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: AZX =MassnumberAtomicnumberX

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

An alpha particle is emitted and an alpha particle is 42α. Atomic number of Thorium is 90.

The given unbalanced nuclear equation is,

  23290Th ? + 42α

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 = Product23290Th =x82Ra+ 4(alpha)x=228

The elemental symbol of the unknown element is 22888Ra. Therefore, the balanced nuclear equation is,

  23290Th 22888Ra + 42α_

  • Step (2): Emission of β particle from 22888Ra

A β particle is emitted and a β particle is 01β.

The given unbalanced nuclear equation is,

  22888Ra 01β + ?

Product formed by β decaying of 22888Ra 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 = Product22888Ra  =x89Ac +  0(β)x=228

The elemental symbol of the unknown element is 22889Ac. Therefore, the balanced nuclear equation is,

  22888Ra 01β + 22889Ac_

  • Step (3): Emission of β particle from 22889Ac

The given unbalanced nuclear equation is,

  22889Ac  01β + ?

Product formed by β decaying of 22889Ac 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 = Product22889Ac  =x90Th +  0(β)x=228

The elemental symbol of the unknown element is 22890Th. Therefore, the balanced nuclear equation is,

  22889Ac  01β + 22890Th_

  • Step (4): Emission of α particle from 22890Th

The given unbalanced nuclear equation is,

  22890Th ? + 42α

Product formed by α decaying of 22890Th 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 = Product22890Th =x88Ra+ 4(alpha)x=224

The elemental symbol of the unknown element is 22488Ra. Therefore, the balanced nuclear equation is,

  22890Th 22488Ra+ 42α_

  • Step (5): Emission of α particle from 22488Ra

The given unbalanced nuclear equation is,

  22488Ra ? + 42α

Product formed by α decaying of 22488Ra 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 = Product22488Ra =x86Rn+ 4(alpha)x=220

The elemental symbol of the unknown element is 22086Rn. Therefore, the balanced nuclear equation is,

  22488Ra 22086Rn + 42α_

  • Step (6): Emission of α particle from 22086Rn

The given unbalanced nuclear equation is,

  22086Rn ? + 42α

Product formed by α decaying of 22086Rn 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 = Product22086Rn =x84Po+ 4(alpha)x=216

The elemental symbol of the unknown element is 21684Po. Therefore, the balanced nuclear equation is,

  22086Rn 21684Po + 42α_

  • Step (7): Emission of α particle from 21684Po

The given unbalanced nuclear equation is,

  21684Po ? + 42α

Product formed by α decaying of 21684Po 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 = Product21684Po =x82Pb+ 4(alpha)x=212

The elemental symbol of the unknown element is 21282Pb. Therefore, the balanced nuclear equation is,

  21684Po 21282Pb + 42α_

  • Step (8): Emission of β particle from 21282Pb

The given unbalanced nuclear equation is,

  21282Pb 01β + ? 

Product formed by β decaying of 21282Pb 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 = Product21282Pb  =x83Bi +  0(β)x=212

The elemental symbol of the unknown element is 21283Bi. Therefore, the balanced nuclear equation is,

  20982Pb  01β + 21283Bi_

  • Step (9): Emission of β particle from 21283Bi

The given unbalanced nuclear equation is,

  21283Bi  01β + ? 

Product formed by β decaying of 21283Bi 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 = Product21283Bi  =x84Po +  0(β)x=212

The elemental symbol of the unknown element is 21284Po. Therefore, the balanced nuclear equation is,

  21283Bi  01β + 21284Po_

  • Step (10): Emission of α particle from 21284Po

The given unbalanced nuclear equation is,

  21284Po ? + 42α

Product formed by α decaying of 21284Po 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 = Product21284Po =x82Pb+ 4(alpha)x=208

The elemental symbol of the unknown element is 20882Pb. Therefore, the balanced nuclear equation is,

  21284Po 20882Pb + 42α_

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

  23290Th 22888Ra + 42α22888Ra 01β + 22889Ac22889Ac  01β + 22890Th22890Th 22488Ra+ 42α22488Ra 22086Rn + 42α22086Rn 21684Po + 42α21684Po 21282Pb + 42α20982Pb  01β + 21283Bi21283Bi  01β + 21284Po21284Po 20882Pb + 42α

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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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