is determined as follows,

$Sum of Atomic number of Reactant = Product88(Ra)= x + 2(alpha)x = 86Element with Z = 86 ⇒ Radon(Rn)Sum of Mass number of Reactant = Product88224Ra = 86xRn + 4(alpha)x = 220$

The elemental symbol of the unknown element is $86220Rn$ . Therefore, the balanced nuclear equation is,

$88224Ra→ 86220Rn + 24α_$

Step (6): Emission of $α$ particle from $86220Rn$

The given unbalanced nuclear equation is,

$86220Rn→ ? + 24α$

Product formed by $α$ decaying of $86220Rn$ is determined as follows,

$Sum of Atomic number of Reactant = Product86(Rn)= x + 2(alpha)x = 84Element with Z = 84 ⇒ Polonium(Po)Sum of Mass number of Reactant = Product86220Rn = 84xPo + 4(alpha)x = 216$

The elemental symbol of the unknown element is $84216Po$ . Therefore, the balanced nuclear equation is,

$86220Rn→ 84216Po + 24α_$

Step (7): Emission of $α$ particle from $84216Po$

The given unbalanced nuclear equation is,

$84216Po→ ? + 24α$

Product formed by $α$ decaying of $84216Po$ is determined as follows,

$Sum of Atomic number of Reactant = Product84(Po)= x + 2(alpha)x = 82Element with Z = 82 ⇒ Lead(Pb)Sum of Mass number of Reactant = Product84216Po = 82xPb + 4(alpha)x = 212$

The elemental symbol of the unknown element is $82212Pb$ . Therefore, the balanced nuclear equation is,

$84216Po→ 82212Pb + 24α_$

Step (8): Emission of $β−$ particle from $82212Pb$

The given unbalanced nuclear equation is,

$82212Pb→ −10β + ?$

Product formed by $β−$ decaying of $82212Pb$ is determined as follows,

$Sum of Atomic number of Reactant = Product82(Pb) = x+ −1(β−) x = 83Element with Z = 83 ⇒ Bismuth(Bi)Sum of Atomic number of Reactant = Product82212Pb = 83xBi + 0(β−)x = 212$

The elemental symbol of the unknown element is $83212Bi$ . Therefore, the balanced nuclear equation is,

$82209Pb → −10β + 83212Bi_$

Step (9): Emission of $β−$ particle from $83212Bi$

The given unbalanced nuclear equation is,

$83212Bi → −10β + ?$

Product formed by $β−$ decaying of $83212Bi$ is determined as follows,

$Sum of Atomic number of Reactant = Product83(Bi) = x+ −1(β−) x = 84Element with Z = 84 ⇒ Polonium(Po)Sum of Atomic number of Reactant = Product83212Bi = 84xPo + 0(β−)x = 212$

The elemental symbol of the unknown element is $84212Po$ . Therefore, the balanced nuclear equation is,

$83212Bi → −10β + 84212Po_$

Step (10): Emission of $α$ particle from $84212Po$

The given unbalanced nuclear equation is,

$84212Po→ ? + 24α$

Product formed by $α$ decaying of $84212Po$ is determined as follows,

$Sum of Atomic number of Reactant = Product84(Po)= x + 2(alpha)x = 82Element with Z = 82 ⇒ Lead(Pb)Sum of Mass number of Reactant = Product84212Po = 82xPb + 4(alpha)x = 208$

The elemental symbol of the unknown element is $82208Pb$ . Therefore, the balanced nuclear equation is,

$84212Po→ 82208Pb + 24α_$

Therefore, balanced nuclear equations for decay series of thorium-232 follows as,

$90232Th→ 88228Ra + 24α88228Ra→ −10β + 89228Ac89228Ac → −10β + 90228Th90228Th→ 88224Ra+ 24α88224Ra→ 86220Rn + 24α86220Rn→ 84216Po + 24α84216Po→ 82212Pb + 24α82209Pb → −10β + 83212Bi83212Bi → −10β + 84212Po84212Po→ 82208Pb + 24α$

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Answer 3

Question

Chapter 24, Problem 24.132P

Interpretation Introduction

Interpretation:

Balanced nuclear equations for decay series of thorium-232 has to be written from the given data.

Concept Introduction:

Nuclear reaction: A nuclear reaction in which a lighter nucleus fuses together into new stable nuclei or a heavier nucleus split into stable daughter nuclei with the release of large amount of energy.

Balancing nuclear reaction equation: The balanced nuclear reaction should conserve both mass number and atomic number.

The sum of the mass numbers of the reactants should be equal to the sum of mass numbers of the products in the reaction.

The sum of atomic numbers (or the atomic charge) of the reactants should be equal to the sum of atomic numbers (or the atomic charge) of the products in the reaction.

Expert Solution & Answer

Explanation of Solution

In the decay series of $232Th$ , the sequence of particles emitted in each step follows as,

$α, β−, β−, α, α, α, α, β−, β−, α$

According to the law of conservation of atomic mass and number, the unknown element can be predicted.

The elemental symbol of an element: $ZAX = Atomic numberMass numberX$

Step (1): Emission of $α$ particle from $232Th$

An alpha particle is emitted and an alpha particle is $24α.$ Atomic number of Thorium is 90.

The given unbalanced nuclear equation is,

$90232Th→ ? + 24α$

Product formed by $α$ decaying of $232Th$ is determined as follows,

$Sum of Atomic number of Reactant = Product90(Th)= x + 2(alpha)x = 88Element with Z = 88 ⇒ Radium(Ra)Sum of Mass number of Reactant = Product90232Th = 82xRa + 4(alpha)x = 228$

The elemental symbol of the unknown element is $88228Ra$ . Therefore, the balanced nuclear equation is,

$90232Th→ 88228Ra + 24α_$

Step (2): Emission of $β−$ particle from $88228Ra$

A $β−$ particle is emitted and a $β−$ particle is $−10β$ .

The given unbalanced nuclear equation is,

$88228Ra→ −10β + ?$

Product formed by $β−$ decaying of $88228Ra$ is determined as follows,

$Sum of Atomic number of Reactant = Product88(Ra) = x+ −1(β−) x = 89Element with Z = 89 ⇒ Actinium(Ac)Sum of Atomic number of Reactant = Product88228Ra = 89xAc + 0(β−)x = 228$

The elemental symbol of the unknown element is $89228Ac$ . Therefore, the balanced nuclear equation is,

$88228Ra→ −10β + 89228Ac_$

Step (3): Emission of $β−$ particle from $89228Ac$

The given unbalanced nuclear equation is,

$89228Ac → −10β + ?$

Product formed by $β−$ decaying of $89228Ac$ is determined as follows,

$Sum of Atomic number of Reactant = Product89(Ac) = x+ −1(β−) x = 90Element with Z = 90 ⇒ Thorium(Th)Sum of Atomic number of Reactant = Product89228Ac = 90xTh + 0(β−)x = 228$

The elemental symbol of the unknown element is $90228Th$ . Therefore, the balanced nuclear equation is,

$89228Ac → −10β + 90228Th_$

Step (4): Emission of $α$ particle from $90228Th$

The given unbalanced nuclear equation is,

$90228Th→ ? + 24α$

Product formed by $α$ decaying of $90228Th$ is determined as follows,

$Sum of Atomic number of Reactant = Product90(Th)= x + 2(alpha)x = 88Element with Z = 88 ⇒ Radium(Ra)Sum of Mass number of Reactant = Product90228Th = 88xRa + 4(alpha)x = 224$

The elemental symbol of the unknown element is $88224Ra$ . Therefore, the balanced nuclear equation is,

$90228Th→ 88224Ra+ 24α_$

Step (5): Emission of $α$ particle from $88224Ra$

The given unbalanced nuclear equation is,

$88224Ra→ ? + 24α$

Product formed by $α$ decaying of $88224Ra$ is determined as follows,

$Sum of Atomic number of Reactant = Product88(Ra)= x + 2(alpha)x = 86Element with Z = 86 ⇒ Radon(Rn)Sum of Mass number of Reactant = Product88224Ra = 86xRn + 4(alpha)x = 220$

The elemental symbol of the unknown element is $86220Rn$ . Therefore, the balanced nuclear equation is,

$88224Ra→ 86220Rn + 24α_$

Step (6): Emission of $α$ particle from $86220Rn$

The given unbalanced nuclear equation is,

$86220Rn→ ? + 24α$

Product formed by $α$ decaying of $86220Rn$ is determined as follows,

$Sum of Atomic number of Reactant = Product86(Rn)= x + 2(alpha)x = 84Element with Z = 84 ⇒ Polonium(Po)Sum of Mass number of Reactant = Product86220Rn = 84xPo + 4(alpha)x = 216$

The elemental symbol of the unknown element is $84216Po$ . Therefore, the balanced nuclear equation is,

$86220Rn→ 84216Po + 24α_$

Step (7): Emission of $α$ particle from $84216Po$

The given unbalanced nuclear equation is,

$84216Po→ ? + 24α$

Product formed by $α$ decaying of $84216Po$ is determined as follows,

$Sum of Atomic number of Reactant = Product84(Po)= x + 2(alpha)x = 82Element with Z = 82 ⇒ Lead(Pb)Sum of Mass number of Reactant = Product84216Po = 82xPb + 4(alpha)x = 212$

The elemental symbol of the unknown element is $82212Pb$ . Therefore, the balanced nuclear equation is,

$84216Po→ 82212Pb + 24α_$

Step (8): Emission of $β−$ particle from $82212Pb$

The given unbalanced nuclear equation is,

$82212Pb→ −10β + ?$

Product formed by $β−$ decaying of $82212Pb$ is determined as follows,

$Sum of Atomic number of Reactant = Product82(Pb) = x+ −1(β−) x = 83Element with Z = 83 ⇒ Bismuth(Bi)Sum of Atomic number of Reactant = Product82212Pb = 83xBi + 0(β−)x = 212$

The elemental symbol of the unknown element is $83212Bi$ . Therefore, the balanced nuclear equation is,

$82209Pb → −10β + 83212Bi_$

Step (9): Emission of $β−$ particle from $83212Bi$

The given unbalanced nuclear equation is,

$83212Bi → −10β + ?$

Product formed by $β−$ decaying of $83212Bi$ is determined as follows,

$Sum of Atomic number of Reactant = Product83(Bi) = x+ −1(β−) x = 84Element with Z = 84 ⇒ Polonium(Po)Sum of Atomic number of Reactant = Product83212Bi = 84xPo + 0(β−)x = 212$

The elemental symbol of the unknown element is $84212Po$ . Therefore, the balanced nuclear equation is,

$83212Bi → −10β + 84212Po_$

Step (10): Emission of $α$ particle from $84212Po$

The given unbalanced nuclear equation is,

$84212Po→ ? + 24α$

Product formed by $α$ decaying of $84212Po$ is determined as follows,

$Sum of Atomic number of Reactant = Product84(Po)= x + 2(alpha)x = 82Element with Z = 82 ⇒ Lead(Pb)Sum of Mass number of Reactant = Product84212Po = 82xPb + 4(alpha)x = 208$

