Chapter 11, Problem 5PEA

(a)

To determine

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

Answer to Problem 5PEA

The nuclear equation for the alpha emission decay for $\text{U}$ is $\text{U}\to \text{T}\text{h}+\text{H}\text{e}$.

Explanation of Solution

The nuclear equation for the alpha emission decay reaction for the given element is given as,

    $\text{X}\to \text{Y}+\text{H}\text{e}$

Here, $\text{X}$ is the given element which has $p$ number of proton and has $q$ atomic mass, $\text{Y}$ is the element which has $\left(p-2\right)$ number of protons and has $\left(q-4\right)$ atomic mass after alpha emission decay reaction of the given element and $\text{He}$ is the helium released.

Thus the nuclear equation for the alpha emission decay reaction of $\text{U}$ is given as,

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

(b)

To determine

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

Answer to Problem 5PEA

The nuclear equation for the alpha emission decay for $\text{R}\text{a}$ is $\text{R}\text{a}\to \text{R}\text{n}+\text{H}\text{e}$.

Explanation of Solution

The nuclear equation for the alpha emission decay reaction for the given element is given as,

    $\text{X}\to \text{Y}+\text{H}\text{e}$

Here, $\text{X}$ is the given element which has $p$ number of proton and has $q$ atomic mass, $\text{Y}$ is the element which has $\left(p-2\right)$ number of protons and has $\left(q-4\right)$ atomic mass after alpha emission decay reaction of the given element and $\text{He}$ is the helium released.

Thus the nuclear equation for the alpha emission decay reaction of $\text{R}\text{a}$ is given as,

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

(c)

To determine

The nuclear equation for the alpha emission decay reaction for $\text{P}\text{u}$.

Answer to Problem 5PEA

The nuclear equation for the alpha emission decay for $\text{P}\text{u}$ is $\text{P}\text{u}\to \text{U}+\text{H}\text{e}$.

Explanation of Solution

The nuclear equation for the alpha emission decay reaction for the given element is given as,

    $\text{X}\to \text{Y}+\text{H}\text{e}$

Here, $\text{X}$ is the given element which has $p$ number of proton and has $q$ atomic mass, $\text{Y}$ is the element which has $\left(p-2\right)$ number of protons and has $\left(q-4\right)$ atomic mass after alpha emission decay reaction of the given element and $\text{He}$ is the helium released.

Thus the nuclear equation for the alpha emission decay reaction of $\text{P}\text{u}$ is given as,

    $\text{P}\text{u}\to \text{U}+\text{H}\text{e}$

(d)

To determine

The nuclear equation for the alpha emission decay reaction for $\text{B}\text{i}$.

Answer to Problem 5PEA

The nuclear equation for the alpha emission decay for $\text{B}\text{i}$ is $\text{B}\text{i}\to \text{T}\text{l}+\text{H}\text{e}$.

Explanation of Solution

The nuclear equation for the alpha emission decay reaction for the given element is given as,

    $\text{X}\to \text{Y}+\text{H}\text{e}$

Here, $\text{X}$ is the given element which has $p$ number of proton and has $q$ atomic mass, $\text{Y}$ is the element which has $\left(p-2\right)$ number of protons and has $\left(q-4\right)$ atomic mass after alpha emission decay reaction of the given element and $\text{He}$ is the helium released.

Thus the nuclear equation for the alpha emission decay reaction of $\text{B}\text{i}$ is given as,

    $\text{B}\text{i}\to \text{T}\text{l}+\text{H}\text{e}$

(e)

To determine

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

Answer to Problem 5PEA

The nuclear equation for the alpha emission decay for $\text{T}\text{h}$ is $\text{T}\text{h}\to \text{R}\text{a}+\text{H}\text{e}$.

Explanation of Solution

The nuclear equation for the alpha emission decay reaction for the given element is given as,

    $\text{X}\to \text{Y}+\text{H}\text{e}$

Here, $\text{X}$ is the given element which has $p$ number of proton and has $q$ atomic mass, $\text{Y}$ is the element which has $\left(p-2\right)$ number of protons and has $\left(q-4\right)$ atomic mass after alpha emission decay reaction of the given element and $\text{He}$ is the helium released.

Thus the nuclear equation for the alpha emission decay reaction of $\text{T}\text{h}$ is given as,

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

(f)

To determine

The nuclear equation for the alpha emission decay reaction for $\text{P}\text{o}$.

Answer to Problem 5PEA

The nuclear equation for the alpha emission decay for $\text{P}\text{o}$ is $\text{P}\text{o}\to \text{P}\text{b}+\text{H}\text{e}$.

Explanation of Solution

The nuclear equation for the alpha emission decay reaction for the given element is given as,

    $\text{X}\to \text{Y}+\text{H}\text{e}$

Here, $\text{X}$ is the given element which has $p$ number of proton and has $q$ atomic mass, $\text{Y}$ is the element which has $\left(p-2\right)$ number of protons and has $\left(q-4\right)$ atomic mass after alpha emission decay reaction of the given element and $\text{He}$ is the helium released.

Thus the nuclear equation for the alpha emission decay reaction of $\text{P}\text{o}$ is given as,

    $\text{P}\text{o}\to \text{P}\text{b}+\text{H}\text{e}$