bartleby

Concept explainers

Question
Book Icon
Chapter 11, Problem 11.21EP

(a)

Interpretation Introduction

Interpretation:

Balanced nuclear equation for decay reaction of polonium-210 to lead-206 has to be written.

Concept Introduction:

If the reaction occurs in the nucleus of an atom then it is known as nuclear reaction.  These reactions are not considered as ordinary chemical reactions because the electrons do not take part in reaction while the particle inside the nucleus does.  Isotope and nuclide are almost similar terms.  Isotopes refer to the same element that has different mass number while nuclide refers to atoms of same or different elements with specific atomic number and mass number.

This nuclear reaction can be represented by nuclear equation.  This is not a normal chemical equation.  Nuclear equation considers the mass number and atomic number of the reactants and products.  Unstable nucleus tends to emit radiation spontaneously.  During this process the nuclide is transformed into nuclide of another element.  Parent nuclide is the one which undergoes the radioactive decay.  Daughter nuclide is the one that is formed from parent nuclide after radioactive decay.

The radioactive decay can take place by emission of alpha particle, beta particle or gamma ray emission.  Alpha particle decay is a process in which an alpha particle is emitted.  This results in the formation of nuclide of different element that has atomic number that is 2 less and mass number that is 4 less than the original nucleus.  Beta particle decay is a process in which a beta particle is emitted.  This produces a nuclide of different element similar to that of alpha particle decay.  The mass number is same as that of parent nuclide while the atomic number increases by 1 unit.  Gamma ray emission is a process in which the unstable nucleus emits gamma ray.  This occurs along with alpha or beta particle emission.  The gamma rays are not shown in the nuclear equation because they do not affect balancing the nuclear equation.

(a)

Expert Solution
Check Mark

Answer to Problem 11.21EP

Balanced nuclear equation is,

P84210o α24 + P82206b

Explanation of Solution

Given decay reaction is polonium-210 to lead-206.  The atomic number of polonium is 82.  Atomic number of lead is 82.  Therefore, the nuclear equation for this decay reaction can be given as,

P84210o ? + P82206b

The sum of subscript on both sides has to be equal and the sum of superscript on both sides has to be equal.  Considering this, the particle that is emitted is found to contain 4 as superscript and 2 as subscript.  This means it is an alpha particle.  Therefore, the given decay reaction is classified as alpha decay.  The balanced nuclear equation can be given as shown below,

P84210o α24 + P82206b

(b)

Interpretation Introduction

Interpretation:

Balanced nuclear equation for decay reaction of thorium-225 to protactinium-225 has to be written.

Concept Introduction:

If the reaction occurs in the nucleus of an atom then it is known as nuclear reaction.  These reactions are not considered as ordinary chemical reactions because the electrons do not take part in reaction while the particle inside the nucleus does.  Isotope and nuclide are almost similar terms.  Isotopes refer to the same element that has different mass number while nuclide refers to atoms of same or different elements with specific atomic number and mass number.

This nuclear reaction can be represented by nuclear equation.  This is not a normal chemical equation.  Nuclear equation considers the mass number and atomic number of the reactants and products.  Unstable nucleus tends to emit radiation spontaneously.  During this process the nuclide is transformed into nuclide of another element.  Parent nuclide is the one which undergoes the radioactive decay.  Daughter nuclide is the one that is formed from parent nuclide after radioactive decay.

The radioactive decay can take place by emission of alpha particle, beta particle or gamma ray emission.  Alpha particle decay is a process in which an alpha particle is emitted.  This results in the formation of nuclide of different element that has atomic number that is 2 less and mass number that is 4 less than the original nucleus.  Beta particle decay is a process in which a beta particle is emitted.  This produces a nuclide of different element similar to that of alpha particle decay.  The mass number is same as that of parent nuclide while the atomic number increases by 1 unit.  Gamma ray emission is a process in which the unstable nucleus emits gamma ray.  This occurs along with alpha or beta particle emission.  The gamma rays are not shown in the nuclear equation because they do not affect balancing the nuclear equation.

(b)

Expert Solution
Check Mark

Answer to Problem 11.21EP

Balanced nuclear equation is,

T90225h β10 + P91225a

Explanation of Solution

Given decay reaction is thorium-225 to protactinium-225.  The atomic number of thorium is 90.  Atomic number of protactinium is 91.  Therefore, the nuclear equation for this decay reaction can be given as,

T90225h ? + P91225a

The sum of subscript on both sides has to be equal and the sum of superscript on both sides has to be equal.  Considering this, the particle that is emitted is found to contain 0 as superscript and -1 as subscript.  This means it is a beta particle.  Therefore, the given decay reaction is classified as beta decay.  The balanced nuclear equation can be given as shown below,

T90225h β10 + P91225a

(c)

Interpretation Introduction

Interpretation:

Balanced nuclear equation for decay reaction of Pt-190 to Os-186 has to be written.