The elemental symbol of the unknown element is $82208Pb$ . Therefore, the balanced nuclear equation is,

$84212Po→ 82208Pb + 24α_$

Therefore, balanced nuclear equations for decay series of thorium-232 follows as,

$90232Th→ 88228Ra + 24α88228Ra→ −10β + 89228Ac89228Ac → −10β + 90228Th90228Th→ 88224Ra+ 24α88224Ra→ 86220Rn + 24α86220Rn→ 84216Po + 24α84216Po→ 82212Pb + 24α82209Pb → −10β + 83212Bi83212Bi → −10β + 84212Po84212Po→ 82208Pb + 24α$

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Answer 4

Question

Chapter 24, Problem 24.132P

Interpretation Introduction

Interpretation:

Balanced nuclear equations for decay series of thorium-232 has to be written from the given data.

Concept Introduction:

Nuclear reaction: A nuclear reaction in which a lighter nucleus fuses together into new stable nuclei or a heavier nucleus split into stable daughter nuclei with the release of large amount of energy.

Balancing nuclear reaction equation: The balanced nuclear reaction should conserve both mass number and atomic number.

The sum of the mass numbers of the reactants should be equal to the sum of mass numbers of the products in the reaction.

The sum of atomic numbers (or the atomic charge) of the reactants should be equal to the sum of atomic numbers (or the atomic charge) of the products in the reaction.

Expert Solution & Answer

Explanation of Solution

In the decay series of $232Th$ , the sequence of particles emitted in each step follows as,

$α, β−, β−, α, α, α, α, β−, β−, α$

According to the law of conservation of atomic mass and number, the unknown element can be predicted.

The elemental symbol of an element: $ZAX = Atomic numberMass numberX$

Step (1): Emission of $α$ particle from $232Th$

An alpha particle is emitted and an alpha particle is $24α.$ Atomic number of Thorium is 90.

The given unbalanced nuclear equation is,

$90232Th→ ? + 24α$

Product formed by $α$ decaying of $232Th$ is determined as follows,

$Sum of Atomic number of Reactant = Product90(Th)= x + 2(alpha)x = 88Element with Z = 88 ⇒ Radium(Ra)Sum of Mass number of Reactant = Product90232Th = 82xRa + 4(alpha)x = 228$

The elemental symbol of the unknown element is $88228Ra$ . Therefore, the balanced nuclear equation is,

$90232Th→ 88228Ra + 24α_$

Step (2): Emission of $β−$ particle from $88228Ra$

A $β−$ particle is emitted and a $β−$ particle is $−10β$ .

The given unbalanced nuclear equation is,

$88228Ra→ −10β + ?$

Product formed by $β−$ decaying of $88228Ra$ is determined as follows,

$Sum of Atomic number of Reactant = Product88(Ra) = x+ −1(β−) x = 89Element with Z = 89 ⇒ Actinium(Ac)Sum of Atomic number of Reactant = Product88228Ra = 89xAc + 0(β−)x = 228$

The elemental symbol of the unknown element is $89228Ac$ . Therefore, the balanced nuclear equation is,

$88228Ra→ −10β + 89228Ac_$

Step (3): Emission of $β−$ particle from $89228Ac$

The given unbalanced nuclear equation is,

$89228Ac → −10β + ?$

Product formed by $β−$ decaying of $89228Ac$ is determined as follows,

$Sum of Atomic number of Reactant = Product89(Ac) = x+ −1(β−) x = 90Element with Z = 90 ⇒ Thorium(Th)Sum of Atomic number of Reactant = Product89228Ac = 90xTh + 0(β−)x = 228$

The elemental symbol of the unknown element is $90228Th$ . Therefore, the balanced nuclear equation is,

$89228Ac → −10β + 90228Th_$

Step (4): Emission of $α$ particle from $90228Th$

The given unbalanced nuclear equation is,

$90228Th→ ? + 24α$

Product formed by $α$ decaying of $90228Th$ is determined as follows,

$Sum of Atomic number of Reactant = Product90(Th)= x + 2(alpha)x = 88Element with Z = 88 ⇒ Radium(Ra)Sum of Mass number of Reactant = Product90228Th = 88xRa + 4(alpha)x = 224$

The elemental symbol of the unknown element is $88224Ra$ . Therefore, the balanced nuclear equation is,

$90228Th→ 88224Ra+ 24α_$

Step (5): Emission of $α$ particle from $88224Ra$

The given unbalanced nuclear equation is,

$88224Ra→ ? + 24α$

Product formed by $α$ decaying of $88224Ra$ is determined as follows,

$Sum of Atomic number of Reactant = Product88(Ra)= x + 2(alpha)x = 86Element with Z = 86 ⇒ Radon(Rn)Sum of Mass number of Reactant = Product88224Ra = 86xRn + 4(alpha)x = 220$

The elemental symbol of the unknown element is $86220Rn$ . Therefore, the balanced nuclear equation is,

$88224Ra→ 86220Rn + 24α_$

Step (6): Emission of $α$ particle from $86220Rn$

The given unbalanced nuclear equation is,

$86220Rn→ ? + 24α$

Product formed by $α$ decaying of $86220Rn$ is determined as follows,

$Sum of Atomic number of Reactant = Product86(Rn)= x + 2(alpha)x = 84Element with Z = 84 ⇒ Polonium(Po)Sum of Mass number of Reactant = Product86220Rn = 84xPo + 4(alpha)x = 216$

