Chapter 18, Problem 12E

Interpretation Introduction

(a)

Interpretation:

The nuclide $\text{X}$ that decays by alpha emission to give $\text{R}\text{a}$ is to be stated.

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Answer to Problem 12E

The nuclide $\text{X}$ that decays by alpha emission to give $\text{R}\text{a}$ is thorium $\left(\text{T}\text{h}\right)$.

Explanation of Solution

It is given that $\text{X}$ undergoes alpha decay to give $\text{R}\text{a}$. An alpha particle is $\text{H}\text{e}$. Therefore, in the case of alpha decay the atomic number of an element decreases by $2$ and the mass number decreases by $4$. The nuclear equation is shown below.

$\text{X}{\to }_{88}^{\text{218}}\text{Ra}{+}_{\text{2}}^{\text{4}}\text{He}$

The mass number of $\text{X}$ can be calculated by the sum of mass number of both of the elements on the right side.

$\begin{array}{rll}\text{m}& =& 218+4\\ & =& 222\end{array}$

The atomic number of $\text{X}$ can be calculated by the sum of an atomic number of both of the elements on the right side.

$\begin{array}{rll}\text{x}& =& 88+2\\ & =& 90\end{array}$

The element having atomic number $90$ is Thorium. Therefore, the nuclide $\text{X}$ is represented as $\text{T}\text{h}$.

The resulting nuclear equation is shown below.

$\text{T}\text{h}{\to }_{88}^{\text{218}}\text{Ra}{+}_{\text{2}}^{\text{4}}\text{He}$

Conclusion

The nuclide $\text{X}$ which on alpha decay gives $\text{R}\text{a}$ is thorium $\left(\text{T}\text{h}\right)$.

Interpretation Introduction

(b)

Interpretation:

The nuclide $\text{X}$ that decays by beta emission to give $\text{F}\text{e}$ is to be stated.

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Answer to Problem 12E

The nuclide $\text{X}$ that decays by beta emission and gives $\text{F}\text{e}$ is $\text{M}\text{n}$.

Explanation of Solution

It is given that $\text{X}$ on beta emission gives $\text{F}\text{e}$. A beta particle is identical to an electron that is $e$. Therefore, in the case of beta decay the atomic number of an element increases by $1$, and there is no change in mass number.

The nuclear equation is shown below.

$\text{X}{\to }_{26}^{56}\text{Fe}+e$

The mass number of $\text{X}$ can be calculated by the sum of mass number of both of the elements on the right side.

$\begin{array}{rll}\text{m}& =& 56+0\\ & =& 56\end{array}$

The atomic number of $\text{X}$ can be calculated by the sum of atomic number of both of the elements on the right side.

$\begin{array}{rll}\text{x}& =& 26+\left(-1\right)\\ & =& 26-1\\ & =& 25\end{array}$

The element having atomic number $25$ is manganese $\left(\text{Mn}\right)$. Therefore, the nuclide $\text{X}$ is represented as $\text{M}\text{n}$.

The resulting nuclear equation is shown below.

$\text{M}\text{n}{\to }_{26}^{56}\text{Fe}+e$

Conclusion

The nuclide $\text{X}$ which on beta decay gives $\text{F}\text{e}$ is $\text{M}\text{n}$.

Interpretation Introduction

(c)

Interpretation:

The nuclide $\text{X}$ that decays by positron emission to give $\text{F}$ is to be stated.

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Answer to Problem 12E

The nuclide $\text{X}$ that decays on positron emission and gives $\text{F}$ is $\text{N}\text{e}$.

Explanation of Solution

It is given that $\text{X}$ on positron emission gives $\text{F}$. A positron is opposite of an electron and represented by $e$. Therefore, in the case of positron the atomic number of an element decreases by $1$ and there is no change in mass number.

The nuclear equation is shown below

$\text{X}{\to }_9^{19}\text{F}+e$

The mass number of $\text{X}$ can be calculated by the sum of mass number of both of the elements on the right side.

$\begin{array}{rll}\text{m}& =& 19+0\\ & =& 19\end{array}$

The atomic number of $\text{X}$ can be calculated by the sum of atomic number of both of the elements on the right side.

$\begin{array}{rll}\text{x}& =& 9+1\\ & =& 10\end{array}$

The element having atomic number $10$ is neon $\left(\text{Ne}\right)$. Therefore, the nuclide $\text{X}$ is represented as $\text{N}\text{e}$.

The resulting nuclear equation is shown below.

$\text{N}\text{e}{\to }_9^{19}\text{F}+e$

Conclusion

The nuclide $\text{X}$ which on positron decay gives $\text{F}$ is $\text{N}\text{e}$.

Interpretation Introduction

(d)

Interpretation:

The nuclide $\text{X}$ that decays by electron capture to give $\text{C}\text{l}$ is to be stated.

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Answer to Problem 12E

The nuclide $\text{X}$ that decays on electron capture and gives $\text{C}\text{l}$ is $\text{A}\text{r}$.

Explanation of Solution

It is given that $\text{X}$ on electron capture gives $\text{C}\text{l}$. Electron capture means that when an electron is used to make any unstable atom stable. Therefore, in the case of electron capture the atomic number of an element decreases by $1$, and there is no change in mass number.

The nuclear equation is shown below

$\text{X}+e{\to }_{17}^{37}\text{Cl}$

The mass number of $\text{X}$ can be calculated by the sum of mass number of both of the elements on the left side.

$\begin{array}{rll}\text{m}+\text{0}& =& 37\\ \text{m}& =& 37\end{array}$

The atomic number of $\text{X}$ can be calculated by the sum of atomic number of both of the elements on the left side.

$\begin{array}{rll}\text{x+}\left(-\text{1}\right)& =& 17\\ \text{x}-1& =& 17\end{array}$

Rearrange the above equation.

$\begin{array}{rll}\text{x}& =& 17+1\\ & =& 18\end{array}$

The element having atomic number $18$ is argon $\left(\text{Ar}\right)$. Therefore, the nuclide $\text{X}$ is represented as $\text{A}\text{r}$.

The resulting nuclear equation is shown below.

$\text{A}\text{r}+e{\to }_{17}^{37}\text{Cl}$

Conclusion

The nuclide $\text{X}$ which on electron capture gives $\text{C}\text{l}$ is $\text{A}\text{r}$.