Chapter 11, Problem 11.23EP

(a)

Interpretation Introduction

Interpretation:

Nuclear equation for beta emission that produces mercury-199 as product 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.

Answer to Problem 11.23EP

Nuclear equation is,

$\text{A}\text{u}\stackrel{}{\to }\beta +\text{H}\text{g}$

Explanation of Solution

Given nuclear reaction is beta emission produces mercury-199 as product.  This can be represented as,

$?\stackrel{}{\to }\beta +\text{H}\text{g}$

The sum of subscript on both sides has to be equal and the sum of superscript on both sides has to be equal.  In the problem statement it is given that beta emission takes place.  Therefore, there won’t be any change in mass number but the atomic number of the parent nuclide will be 1 less than the daughter nuclide.  This means the atomic number of the parent nuclide has to be 79.  The element that has atomic number 79 is gold.  Therefore, the nuclear equation can be written as,

$\text{A}\text{u}\stackrel{}{\to }\beta +\text{H}\text{g}$

(b)

Interpretation Introduction

Interpretation:

Nuclear equation for palladium-109 undergoing beta emission 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.

Answer to Problem 11.23EP

Nuclear equation is,

$\text{P}\text{d}\stackrel{}{\to }\beta +\text{A}\text{g}$

Explanation of Solution

Given nuclear reaction is palladium-109 undergoes a beta emission.  This can be represented as,

$\text{P}\text{d}\stackrel{}{\to }\beta +?$

The sum of subscript on both sides has to be equal and the sum of superscript on both sides has to be equal.  In the problem statement it is given that beta emission takes place.  Therefore, there won’t be any change in mass number but the atomic number of the daughter nuclide nucleus will be 1 more than the parent nuclide.  This means the atomic number of the daughter nuclide has to be 47.  The element that has atomic number 47 is silver.  Therefore, the nuclear equation can be written as,

$\text{P}\text{d}\stackrel{}{\to }\beta +\text{A}\text{g}$

(c)

Interpretation Introduction

Interpretation:

Nuclear equation for alpha emission that produces terbium-148 as product 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.

Answer to Problem 11.23EP

Nuclear equation is,

$\text{H}\text{o}\stackrel{}{\to }\alpha +\text{T}\text{b}$

Explanation of Solution

Given nuclear reaction is alpha emission produces terbium-148 as product.  This can be represented as,

$?\stackrel{}{\to }\alpha +\text{T}\text{b}$

The sum of subscript on both sides has to be equal and the sum of superscript on both sides has to be equal.  In the problem statement it is given that alpha emission takes place.  When alpha emission takes place the mass number of the formed daughter nuclide will be 4 less than the parent nuclide and atomic number will be 2 less than the parent nuclide.  This means the atomic number of the parent nuclide has to be 67 and mass number has to be 152.  The element that has atomic number 67 is holmium.  Therefore, the nuclear equation can be written as,

$\text{H}\text{o}\stackrel{}{\to }\alpha +\text{T}\text{b}$

(d)

Interpretation Introduction

Interpretation:

Nuclear equation for fermium-249 undergoing alpha emission 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.

Answer to Problem 11.23EP

Nuclear equation is,

$\text{F}\text{m}\stackrel{}{\to }\alpha +\text{C}\text{f}$

Explanation of Solution

Given nuclear reaction is alpha emission of fermium-249.  This can be represented as,

$\text{F}\text{m}\stackrel{}{\to }\alpha +?$

The sum of subscript on both sides has to be equal and the sum of superscript on both sides has to be equal.  In the problem statement it is given that alpha emission takes place.  When alpha emission takes place the mass number of the formed daughter nuclide will be 4 less than the parent nuclide and atomic number will be 2 less than the parent nuclide.  This means the atomic number of the daughter nuclide has to be 98 and mass number has to be 245.  The element that has atomic number 98 is californium.  Therefore, the nuclear equation can be written as,

$\text{F}\text{m}\stackrel{}{\to }\alpha +\text{C}\text{f}$