The elemental symbol of the unknown element is $84216Po$ . Therefore, the balanced nuclear equation is,

$86220Rn→ 84216Po + 24α_$

Step (7): Emission of $α$ particle from $84216Po$

The given unbalanced nuclear equation is,

$84216Po→ ? + 24α$

Product formed by $α$ decaying of $84216Po$ is determined as follows,

$Sum of Atomic number of Reactant = Product84(Po)= x + 2(alpha)x = 82Element with Z = 82 ⇒ Lead(Pb)Sum of Mass number of Reactant = Product84216Po = 82xPb + 4(alpha)x = 212$

The elemental symbol of the unknown element is $82212Pb$ . Therefore, the balanced nuclear equation is,

$84216Po→ 82212Pb + 24α_$

Step (8): Emission of $β−$ particle from $82212Pb$

The given unbalanced nuclear equation is,

$82212Pb→ −10β + ?$

Product formed by $β−$ decaying of $82212Pb$ is determined as follows,

$Sum of Atomic number of Reactant = Product82(Pb) = x+ −1(β−) x = 83Element with Z = 83 ⇒ Bismuth(Bi)Sum of Atomic number of Reactant = Product82212Pb = 83xBi + 0(β−)x = 212$

The elemental symbol of the unknown element is $83212Bi$ . Therefore, the balanced nuclear equation is,

$82209Pb → −10β + 83212Bi_$

Step (9): Emission of $β−$ particle from $83212Bi$

The given unbalanced nuclear equation is,

$83212Bi → −10β + ?$

Product formed by $β−$ decaying of $83212Bi$ is determined as follows,

$Sum of Atomic number of Reactant = Product83(Bi) = x+ −1(β−) x = 84Element with Z = 84 ⇒ Polonium(Po)Sum of Atomic number of Reactant = Product83212Bi = 84xPo + 0(β−)x = 212$

The elemental symbol of the unknown element is $84212Po$ . Therefore, the balanced nuclear equation is,

$83212Bi → −10β + 84212Po_$

Step (10): Emission of $α$ particle from $84212Po$

The given unbalanced nuclear equation is,

$84212Po→ ? + 24α$

Product formed by $α$ decaying of $84212Po$ is determined as follows,

$Sum of Atomic number of Reactant = Product84(Po)= x + 2(alpha)x = 82Element with Z = 82 ⇒ Lead(Pb)Sum of Mass number of Reactant = Product84212Po = 82xPb + 4(alpha)x = 208$

The elemental symbol of the unknown element is $82208Pb$ . Therefore, the balanced nuclear equation is,

$84212Po→ 82208Pb + 24α_$

Therefore, balanced nuclear equations for decay series of thorium-232 follows as,

$90232Th→ 88228Ra + 24α88228Ra→ −10β + 89228Ac89228Ac → −10β + 90228Th90228Th→ 88224Ra+ 24α88224Ra→ 86220Rn + 24α86220Rn→ 84216Po + 24α84216Po→ 82212Pb + 24α82209Pb → −10β + 83212Bi83212Bi → −10β + 84212Po84212Po→ 82208Pb + 24α$

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Answer 5

Chapter 24, Problem 24.132P

Interpretation Introduction

Interpretation:

Balanced nuclear equations for decay series of thorium-232 has to be written from the given data.

Concept Introduction:

Nuclear reaction: A nuclear reaction in which a lighter nucleus fuses together into new stable nuclei or a heavier nucleus split into stable daughter nuclei with the release of large amount of energy.

Balancing nuclear reaction equation: The balanced nuclear reaction should conserve both mass number and atomic number.

The sum of the mass numbers of the reactants should be equal to the sum of mass numbers of the products in the reaction.

The sum of atomic numbers (or the atomic charge) of the reactants should be equal to the sum of atomic numbers (or the atomic charge) of the products in the reaction.

Expert Solution & Answer

Explanation of Solution

In the decay series of $232Th$ , the sequence of particles emitted in each step follows as,

$α, β−, β−, α, α, α, α, β−, β−, α$

According to the law of conservation of atomic mass and number, the unknown element can be predicted.

The elemental symbol of an element: $ZAX = Atomic numberMass numberX$

Step (1): Emission of $α$ particle from $232Th$

An alpha particle is emitted and an alpha particle is $24α.$ Atomic number of Thorium is 90.

The given unbalanced nuclear equation is,

$90232Th→ ? + 24α$

Product formed by $α$ decaying of $232Th$ is determined as follows,

$Sum of Atomic number of Reactant = Product90(Th)= x + 2(alpha)x = 88Element with Z = 88 ⇒ Radium(Ra)Sum of Mass number of Reactant = Product90232Th = 82xRa + 4(alpha)x = 228$

The elemental symbol of the unknown element is $88228Ra$ . Therefore, the balanced nuclear equation is,

$90232Th→ 88228Ra + 24α_$

Step (2): Emission of $β−$ particle from $88228Ra$

A $β−$ particle is emitted and a $β−$ particle is $−10β$ .

The given unbalanced nuclear equation is,

$88228Ra→ −10β + ?$

Product formed by $β−$ decaying of $88228Ra$ is determined as follows,

$Sum of Atomic number of Reactant = Product88(Ra) = x+ −1(β−) x = 89Element with Z = 89 ⇒ Actinium(Ac)Sum of Atomic number of Reactant = Product88228Ra = 89xAc + 0(β−)x = 228$

The elemental symbol of the unknown element is $89228Ac$ . Therefore, the balanced nuclear equation is,

$88228Ra→ −10β + 89228Ac_$

Step (3): Emission of $β−$ particle from $89228Ac$

The given unbalanced nuclear equation is,

$89228Ac → −10β + ?$

Product formed by $β−$ decaying of $89228Ac$ is determined as follows,

$Sum of Atomic number of Reactant = Product89(Ac) = x+ −1(β−) x = 90Element with Z = 90 ⇒ Thorium(Th)Sum of Atomic number of Reactant = Product89228Ac = 90xTh + 0(β−)x = 228$

The elemental symbol of the unknown element is $90228Th$ . Therefore, the balanced nuclear equation is,

$89228Ac → −10β + 90228Th_$

Step (4): Emission of $α$ particle from $90228Th$

The given unbalanced nuclear equation is,

$90228Th→ ? + 24α$

Product formed by $α$ decaying of $90228Th$ is determined as follows,

$Sum of Atomic number of Reactant = Product90(Th)= x + 2(alpha)x = 88Element with Z = 88 ⇒ Radium(Ra)Sum of Mass number of Reactant = Product90228Th = 88xRa + 4(alpha)x = 224$

The elemental symbol of the unknown element is $88224Ra$ . Therefore, the balanced nuclear equation is,

$90228Th→ 88224Ra+ 24α_$

Step (5): Emission of $α$ particle from $88224Ra$

The given unbalanced nuclear equation is,

$88224Ra→ ? + 24α$

Product formed by $α$ decaying of $88224Ra$ is determined as follows,

$Sum of Atomic number of Reactant = Product88(Ra)= x + 2(alpha)x = 86Element with Z = 86 ⇒ Radon(Rn)Sum of Mass number of Reactant = Product88224Ra = 86xRn + 4(alpha)x = 220$

The elemental symbol of the unknown element is $86220Rn$ . Therefore, the balanced nuclear equation is,

$88224Ra→ 86220Rn + 24α_$

Step (6): Emission of $α$ particle from $86220Rn$

The given unbalanced nuclear equation is,

$86220Rn→ ? + 24α$

Product formed by $α$ decaying of $86220Rn$ is determined as follows,

$Sum of Atomic number of Reactant = Product86(Rn)= x + 2(alpha)x = 84Element with Z = 84 ⇒ Polonium(Po)Sum of Mass number of Reactant = Product86220Rn = 84xPo + 4(alpha)x = 216$

The elemental symbol of the unknown element is $84216Po$ . Therefore, the balanced nuclear equation is,

$86220Rn→ 84216Po + 24α_$

Step (7): Emission of $α$ particle from $84216Po$

The given unbalanced nuclear equation is,

$84216Po→ ? + 24α$

Product formed by $α$ decaying of $84216Po$ is determined as follows,

$Sum of Atomic number of Reactant = Product84(Po)= x + 2(alpha)x = 82Element with Z = 82 ⇒ Lead(Pb)Sum of Mass number of Reactant = Product84216Po = 82xPb + 4(alpha)x = 212$

The elemental symbol of the unknown element is $82212Pb$ . Therefore, the balanced nuclear equation is,

$84216Po→ 82212Pb + 24α_$

Step (8): Emission of $β−$ particle from $82212Pb$

The given unbalanced nuclear equation is,

$82212Pb→ −10β + ?$

Product formed by $β−$ decaying of $82212Pb$ is determined as follows,

$Sum of Atomic number of Reactant = Product82(Pb) = x+ −1(β−) x = 83Element with Z = 83 ⇒ Bismuth(Bi)Sum of Atomic number of Reactant = Product82212Pb = 83xBi + 0(β−)x = 212$

The elemental symbol of the unknown element is $83212Bi$ . Therefore, the balanced nuclear equation is,

$82209Pb → −10β + 83212Bi_$

Step (9): Emission of $β−$ particle from $83212Bi$

The given unbalanced nuclear equation is,

$83212Bi → −10β + ?$

Product formed by $β−$ decaying of $83212Bi$ is determined as follows,

$Sum of Atomic number of Reactant = Product83(Bi) = x+ −1(β−) x = 84Element with Z = 84 ⇒ Polonium(Po)Sum of Atomic number of Reactant = Product83212Bi = 84xPo + 0(β−)x = 212$

The elemental symbol of the unknown element is $84212Po$ . Therefore, the balanced nuclear equation is,

$83212Bi → −10β + 84212Po_$

Step (10): Emission of $α$ particle from $84212Po$

The given unbalanced nuclear equation is,

$84212Po→ ? + 24α$

Product formed by $α$ decaying of $84212Po$ is determined as follows,

$Sum of Atomic number of Reactant = Product84(Po)= x + 2(alpha)x = 82Element with Z = 82 ⇒ Lead(Pb)Sum of Mass number of Reactant = Product84212Po = 82xPb + 4(alpha)x = 208$

The elemental symbol of the unknown element is $82208Pb$ . Therefore, the balanced nuclear equation is,

$84212Po→ 82208Pb + 24α_$

Therefore, balanced nuclear equations for decay series of thorium-232 follows as,

$90232Th→ 88228Ra + 24α88228Ra→ −10β + 89228Ac89228Ac → −10β + 90228Th90228Th→ 88224Ra+ 24α88224Ra→ 86220Rn + 24α86220Rn→ 84216Po + 24α84216Po→ 82212Pb + 24α82209Pb → −10β + 83212Bi83212Bi → −10β + 84212Po84212Po→ 82208Pb + 24α$

Answer 6

Chapter 24, Problem 24.132P

Interpretation Introduction

Interpretation:

Balanced nuclear equations for decay series of thorium-232 has to be written from the given data.

