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Chapter 11, Problem 11.22EP

(a)

Interpretation Introduction

Interpretation:

Balanced nuclear equation for decay reaction of potassium-40 to calcium-40 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 11.22EP

Balanced nuclear equation is,

K1940 β10 + C2040a

Explanation of Solution

Given decay reaction is potassium-40 to calcium-40.  The atomic number of potassium is 19.  Atomic number of calcium is 20.  Therefore, the nuclear equation for this decay reaction can be given as,

K1940 ? + C2040a

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,

K1940 β10 + C2040a

(b)

Interpretation Introduction

Interpretation:

Balanced nuclear equation for decay reaction of lead-204 to mercury-200 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 11.22EP

Balanced nuclear equation is,

P82204b α24 + H80200g

Explanation of Solution

Given decay reaction is lead-204 to mercury-200.  The atomic number of lead is 82.  Atomic number of mercury is 80.  Therefore, the nuclear equation for this decay reaction can be given as,

P82204b ? + H80200g

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,

P82204b α24 + H80200g

(c)

Interpretation Introduction

Interpretation:

Balanced nuclear equation for decay reaction of U-238 to Th-234 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 11.22EP

Balanced nuclear equation is,

U92238 α24 + T90234h

Explanation of Solution

Given decay reaction is U-238 to Th-234.  The atomic number of uranium is 92.  Atomic number of thorium is 90.  Therefore, the nuclear equation for this decay reaction can be given as,

U92238 ? + T90234h

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,

U92238 α24 + T90234h

(d)

Interpretation Introduction

Interpretation:

Balanced nuclear equation for decay reaction of Rh-104 to Pd-104 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 11.22EP

Balanced nuclear equation is,

R45104h β10 + P46104d

Explanation of Solution

Given decay reaction is Rh-104 to Pd-104.  The atomic number of rhodium is 45.  Atomic number of palladium is 46.  Therefore, the nuclear equation for this decay reaction can be given as,

R45104h ? + P46104d

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,

R45104h β10 + P46104d

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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
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