Concept Introduction:

If the reaction occurs in the nucleus of an atom then it is known as nuclear reaction.  These reactions are not considered as ordinary chemical reactions because the electrons do not take part in reaction while the particle inside the nucleus does.  Isotope and nuclide are almost similar terms.  Isotopes refer to the same element that has different mass number while nuclide refers to atoms of same or different elements with specific atomic number and mass number.

This nuclear reaction can be represented by nuclear equation.  This is not a normal chemical equation.  Nuclear equation considers the mass number and atomic number of the reactants and products.  Unstable nucleus tends to emit radiation spontaneously.  During this process the nuclide is transformed into nuclide of another element.  Parent nuclide is the one which undergoes the radioactive decay.  Daughter nuclide is the one that is formed from parent nuclide after radioactive decay.

The radioactive decay can take place by emission of alpha particle, beta particle or gamma ray emission.  Alpha particle decay is a process in which an alpha particle is emitted.  This results in the formation of nuclide of different element that has atomic number that is 2 less and mass number that is 4 less than the original nucleus.  Beta particle decay is a process in which a beta particle is emitted.  This produces a nuclide of different element similar to that of alpha particle decay.  The mass number is same as that of parent nuclide while the atomic number increases by 1 unit.  Gamma ray emission is a process in which the unstable nucleus emits gamma ray.  This occurs along with alpha or beta particle emission.  The gamma rays are not shown in the nuclear equation because they do not affect balancing the nuclear equation.

(c)

Expert Solution
Check Mark

Answer to Problem 11.21EP

Balanced nuclear equation is,

P78190t α24 + O76186s

Explanation of Solution

Given decay reaction is Pt-190 to Os-186.  The atomic number of platinum is 78.  Atomic number of osmium is 76.  Therefore, the nuclear equation for this decay reaction can be given as,

P78190t ? + O76186s

The sum of subscript on both sides has to be equal and the sum of superscript on both sides has to be equal.  Considering this, the particle that is emitted is found to contain 4 as superscript and 2 as subscript.  This means it is an alpha particle.  Therefore, the given decay reaction is classified as alpha decay.  The balanced nuclear equation can be given as shown below,

P78190t α24 + O76186s

(d)

Interpretation Introduction

Interpretation:

Balanced nuclear equation for decay reaction of O-19 to F-19 has to be written.

Concept Introduction:

If the reaction occurs in the nucleus of an atom then it is known as nuclear reaction.  These reactions are not considered as ordinary chemical reactions because the electrons do not take part in reaction while the particle inside the nucleus does.  Isotope and nuclide are almost similar terms.  Isotopes refer to the same element that has different mass number while nuclide refers to atoms of same or different elements with specific atomic number and mass number.

This nuclear reaction can be represented by nuclear equation.  This is not a normal chemical equation.  Nuclear equation considers the mass number and atomic number of the reactants and products.  Unstable nucleus tends to emit radiation spontaneously.  During this process the nuclide is transformed into nuclide of another element.  Parent nuclide is the one which undergoes the radioactive decay.  Daughter nuclide is the one that is formed from parent nuclide after radioactive decay.

The radioactive decay can take place by emission of alpha particle, beta particle or gamma ray emission.  Alpha particle decay is a process in which an alpha particle is emitted.  This results in the formation of nuclide of different element that has atomic number that is 2 less and mass number that is 4 less than the original nucleus.  Beta particle decay is a process in which a beta particle is emitted.  This produces a nuclide of different element similar to that of alpha particle decay.  The mass number is same as that of parent nuclide while the atomic number increases by 1 unit.  Gamma ray emission is a process in which the unstable nucleus emits gamma ray.  This occurs along with alpha or beta particle emission.  The gamma rays are not shown in the nuclear equation because they do not affect balancing the nuclear equation.

(d)

Expert Solution
Check Mark

Answer to Problem 11.21EP

Balanced nuclear equation is,

O819 β10 + F919

Explanation of Solution

Given decay reaction is O-19 to F-19.  The atomic number of oxygen is 8.  Atomic number of fluorine is 9.  Therefore, the nuclear equation for this decay reaction can be given as,

O819 ? + F919

The sum of subscript on both sides has to be equal and the sum of superscript on both sides has to be equal.  Considering this, the particle that is emitted is found to contain 0 as superscript and -1 as subscript.  This means it is a beta particle.  Therefore, the given decay reaction is classified as beta decay.  The balanced nuclear equation can be given as shown below,