Concept Introduction:

Nuclear reaction: A nuclear reaction in which a lighter nucleus fuses together into new stable nuclei or a heavier nucleus split into stable daughter nuclei with the release of large amount of energy.

Balancing nuclear reaction equation: The balanced nuclear reaction should conserve both mass number and atomic number.

The sum of the mass numbers of the reactants should be equal to the sum of mass numbers of the products in the reaction.

The sum of atomic numbers (or the atomic charge) of the reactants should be equal to the sum of atomic numbers (or the atomic charge) of the products in the reaction.

Expert Solution & Answer

Explanation of Solution

In the decay series of $232Th$ , the sequence of particles emitted in each step follows as,

$α, β−, β−, α, α, α, α, β−, β−, α$

According to the law of conservation of atomic mass and number, the unknown element can be predicted.

The elemental symbol of an element: $ZAX = Atomic numberMass numberX$

Step (1): Emission of $α$ particle from $232Th$

An alpha particle is emitted and an alpha particle is $24α.$ Atomic number of Thorium is 90.

The given unbalanced nuclear equation is,

$90232Th→ ? + 24α$

Product formed by $α$ decaying of $232Th$ is determined as follows,

$Sum of Atomic number of Reactant = Product90(Th)= x + 2(alpha)x = 88Element with Z = 88 ⇒ Radium(Ra)Sum of Mass number of Reactant = Product90232Th = 82xRa + 4(alpha)x = 228$

The elemental symbol of the unknown element is $88228Ra$ . Therefore, the balanced nuclear equation is,

$90232Th→ 88228Ra + 24α_$

Step (2): Emission of $β−$ particle from $88228Ra$

A $β−$ particle is emitted and a $β−$ particle is $−10β$ .

The given unbalanced nuclear equation is,

$88228Ra→ −10β + ?$

Product formed by $β−$ decaying of $88228Ra$ is determined as follows,

$Sum of Atomic number of Reactant = Product88(Ra) = x+ −1(β−) x = 89Element with Z = 89 ⇒ Actinium(Ac)Sum of Atomic number of Reactant = Product88228Ra = 89xAc + 0(β−)x = 228$

The elemental symbol of the unknown element is $89228Ac$ . Therefore, the balanced nuclear equation is,

$88228Ra→ −10β + 89228Ac_$

Step (3): Emission of $β−$ particle from $89228Ac$

The given unbalanced nuclear equation is,

$89228Ac → −10β + ?$

Product formed by $β−$ decaying of $89228Ac$ is determined as follows,

$Sum of Atomic number of Reactant = Product89(Ac) = x+ −1(β−) x = 90Element with Z = 90 ⇒ Thorium(Th)Sum of Atomic number of Reactant = Product89228Ac = 90xTh + 0(β−)x = 228$

The elemental symbol of the unknown element is $90228Th$ . Therefore, the balanced nuclear equation is,

$89228Ac → −10β + 90228Th_$

Step (4): Emission of $α$ particle from $90228Th$

The given unbalanced nuclear equation is,

$90228Th→ ? + 24α$

Product formed by $α$ decaying of $90228Th$ is determined as follows,

$Sum of Atomic number of Reactant = Product90(Th)= x + 2(alpha)x = 88Element with Z = 88 ⇒ Radium(Ra)Sum of Mass number of Reactant = Product90228Th = 88xRa + 4(alpha)x = 224$

The elemental symbol of the unknown element is $88224Ra$ . Therefore, the balanced nuclear equation is,

$90228Th→ 88224Ra+ 24α_$

Step (5): Emission of $α$ particle from $88224Ra$

The given unbalanced nuclear equation is,

$88224Ra→ ? + 24α$

Product formed by $α$ decaying of $88224Ra$ is determined as follows,

$Sum of Atomic number of Reactant = Product88(Ra)= x + 2(alpha)x = 86Element with Z = 86 ⇒ Radon(Rn)Sum of Mass number of Reactant = Product88224Ra = 86xRn + 4(alpha)x = 220$

The elemental symbol of the unknown element is $86220Rn$ . Therefore, the balanced nuclear equation is,

$88224Ra→ 86220Rn + 24α_$

Step (6): Emission of $α$ particle from $86220Rn$

The given unbalanced nuclear equation is,

$86220Rn→ ? + 24α$

Product formed by $α$ decaying of $86220Rn$ is determined as follows,

$Sum of Atomic number of Reactant = Product86(Rn)= x + 2(alpha)x = 84Element with Z = 84 ⇒ Polonium(Po)Sum of Mass number of Reactant = Product86220Rn = 84xPo + 4(alpha)x = 216$

The elemental symbol of the unknown element is $84216Po$ . Therefore, the balanced nuclear equation is,

$86220Rn→ 84216Po + 24α_$

Step (7): Emission of $α$ particle from $84216Po$

The given unbalanced nuclear equation is,

$84216Po→ ? + 24α$

Product formed by $α$ decaying of $84216Po$ is determined as follows,

$Sum of Atomic number of Reactant = Product84(Po)= x + 2(alpha)x = 82Element with Z = 82 ⇒ Lead(Pb)Sum of Mass number of Reactant = Product84216Po = 82xPb + 4(alpha)x = 212$

