Chapter 11, Problem 5PEB

(a)

To determine

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

Answer to Problem 5PEB

The nuclear equation for the alpha emission decay is $\text{A}\text{m}\to \text{N}\text{p}+\text{H}\text{e}$.

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 americium is given 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 for $\text{T}\text{h}$.

Answer to Problem 5PEB

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

Answer to Problem 5PEB

The nuclear equation for the alpha emission decay 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 radium is given 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 for $\text{U}$.

Answer to Problem 5PEB

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

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

(e)

To determine

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

Answer to Problem 5PEB

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

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 curium is given as,

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

(f)

To determine

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

Answer to Problem 5PEB

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

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 neptunium is given as,

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