O819 β10 + F919

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 11 Solutions

Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card

Ch. 11.4 - The half-life of cobalt-60 is 5.2 years. This...Ch. 11.4 - Prob. 2QQCh. 11.4 - Prob. 3QQCh. 11.4 - Prob. 4QQCh. 11.4 - Prob. 5QQCh. 11.5 - Prob. 1QQCh. 11.5 - The bombardment reaction involving 1123Na and 12H...Ch. 11.5 - Prob. 3QQCh. 11.5 - Prob. 4QQCh. 11.6 - Prob. 1QQCh. 11.6 - In the 14-step uranium-238 decay series a. all...Ch. 11.7 - Prob. 1QQCh. 11.7 - Prob. 2QQCh. 11.8 - Which of the following is not a form of ionizing...Ch. 11.8 - Prob. 2QQCh. 11.8 - Prob. 3QQCh. 11.8 - Prob. 4QQCh. 11.9 - Prob. 1QQCh. 11.9 - Which of the following correctly orders the three...Ch. 11.10 - Prob. 1QQCh. 11.10 - Prob. 2QQCh. 11.10 - Prob. 3QQCh. 11.11 - Prob. 1QQCh. 11.11 - Prob. 2QQCh. 11.11 - Prob. 3QQCh. 11.12 - Prob. 1QQCh. 11.12 - Prob. 2QQCh. 11.12 - Prob. 3QQCh. 11.12 - Prob. 4QQCh. 11.13 - Prob. 1QQCh. 11.13 - Prob. 2QQCh. 11 - Prob. 11.1EPCh. 11 - Prob. 11.2EPCh. 11 - Prob. 11.3EPCh. 11 - Prob. 11.4EPCh. 11 - Prob. 11.5EPCh. 11 - Prob. 11.6EPCh. 11 - Prob. 11.7EPCh. 11 - Prob. 11.8EPCh. 11 - Prob. 11.9EPCh. 11 - Prob. 11.10EPCh. 11 - Prob. 11.11EPCh. 11 - Prob. 11.12EPCh. 11 - Prob. 11.13EPCh. 11 - Prob. 11.14EPCh. 11 - Prob. 11.15EPCh. 11 - Prob. 11.16EPCh. 11 - Prob. 11.17EPCh. 11 - Prob. 11.18EPCh. 11 - Prob. 11.19EPCh. 11 - Prob. 11.20EPCh. 11 - Prob. 11.21EPCh. 11 - Prob. 11.22EPCh. 11 - Prob. 11.23EPCh. 11 - Prob. 11.24EPCh. 11 - Prob. 11.25EPCh. 11 - Prob. 11.26EPCh. 11 - Prob. 11.27EPCh. 11 - Prob. 11.28EPCh. 11 - Prob. 11.29EPCh. 11 - Fill in the blanks in each line of the following...Ch. 11 - Prob. 11.31EPCh. 11 - Prob. 11.32EPCh. 11 - Prob. 11.33EPCh. 11 - Prob. 11.34EPCh. 11 - Prob. 11.35EPCh. 11 - Prob. 11.36EPCh. 11 - Prob. 11.37EPCh. 11 - Prob. 11.38EPCh. 11 - Prob. 11.39EPCh. 11 - Prob. 11.40EPCh. 11 - Prob. 11.41EPCh. 11 - Prob. 11.42EPCh. 11 - Prob. 11.43EPCh. 11 - Prob. 11.44EPCh. 11 - Prob. 11.45EPCh. 11 - Prob. 11.46EPCh. 11 - Prob. 11.47EPCh. 11 - Prob. 11.48EPCh. 11 - Prob. 11.49EPCh. 11 - Prob. 11.50EPCh. 11 - Prob. 11.51EPCh. 11 - Prob. 11.52EPCh. 11 - Prob. 11.53EPCh. 11 - Prob. 11.54EPCh. 11 - Prob. 11.55EPCh. 11 - Prob. 11.56EPCh. 11 - Prob. 11.57EPCh. 11 - Write a chemical equation that involves water as a...Ch. 11 - Prob. 11.59EPCh. 11 - Prob. 11.60EPCh. 11 - Prob. 11.61EPCh. 11 - Prob. 11.62EPCh. 11 - Prob. 11.63EPCh. 11 - Prob. 11.64EPCh. 11 - Prob. 11.65EPCh. 11 - Prob. 11.66EPCh. 11 - Prob. 11.67EPCh. 11 - Prob. 11.68EPCh. 11 - Prob. 11.69EPCh. 11 - Prob. 11.70EPCh. 11 - Prob. 11.71EPCh. 11 - Prob. 11.72EPCh. 11 - Prob. 11.73EPCh. 11 - Prob. 11.74EPCh. 11 - Prob. 11.75EPCh. 11 - Prob. 11.76EPCh. 11 - Prob. 11.77EPCh. 11 - Prob. 11.78EPCh. 11 - Prob. 11.79EPCh. 11 - Prob. 11.80EPCh. 11 - Prob. 11.81EPCh. 11 - Prob. 11.82EPCh. 11 - Prob. 11.83EPCh. 11 - Prob. 11.84EPCh. 11 - Prob. 11.85EPCh. 11 - Prob. 11.86EPCh. 11 - Prob. 11.87EPCh. 11 - Prob. 11.88EPCh. 11 - Prob. 11.89EPCh. 11 - Prob. 11.90EP
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
General, Organic, and Biological Chemistry
Chemistry
ISBN:9781285853918
Author:H. Stephen Stoker
Publisher:Cengage Learning
Text book image
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
Text book image
Chemistry by OpenStax (2015-05-04)
Chemistry
ISBN:9781938168390
Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
Publisher:OpenStax
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Chemistry for Engineering Students
Chemistry
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Cengage Learning
Text book image
General Chemistry - Standalone book (MindTap Cour...
Chemistry
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Cengage Learning