The elemental symbol of the unknown element is $82212Pb$ . Therefore, the balanced nuclear equation is,

$84216Po→ 82212Pb + 24α_$

Step (8): Emission of $β−$ particle from $82212Pb$

The given unbalanced nuclear equation is,

$82212Pb→ −10β + ?$

Product formed by $β−$ decaying of $82212Pb$ is determined as follows,

$Sum of Atomic number of Reactant = Product82(Pb) = x+ −1(β−) x = 83Element with Z = 83 ⇒ Bismuth(Bi)Sum of Atomic number of Reactant = Product82212Pb = 83xBi + 0(β−)x = 212$

The elemental symbol of the unknown element is $83212Bi$ . Therefore, the balanced nuclear equation is,

$82209Pb → −10β + 83212Bi_$

Step (9): Emission of $β−$ particle from $83212Bi$

The given unbalanced nuclear equation is,

$83212Bi → −10β + ?$

Product formed by $β−$ decaying of $83212Bi$ is determined as follows,

$Sum of Atomic number of Reactant = Product83(Bi) = x+ −1(β−) x = 84Element with Z = 84 ⇒ Polonium(Po)Sum of Atomic number of Reactant = Product83212Bi = 84xPo + 0(β−)x = 212$

The elemental symbol of the unknown element is $84212Po$ . Therefore, the balanced nuclear equation is,

$83212Bi → −10β + 84212Po_$

Step (10): Emission of $α$ particle from $84212Po$

The given unbalanced nuclear equation is,

$84212Po→ ? + 24α$

Product formed by $α$ decaying of $84212Po$ is determined as follows,

$Sum of Atomic number of Reactant = Product84(Po)= x + 2(alpha)x = 82Element with Z = 82 ⇒ Lead(Pb)Sum of Mass number of Reactant = Product84212Po = 82xPb + 4(alpha)x = 208$

The elemental symbol of the unknown element is $82208Pb$ . Therefore, the balanced nuclear equation is,

$84212Po→ 82208Pb + 24α_$

Therefore, balanced nuclear equations for decay series of thorium-232 follows as,

$90232Th→ 88228Ra + 24α88228Ra→ −10β + 89228Ac89228Ac → −10β + 90228Th90228Th→ 88224Ra+ 24α88224Ra→ 86220Rn + 24α86220Rn→ 84216Po + 24α84216Po→ 82212Pb + 24α82209Pb → −10β + 83212Bi83212Bi → −10β + 84212Po84212Po→ 82208Pb + 24α$

Answer 7

Chapter 24, Problem 24.132P

Interpretation Introduction

Interpretation:

Balanced nuclear equations for decay series of thorium-232 has to be written from the given data.

Concept Introduction:

Nuclear reaction: A nuclear reaction in which a lighter nucleus fuses together into new stable nuclei or a heavier nucleus split into stable daughter nuclei with the release of large amount of energy.

Balancing nuclear reaction equation: The balanced nuclear reaction should conserve both mass number and atomic number.

The sum of the mass numbers of the reactants should be equal to the sum of mass numbers of the products in the reaction.

The sum of atomic numbers (or the atomic charge) of the reactants should be equal to the sum of atomic numbers (or the atomic charge) of the products in the reaction.

Answer 8

Expert Solution & Answer

Explanation of Solution

In the decay series of $232Th$ , the sequence of particles emitted in each step follows as,

$α, β−, β−, α, α, α, α, β−, β−, α$

According to the law of conservation of atomic mass and number, the unknown element can be predicted.

The elemental symbol of an element: $ZAX = Atomic numberMass numberX$

Step (1): Emission of $α$ particle from $232Th$

An alpha particle is emitted and an alpha particle is $24α.$ Atomic number of Thorium is 90.

The given unbalanced nuclear equation is,

$90232Th→ ? + 24α$

Product formed by $α$ decaying of $232Th$ is determined as follows,

$Sum of Atomic number of Reactant = Product90(Th)= x + 2(alpha)x = 88Element with Z = 88 ⇒ Radium(Ra)Sum of Mass number of Reactant = Product90232Th = 82xRa + 4(alpha)x = 228$

The elemental symbol of the unknown element is $88228Ra$ . Therefore, the balanced nuclear equation is,

$90232Th→ 88228Ra + 24α_$

Step (2): Emission of $β−$ particle from $88228Ra$

A $β−$ particle is emitted and a $β−$ particle is $−10β$ .

The given unbalanced nuclear equation is,

$88228Ra→ −10β + ?$

Product formed by $β−$ decaying of $88228Ra$ is determined as follows,

$Sum of Atomic number of Reactant = Product88(Ra) = x+ −1(β−) x = 89Element with Z = 89 ⇒ Actinium(Ac)Sum of Atomic number of Reactant = Product88228Ra = 89xAc + 0(β−)x = 228$

The elemental symbol of the unknown element is $89228Ac$ . Therefore, the balanced nuclear equation is,

$88228Ra→ −10β + 89228Ac_$

Step (3): Emission of $β−$ particle from $89228Ac$

The given unbalanced nuclear equation is,

$89228Ac → −10β + ?$

Product formed by $β−$ decaying of $89228Ac$ is determined as follows,

$Sum of Atomic number of Reactant = Product89(Ac) = x+ −1(β−) x = 90Element with Z = 90 ⇒ Thorium(Th)Sum of Atomic number of Reactant = Product89228Ac = 90xTh + 0(β−)x = 228$

