Chapter 13, Problem 5PEB

(a)

To determine

The nuclear equation for the alpha emission decay reaction of $\text{A}\text{m}$.

Answer to Problem 5PEB

Solution: $\text{A}\text{m}\to \text{N}\text{p}+\text{H}\text{e}$

Explanation of Solution

Introduction:

In alpha decay process, the atomic number decreases by two units while mass number decreases by four units.

Explanation:

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

Step 1: The symbol for alpha particle is $\text{H}\text{e}$. The nuclear equation so far is:

$\text{A}\text{m}\to \text{H}\text{e}+?$

Step 2: From the subscripts, atomic number is calculated as: $95-2=93$. Thus, the new nucleus has an atomic number 93. The element with atomic number 93 is $Np$.

Step 3: From the superscripts, mass number of neptunium is calculated as: $241-4=237$. Thus, the product nucleus is $Np$.

Conclusion:

Thus, the complete nuclear equation for the alpha emission decay of the $\text{A}\text{m}$ is written as:

$\text{A}\text{m}\to \text{N}\text{p}+\text{H}\text{e}$.

(b)

To determine

The nuclear equation for the alpha emission decay reaction of $\text{T}\text{h}$.

Answer to Problem 5PEB

Solution: $\text{T}\text{h}\to \text{R}\text{a}+\text{H}\text{e}$

Explanation of Solution

Introduction:

In alpha decay process, the atomic number decreases by two units while mass number decreases by four units.

Explanation:

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

Step 1: The symbol for alpha particle is $\text{H}\text{e}$. The nuclear equation so far is:

$\text{T}\text{h}\to \text{H}\text{e}+?$

Step 2: From the subscripts, atomic number is calculated as: $90-2=88$. Thus, the new nucleus has an atomic number 88. The element with atomic number 88 is $Ra$.

Step 3: From the superscripts, mass number of radium is calculated as: $223-4=219$. Thus, the product nucleus is $Ra$.

Conclusion:

Thus, the complete nuclear equation for the alpha emission decay of $\text{T}\text{h}$ is written as:

$\text{T}\text{h}\to \text{R}\text{a}+\text{H}\text{e}$.

(c)

To determine

The nuclear equation for the alpha emission decay reaction of $\text{R}\text{a}$.

Answer to Problem 5PEB

Solution: $\text{R}\text{a}\to \text{R}\text{n}+\text{H}\text{e}$

Explanation of Solution

Introduction:

In alpha decay process, the atomic number decreases by two units while mass number decreases by four units.

Explanation:

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

Step 1: The symbol for alpha particle is $\text{H}\text{e}$. The nuclear equation so far is:

$\text{R}\text{a}\to \text{H}\text{e}+?$

Step 2: From the subscripts, atomic number is calculated as: $88-2=86$. Thus, the new nucleus has an atomic number 86. The element with atomic number 86 is $Rn$.

Step 3: From the superscripts, mass number of radon is calculated as: $223-4=219$. Thus, the product nucleus is $Rn$.

Conclusion:

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

$\text{R}\text{a}\to \text{R}\text{n}+\text{H}\text{e}$

(d)

To determine

The nuclear equation for the alpha emission decay reaction of $\text{U}$.

Answer to Problem 5PEB

Solution: $\text{U}\to \text{T}\text{h}+\text{H}\text{e}$

Explanation of Solution

Introduction: In alpha decay process, the atomic number decreases by two units while mass number decreases by four units.

Explanation:

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

Step 1: The symbol for alpha particle is $\text{H}\text{e}$. The nuclear equation so far is:

$\text{U}\to \text{H}\text{e}+?$

Step 2: From the subscripts, atomic number is calculated as: $92-2=90$. Thus, the new nucleus has an atomic number 90. The element with atomic number 90 is $Th$.

Step 3: From the superscripts, mass number of thorium is calculated as: $234-4=230$. Thus, the product nucleus is $Th$.

Conclusion:

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

$\text{U}\to \text{T}\text{h}+\text{H}\text{e}$

(e)

To determine

The nuclear equation for the alpha emission decay reaction of $\text{C}\text{m}$.

Answer to Problem 5PEB

Solution: $\text{C}\text{m}\to \text{P}\text{u}+\text{H}\text{e}$

Explanation of Solution

Introduction: In alpha decay process, the atomic number decreases by two units while mass number decreases by four units.

Explanation:

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

Step 1: The symbol for alpha particle is $\text{H}\text{e}$. The nuclear equation so far is:

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

Step 2: From the subscripts, atomic number is calculated as: $96-2=94$. Thus, the new nucleus has an atomic number 94. The element with atomic number 94 is $Pu$.

Step 3:From the superscripts, mass number of plutonium is calculated as: $242-4=238$. Thus, the product nucleus is $Pu$.

Conclusion:

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

$\text{C}\text{m}\to \text{P}\text{u}+\text{H}\text{e}$

(f)

To determine

The nuclear equation for the aplha emission decay reaction of $\text{N}\text{p}$.

Answer to Problem 5PEB

Solution: $\text{N}\text{p}\to \text{P}\text{a}+\text{H}\text{e}$

Explanation of Solution

Introduction: In alpha decay process, the atomic number decreases by two units while mass number decreases by four units.

Explanation:

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

Step 1: The symbol for alpha particle is $\text{H}\text{e}$. The nuclear equation so far is:

$\text{N}\text{p}\to \text{H}\text{e}+?$

Step 2: From the subscripts, atomic number is calculated as: $93-2=91$. Thus, the new nucleus has an atomic number 91. The element with atomic number 91 is $Pa$.

Step 3: From the superscripts, mass number of protactinium is calculated as: $237-4=233$. Thus, the product nucleus is $Pu$.

Conclusion:

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

$\text{N}\text{p}\to \text{P}\text{a}+\text{H}\text{e}$