Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781260151749
Author: Silberberg Dr., Martin; Amateis Professor, Patricia
Publisher: McGraw-Hill Education
Question
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Chapter 24, Problem 24.53P

(a)

Interpretation Introduction

Interpretation:

Missing species in the given transmutations, 10B(α,n)   _ and the full nuclear equation has to be written.

Concept Introduction:

Nuclear reaction can be written in the shorthand notation with the parentheses. Bombarding particle, that is projectile can be represented as first symbol in the parentheses and the emitted can be represented as the second particle in the parentheses.

Parent nucleus and daughter nucleus can be represented in the front part of the parentheses and back part of the parentheses respectively. For example:

  7N15(p,α)6C12Parentnucleus(Projectile,ejectile)Daughternucleus

Common particles in radioactive decay and nuclear transformations are mentioned below,

  ParticleSymbolProton11Hor11PNeutron01nElectron-10eAlphaparticle24Heor24αBetaparticle-10eor-10βPositron10e

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.

(a)

Expert Solution
Check Mark

Explanation of Solution

For the given reaction, short hand notation is 10B(α,n)X. From the notation it is clear that,

  Parentnucleus : 10BProjectile : αDaughternucleus : XEjectile : n

The given unbalanced nuclear equation is,

  10B + 24α  ? + 01n

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

Missing species is determined as follows,

  Sum of Atomic number of Reactant = Product5(B) + 2(alpha) = x + 0(neutron)x = 7Element with Z = 7  Nitrogen(N)Sum of Mass number of Reactant = Product510B  + 4(alpha) = 7xN + 1(neutron)x = 13

The elemental symbol of the missing species is 713N. Therefore, the balanced nuclear equation is,

  10B + 24α  713N + 01n

(b)

Interpretation Introduction

Interpretation:

Missing species in the given transmutations, 28Si(d,  _)29P and the full nuclear equation has to be written.

Concept Introduction:

Nuclear reaction can be written in the shorthand notation with the parentheses. Bombarding particle, that is projectile can be represented as first symbol in the parentheses and the emitted can be represented as the second particle in the parentheses.

Parent nucleus and daughter nucleus can be represented in the front part of the parentheses and back part of the parentheses respectively. For example:

  7N15(p,α)6C12Parentnucleus(Projectile,ejectile)Daughternucleus

Common particles in radioactive decay and nuclear transformations are mentioned below,

  ParticleSymbolProton11Hor11PNeutron01nElectron-10eAlphaparticle24Heor24αBetaparticle-10eor-10βPositron10e

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.

(b)

Expert Solution
Check Mark

Explanation of Solution

For the given reaction, short hand notation is 28Si(d,x)29P. From the notation it is clear that,

  Parentnucleus : 28SiProjectile : 2HDaughternucleus : 29PEjectile : x

The given unbalanced nuclear equation is,

  28Si + 2H  ? + 29P

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

Missing species is determined as follows,

  Sum of Atomic number of Reactant = Product14(Si) + 1(deuteron) = (15)P + xx = 0Atomic number = 0Sum of Mass number of Reactant = Product1428Si  + 2(deuteron) = x + 1529Px = 1Mass number = 1

By analyzing the X, atomic number of X is 0 and the atomic mass is 1. So the missing species is neutron, (01n).

Therefore, the balanced nuclear equation is,

  1428Si + 12H  01n + 1529P

(c)

Interpretation Introduction

Interpretation:

Missing species in the given transmutations,    _(α,2n)244Cf and the full nuclear equation has to be written.

Concept Introduction:

Nuclear reaction can be written in the shorthand notation with the parentheses. Bombarding particle, that is projectile can be represented as first symbol in the parentheses and the emitted can be represented as the second particle in the parentheses.

Parent nucleus and daughter nucleus can be represented in the front part of the parentheses and back part of the parentheses respectively. For example:

  7N15(p,α)6C12Parentnucleus(Projectile,ejectile)Daughternucleus

Common particles in radioactive decay and nuclear transformations are mentioned below,

  ParticleSymbolProton11Hor11PNeutron01nElectron-10eAlphaparticle24Heor24αBetaparticle-10eor-10βPositron10e

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.

(c)

Expert Solution
Check Mark

Explanation of Solution

For the given reaction, short hand notation is X(α,2n)244Cf. From the notation it is clear that,

  Parentnucleus : 10BProjectile : αDaughternucleus : XEjectile : n

The given unbalanced nuclear equation is,

  X + 24α  98244Cf + 201n

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

Missing species is determined as follows,

  Sum of Atomic number of Reactant = Productx + 2(alpha) = 98(Cm) + [2×0(neutron)]x = 7Element with Z = 96  Curium(Cm)Sum of Mass number of Reactant = Product96xCm + 4(alpha) = 98244Cf  + [2×1(neutron)]x = 242

The elemental symbol of the missing species is 96242Cm. Therefore, the balanced nuclear equation is,

  96242Cm + 24α  98244Cf + 201n

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Chapter 24 Solutions

Loose Leaf for 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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