Chemistry: The Molecular Nature of Matter and Change - Standalone book
Chemistry: The Molecular Nature of Matter and Change - Standalone book
7th Edition
ISBN: 9780073511177
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
bartleby

Concept explainers

Question
Book Icon
Chapter 24, Problem 24.54P

(a)

Interpretation Introduction

Interpretation:

Unidentified species and the shorthand notation have to be written in the given transmutations where gamma irradiation of a nuclide yields a proton, a neutron.

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

The given unbalanced nuclear equation is,

  X + γ  1429Si + 01n + 11H

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 + 0(gamma) = 14(Si) + 0(neutron) + 1(proton)x = 15Element with Z = 15  Phosphorus(P)Sum of Mass number of Reactant = Product15xP + 0(gamma) = 1429Si  + 1(neutron) + 1(proton)x = 31

The elemental symbol of the unidentified species is 1531P. Hence, the balanced nuclear equation is,

  96242Cm + 24α  98244Cf + 201n

Therefore, the shorthand notation of the given transmutation is 31P(γ,p,n)29Si.

(b)

Interpretation Introduction

Interpretation:

Unidentified species and the shorthand notation have to be written in the given transmutations where bombardment of 252Cf with 10B yields five neutrons and a nuclide.

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

The given unbalanced nuclear equation is,

  252Cf + 10B  ? + 501n

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) + 98(Cf) = x + [5×0(neutron)]x = 103Element with Z = 103  Lawrencium(Lr)Sum of Mass number of Reactant = Product510B  + 98252Cf = 103xLr + [5×1(neutron)]x = 257

The elemental symbol of the unidentified species is 103257Lr. Hence, the balanced nuclear equation is,

  252Cf + 10B  103257Lr + 501n

Therefore, the shorthand notation of the given transmutation is 252Cf(10B,5n)257Lr.

(c)

Interpretation Introduction

Interpretation:

Unidentified species and the shorthand notation have to be written in the given transmutations where bombardment of 238U with a particle yields three neutrons and 239Pu .

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

The given unbalanced nuclear equation is,

  92238U + ?  301n + 94239Pu

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 = Product92(U) + x = (94)Pu + [3×0(neutron)]x = 2 Atomic number = 2Sum of Mass number of Reactant = Product92238U   + x = [3×1(neutron)] + 94239Pux = 4Mass number = 4

By analyzing the X, atomic number of X is 2 and the atomic mass is 4. So the unidentified species is an alpha (24α). Hence, the balanced nuclear equation is,

  96242Cm + 24α  98244Cf + 201n

Therefore, the shorthand notation of the given transmutation is 238U(α,3n)239Pu.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 24 Solutions

Chemistry: The Molecular Nature of Matter and Change - Standalone book

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
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY