Chapter 10, Problem 10E

(a)

To determine

The decay mode for each of the sequential decays.

Answer to Problem 10E

The decay mode for each of the sequential decays is,

Explanation of Solution

Neptunium $\left(\text{N}\text{p}\right)$ decays through alpha decay to form protactinium $\left(\text{P}\text{a}\right)$.

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

Protactinium $\left(\text{P}\text{a}\right)$ decays through beta decay to form uranium $\left(\text{U}\right)$.

$\text{P}\text{a}\to \text{U}+\text{e}$

Uranium $\left(\text{U}\right)$ decays through alpha decay to form thorium $\left(\text{T}\text{h}\right)$.

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

Thorium $\left(\text{T}\text{h}\right)$ decays through alpha decay to form radium $\left(\text{R}\text{a}\right)$.

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

Radium $\left(\text{R}\text{a}\right)$ decays through beta decay to form actinium $\left(\text{A}\text{c}\right)$.

$\text{R}\text{a}\to \text{A}\text{c}+\text{e}$

Actinium $\left(\text{A}\text{c}\right)$ decays through alpha decay to form francium $\left(\text{F}\text{r}\right)$.

$\text{A}\text{c}\to \text{F}\text{r}+\text{H}\text{e}$

Francium $\left(\text{F}\text{r}\right)$ decays through alpha decay to form astatine $\left(\text{A}\text{t}\right)$.

$\text{F}\text{r}\to \text{A}\text{t}+\text{H}\text{e}$

Astatine $\left(\text{A}\text{t}\right)$ decays through alpha decay to form bismuth $\left(\text{B}\text{i}\right)$.

$\text{A}\text{t}\to \text{B}\text{i}+\text{H}\text{e}$

Bismuth $\left(\text{B}\text{i}\right)$ decays through both alpha and beta decay to form thallium $\left(\text{T}\text{l}\right)$ and polonium $\left(\text{P}\text{o}\right)$ respectively.

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

Thallium $\left(\text{T}\text{l}\right)$ decays through beta decay to form lead $\left(\text{P}\text{b}\right)$.

$\text{T}\text{l}\to \text{P}\text{b}+\text{e}$

Lead $\left(\text{P}\text{b}\right)$ decays through beta decay to form stable bismuth $\left(\text{B}\text{i}\right)$.

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

Conclusion:

Therefore, the decay mode for each of the sequential decays is,

(b)

To determine

The daughter nucleus at the end of each decay.

Answer to Problem 10E

The daughter nucleus at the end of each decay is,

Explanation of Solution

The daughter nucleus after alpha decay of Neptunium $\left(\text{N}\text{p}\right)$ is Protactinium $\left(\text{P}\text{a}\right)$.

The daughter nucleus after beta decay of Protactinium $\left(\text{P}\text{a}\right)$ is Uranium $\left(\text{U}\right)$.

The daughter nucleus after alpha decay of Uranium $\left(\text{U}\right)$ is Thorium $\left(\text{T}\text{h}\right)$.

The daughter nucleus after alpha decay of Thorium $\left(\text{T}\text{h}\right)$ is Radium $\left(\text{R}\text{a}\right)$.

The daughter nucleus after beta decay of Radium $\left(\text{R}\text{a}\right)$ is Actinium $\left(\text{A}\text{c}\right)$.

The daughter nucleus after alpha decay of Actinium $\left(\text{A}\text{c}\right)$ is Francium $\left(\text{F}\text{r}\right)$.

The daughter nucleus after alpha decay of Francium $\left(\text{F}\text{r}\right)$ is Astatine $\left(\text{A}\text{t}\right)$.

The daughter nucleus after alpha decay of Astatine $\left(\text{A}\text{t}\right)$ is Bismuth $\left(\text{B}\text{i}\right)$.

The daughter nucleus after alpha decay and beta decay of Bismuth $\left(\text{B}\text{i}\right)$ is Thallium $\left(\text{T}\text{l}\right)$ and Polonium $\left(\text{P}\text{o}\right)$ respectively.

The daughter nucleus after beta decay of Thallium $\left(\text{T}\text{l}\right)$ is Lead $\left(\text{P}\text{b}\right)$.

The daughter nucleus after beta decay of Lead $\left(\text{P}\text{b}\right)$ is stable Bismuth $\left(\text{B}\text{i}\right)$.

Conclusion:

Therefore, the daughter nucleus at the end of each decay is,