The elemental symbol of the unknown element is $90228Th$ . Therefore, the balanced nuclear equation is,

$89228Ac → −10β + 90228Th_$

Step (4): Emission of $α$ particle from $90228Th$

The given unbalanced nuclear equation is,

$90228Th→ ? + 24α$

Product formed by $α$ decaying of $90228Th$ is determined as follows,

$Sum of Atomic number of Reactant = Product90(Th)= x + 2(alpha)x = 88Element with Z = 88 ⇒ Radium(Ra)Sum of Mass number of Reactant = Product90228Th = 88xRa + 4(alpha)x = 224$

The elemental symbol of the unknown element is $88224Ra$ . Therefore, the balanced nuclear equation is,

$90228Th→ 88224Ra+ 24α_$

Step (5): Emission of $α$ particle from $88224Ra$

The given unbalanced nuclear equation is,

$88224Ra→ ? + 24α$

Product formed by $α$ decaying of $88224Ra$ is determined as follows,

$Sum of Atomic number of Reactant = Product88(Ra)= x + 2(alpha)x = 86Element with Z = 86 ⇒ Radon(Rn)Sum of Mass number of Reactant = Product88224Ra = 86xRn + 4(alpha)x = 220$

The elemental symbol of the unknown element is $86220Rn$ . Therefore, the balanced nuclear equation is,

$88224Ra→ 86220Rn + 24α_$

Step (6): Emission of $α$ particle from $86220Rn$

The given unbalanced nuclear equation is,

$86220Rn→ ? + 24α$

Product formed by $α$ decaying of $86220Rn$ is determined as follows,

$Sum of Atomic number of Reactant = Product86(Rn)= x + 2(alpha)x = 84Element with Z = 84 ⇒ Polonium(Po)Sum of Mass number of Reactant = Product86220Rn = 84xPo + 4(alpha)x = 216$

The elemental symbol of the unknown element is $84216Po$ . Therefore, the balanced nuclear equation is,

$86220Rn→ 84216Po + 24α_$

Step (7): Emission of $α$ particle from $84216Po$

The given unbalanced nuclear equation is,

$84216Po→ ? + 24α$

Product formed by $α$ decaying of $84216Po$ is determined as follows,

$Sum of Atomic number of Reactant = Product84(Po)= x + 2(alpha)x = 82Element with Z = 82 ⇒ Lead(Pb)Sum of Mass number of Reactant = Product84216Po = 82xPb + 4(alpha)x = 212$

The elemental symbol of the unknown element is $82212Pb$ . Therefore, the balanced nuclear equation is,

$84216Po→ 82212Pb + 24α_$

Step (8): Emission of $β−$ particle from $82212Pb$

The given unbalanced nuclear equation is,

$82212Pb→ −10β + ?$

Product formed by $β−$ decaying of $82212Pb$ is determined as follows,

$Sum of Atomic number of Reactant = Product82(Pb) = x+ −1(β−) x = 83Element with Z = 83 ⇒ Bismuth(Bi)Sum of Atomic number of Reactant = Product82212Pb = 83xBi + 0(β−)x = 212$

The elemental symbol of the unknown element is $83212Bi$ . Therefore, the balanced nuclear equation is,

$82209Pb → −10β + 83212Bi_$

Step (9): Emission of $β−$ particle from $83212Bi$

The given unbalanced nuclear equation is,

$83212Bi → −10β + ?$

Product formed by $β−$ decaying of $83212Bi$ is determined as follows,

$Sum of Atomic number of Reactant = Product83(Bi) = x+ −1(β−) x = 84Element with Z = 84 ⇒ Polonium(Po)Sum of Atomic number of Reactant = Product83212Bi = 84xPo + 0(β−)x = 212$

The elemental symbol of the unknown element is $84212Po$ . Therefore, the balanced nuclear equation is,

$83212Bi → −10β + 84212Po_$

Step (10): Emission of $α$ particle from $84212Po$

The given unbalanced nuclear equation is,

$84212Po→ ? + 24α$

Product formed by $α$ decaying of $84212Po$ is determined as follows,

$Sum of Atomic number of Reactant = Product84(Po)= x + 2(alpha)x = 82Element with Z = 82 ⇒ Lead(Pb)Sum of Mass number of Reactant = Product84212Po = 82xPb + 4(alpha)x = 208$

The elemental symbol of the unknown element is $82208Pb$ . Therefore, the balanced nuclear equation is,

$84212Po→ 82208Pb + 24α_$

Therefore, balanced nuclear equations for decay series of thorium-232 follows as,

$90232Th→ 88228Ra + 24α88228Ra→ −10β + 89228Ac89228Ac → −10β + 90228Th90228Th→ 88224Ra+ 24α88224Ra→ 86220Rn + 24α86220Rn→ 84216Po + 24α84216Po→ 82212Pb + 24α82209Pb → −10β + 83212Bi83212Bi → −10β + 84212Po84212Po→ 82208Pb + 24α$

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Ch. 24.1 - Prob. 24.1AFP Ch. 24.1 - Prob. 24.1BFP Ch. 24.1 - Prob. 24.2AFP Ch. 24.1 - Why is stable but unstable? Ch. 24.1 - Prob. 24.3AFP Ch. 24.1 - Prob. 24.3BFP Ch. 24.2 - Prob. 24.4AFP Ch. 24.2 - Prob. 24.4BFP Ch. 24.2 - Prob. 24.5AFP Ch. 24.2 - Prob. 24.5BFP

Explanation of Solution

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