Chemistry: The Molecular Nature of Matter and Change
Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781259631757
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
Question
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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
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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 =AtomicnumberMassnumberX

  • 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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Chapter 24 Solutions

Chemistry: The Molecular Nature of Matter and Change

Ch. 24.2 - Prob. 24.6AFPCh. 24.2 - Prob. 24.6BFPCh. 24.6 - Prob. 24.7AFPCh. 24.6 - Prob. 24.7BFPCh. 24.7 - Prob. B24.1PCh. 24.7 - Prob. B24.2PCh. 24.7 - Prob. B24.3PCh. 24.7 - Prob. B24.4PCh. 24 - Prob. 24.1PCh. 24 - Prob. 24.2PCh. 24 - Prob. 24.3PCh. 24 - Prob. 24.4PCh. 24 - Prob. 24.5PCh. 24 - Prob. 24.6PCh. 24 - Prob. 24.7PCh. 24 - Prob. 24.8PCh. 24 - Prob. 24.9PCh. 24 - Prob. 24.10PCh. 24 - Prob. 24.11PCh. 24 - Prob. 24.12PCh. 24 - Prob. 24.13PCh. 24 - Prob. 24.14PCh. 24 - Prob. 24.15PCh. 24 - Prob. 24.16PCh. 24 - Prob. 24.17PCh. 24 - Prob. 24.18PCh. 24 - Prob. 24.19PCh. 24 - Prob. 24.20PCh. 24 - Prob. 24.21PCh. 24 - Prob. 24.22PCh. 24 - Prob. 24.23PCh. 24 - Prob. 24.24PCh. 24 - Prob. 24.25PCh. 24 - Prob. 24.26PCh. 24 - Prob. 24.27PCh. 24 - Prob. 24.28PCh. 24 - Prob. 24.29PCh. 24 - Prob. 24.30PCh. 24 - Prob. 24.31PCh. 24 - Prob. 24.32PCh. 24 - Prob. 24.33PCh. 24 - Prob. 24.34PCh. 24 - Prob. 24.35PCh. 24 - Prob. 24.36PCh. 24 - Prob. 24.37PCh. 24 - Prob. 24.38PCh. 24 - Prob. 24.39PCh. 24 - Prob. 24.40PCh. 24 - Prob. 24.41PCh. 24 - Prob. 24.42PCh. 24 - Prob. 24.43PCh. 24 - Prob. 24.44PCh. 24 - Prob. 24.45PCh. 24 - Prob. 24.46PCh. 24 - Prob. 24.47PCh. 24 - Prob. 24.48PCh. 24 - Prob. 24.49PCh. 24 - Prob. 24.50PCh. 24 - Prob. 24.51PCh. 24 - Prob. 24.52PCh. 24 - Prob. 24.53PCh. 24 - Prob. 24.54PCh. 24 - Prob. 24.55PCh. 24 - Prob. 24.56PCh. 24 - Prob. 24.57PCh. 24 - Prob. 24.58PCh. 24 - Prob. 24.59PCh. 24 - Prob. 24.60PCh. 24 - Prob. 24.61PCh. 24 - Prob. 24.62PCh. 24 - Prob. 24.63PCh. 24 - Prob. 24.64PCh. 24 - Prob. 24.65PCh. 24 - Prob. 24.66PCh. 24 - Prob. 24.67PCh. 24 - Prob. 24.68PCh. 24 - Prob. 24.69PCh. 24 - Prob. 24.70PCh. 24 - Prob. 24.71PCh. 24 - Prob. 24.72PCh. 24 - Prob. 24.73PCh. 24 - Prob. 24.74PCh. 24 - Prob. 24.75PCh. 24 - Prob. 24.76PCh. 24 - Prob. 24.77PCh. 24 - Prob. 24.78PCh. 24 - Prob. 24.79PCh. 24 - Prob. 24.80PCh. 24 - Prob. 24.81PCh. 24 - Prob. 24.82PCh. 24 - Prob. 24.83PCh. 24 - Prob. 24.84PCh. 24 - Prob. 24.85PCh. 24 - Prob. 24.86PCh. 24 - Prob. 24.87PCh. 24 - Prob. 24.88PCh. 24 - Prob. 24.89PCh. 24 - Prob. 24.90PCh. 24 - Prob. 24.91PCh. 24 - Prob. 24.92PCh. 24 - Prob. 24.93PCh. 24 - Prob. 24.94PCh. 24 - Prob. 24.95PCh. 24 - Prob. 24.96PCh. 24 - Prob. 24.97PCh. 24 - Prob. 24.98PCh. 24 - Prob. 24.99PCh. 24 - Prob. 24.100PCh. 24 - Prob. 24.101PCh. 24 - Prob. 24.102PCh. 24 - Prob. 24.103PCh. 24 - Prob. 24.104PCh. 24 - Prob. 24.105PCh. 24 - Prob. 24.106PCh. 24 - Prob. 24.107PCh. 24 - Prob. 24.108PCh. 24 - Prob. 24.109PCh. 24 - Prob. 24.110PCh. 24 - Prob. 24.111PCh. 24 - Prob. 24.112PCh. 24 - Prob. 24.113PCh. 24 - Prob. 24.114PCh. 24 - Prob. 24.115PCh. 24 - Prob. 24.116PCh. 24 - Prob. 24.117PCh. 24 - Prob. 24.118PCh. 24 - Prob. 24.119PCh. 24 - Prob. 24.120PCh. 24 - Prob. 24.121PCh. 24 - Prob. 24.122PCh. 24 - Prob. 24.123PCh. 24 - Prob. 24.124PCh. 24 - Prob. 24.125PCh. 24 - Prob. 24.126PCh. 24 - Prob. 24.127PCh. 24 - Prob. 24.128PCh. 24 - Prob. 24.129PCh. 24 - Prob. 24.130PCh. 24 - Prob. 24.131PCh. 24 - Prob. 24.132PCh. 24 - Prob. 24.133PCh. 24 - Prob. 24.134PCh. 24 - Prob. 24.135PCh. 24 - Prob. 24.136PCh. 24 - Prob. 24.137PCh. 24 - Prob. 24.138PCh. 24 - Prob. 24.139PCh. 24 - Prob. 24.140PCh. 24 - Prob. 24.141PCh. 24 - Prob. 24.142PCh. 24 - Prob. 24.143PCh. 24 - Prob. 24.144P
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