Chapter 13, Problem 4PEB

(a)

To determine

To write: The nuclear equation for the beta emission decay of $\text{C}$.

Answer to Problem 4PEB

Solution: $\text{C}\to \text{N}+\text{e}$

Explanation of Solution

Introduction:

In beta decay process, the atomic number increases by one while there is no effect on mass number..

Explanation:

In order to write the nuclear equation for the beta emission decay of $\text{C}$ follow the following steps:

Step 1: The symbol for beta particle is $e$. The nuclear equation so far is:

$\text{C}\to \text{e}+?$

Step 2: From the subscripts, atomic number is calculated as: $6=-1+7$. Thus, the new nucleus has an atomic number 7. The element with atomic number 7 is $N$.

Step 3:From the superscripts, mass number of nitrogen is calculated as: $14-0=14$. Thus, the product nucleus is $N$.

Conclusion:

Thus, the complete nuclear equation for the beta emission decay of $\text{C}$ is written as:

$\text{C}\to \text{N}+\text{e}$

(b)

To determine

To write: The nuclear equation for the beta emission decay of $\text{C}\text{o}$.

Answer to Problem 4PEB

Solution: $\text{C}\text{o}\to \text{N}\text{i}+\text{e}$

Explanation of Solution

Introduction:

In beta decay process, the atomic number increases by one while there is no effect on mass number.

Explanation:

In order to write the nuclear equation for the beta emission decay of $\text{C}\text{o}$ follow the following steps:

Step 1: The symbol for beta particle is $e$. The nuclear equation so far is:

$\text{C}\text{o}\to \text{e}+?$

Step 2: From the subscripts, atomic number is calculated as: $27=-1+28$. Thus, the new nucleus has an atomic number 28. The element with atomic number 28 is $Ni$.

Step 3: From the superscripts, mass number of nickel is calculated as: $60-0=60$. Thus, the product nucleus is $Ni$.

Conclusion:

Thus, the complete nuclear equation for the beta emission decay of $\text{C}\text{o}$ is written as:

$\text{C}\text{o}\to \text{N}\text{i}+\text{e}$.

(c)

To determine

To write: The nuclear equation for the beta emission decay of $\text{N}\text{a}$.

Answer to Problem 4PEB

Solution: $\text{N}\text{a}\to \text{M}\text{g}+\text{e}$

Explanation of Solution

Introduction:

In beta decay process, the atomic number increases by one while there is no effect on mass number.

Explanation:

In order to write the nuclear equation for the beta emission decay of $\text{N}\text{a}$ follow the following steps:

Step 1: The symbol for beta particle is $e$. The nuclear equation so far is:

$\text{N}\text{a}\to \text{e}+?$

Step 2: From the subscripts, atomic number is calculated as: $11=-1+12$. Thus, the new nucleus has an atomic number 12. The element with atomic number 12 is $Mg$.

Step 3: From the superscripts, mass number of magnesium is calculated as: $24-0=24$. Thus, the product nucleus is $Mg$.

Conclusion:

Thus, the complete nuclear equation for the beta emission decay of $\text{N}\text{a}$ is written as:

$\text{N}\text{a}\to \text{M}\text{g}+\text{e}$.

(d)

To determine

To write: The nuclear equation for the beta emission decay of $\text{P}\text{u}$.

Answer to Problem 4PEB

Solution: $\text{P}\text{u}\to \text{A}\text{c}+\text{e}$

Explanation of Solution

Introduction:

In beta decay process, the atomic number increases by one while there is no effect on mass number.

Explanation:

In order to write the nuclear equation for the beta emission decay of $\text{P}\text{u}$ follow the following steps:

Step 1: The symbol for beta particle is $e$. The nuclear equation so far is:

$\text{P}\text{u}\to \text{e}+?$

Step 2: From the subscripts, atomic number is calculated as: $94=-1+95$. Thus, the new nucleus has an atomic number 95. The element with atomic number 95 is $Am$.

Step 3: From the superscripts, mass number of americium is calculated as: $214-0=214$. Thus, the product nucleus is $Am$.

Conclusion:

Thus, the complete nuclear equation for the beta emission decay of $\text{P}\text{u}$ is written as:

$\text{P}\text{u}\to \text{A}\text{m}+\text{e}$.

(e)

To determine

To determine/To Check: The nuclear equation for the beta emission decay of $\text{I}$.

Answer to Problem 4PEB

Solution: $\text{I}\to \text{X}\text{e}+\text{e}$

Explanation of Solution

Introduction:

In beta decay process, the atomic number increases by one while there is no effect on mass number.

Explanation:

In order to write the nuclear equation for the beta emission decay of $\text{I}$ follow the following steps:

Step 1: The symbol for beta particle is $e$. The nuclear equation so far is:

$\text{I}\to \text{e}+?$

Step 2: From the subscripts, atomic number is calculated as: $53=-1+54$. Thus, the new nucleus has an atomic number 54. The element with atomic number 54 is $Xe$.

Step 3: From the superscripts, mass number of xenon is calculated as: $131-0=131$. Thus, the product nucleus is $Xe$.

Conclusion:

Thus, the complete nuclear equation for the beta emission decay of $\text{I}$ is written as:

$\text{I}\to \text{X}\text{e}+\text{e}$

(f)

To determine

To write: The nuclear equation for the beta emission decay of $\text{P}\text{b}$.

Answer to Problem 4PEB

Solution: $\text{P}\text{b}\to \text{B}\text{i}+\text{e}$

Explanation of Solution

Introduction:

In beta decay process, the atomic number increases by one while there is no effect on mass number.

Explanation:

In order to write the nuclear equation for the beta emission decay of $\text{P}\text{b}$ follow the following steps:

Step 1: The symbol for beta particle is $e$. The nuclear equation so far is:

$\text{P}\text{b}\to \text{e}+?$

Step 2: From the subscripts, atomic number is calculated as: $82=-1+83$. Thus, the new nucleus has an atomic number 83. The element with atomic number 83 is $Bi$.

Step 3: From the superscripts, mass number of bismuth is calculated as: $210-0=210$. Thus, the product nucleus is $\text{B}\text{i}$.

Conclusion:

Thus, the complete nuclear equation for the beta emission decay of $\text{P}\text{b}$ is written as:

$\text{P}\text{b}\to \text{B}\text{i}+\text{e}$