Chapter 18, Problem 11E

Interpretation Introduction

(a)

Interpretation:

The nuclide $\text{X}$ that decays by alpha emission to give $\text{R}\text{n}$ 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 11E

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

Explanation of Solution

It is given that $\text{X}$ undergoes alpha decay to give $\text{R}\text{n}$. 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.

$\text{X}{\to }_{86}^{\text{217}}\text{Rn}{+}_{\text{2}}^{\text{4}}\text{He}$

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

$\begin{array}{rll}\text{m}& =& 217+4\\ & =& 221\end{array}$

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

$\begin{array}{rll}\text{x}& =& 86+2\\ & =& 88\end{array}$

The element having atomic number $88$ is radium. Therefore, the nuclide $\text{X}$ is represented as $\text{R}\text{a}$.

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 $\text{R}\text{n}$ is radium $\left(\text{R}\text{a}\right)$.

Interpretation Introduction

(b)

Interpretation:

The nuclide $\text{X}$ that decays by beta emission to give $\text{C}\text{a}$ 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 11E

The nuclide $\text{X}$ that decays by beta emission and gives $\text{C}\text{a}$ is $\text{K}$.

Explanation of Solution

It is given that $\text{X}$ on beta emission gives $\text{C}\text{a}$. 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.

$\text{X}{\to }_{20}^{43}\text{Ca}+e$

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

$\begin{array}{rll}\text{m}& =& 43+0\\ & =& 43\end{array}$

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

$\begin{array}{rll}\text{x}& =& 20+\left(-1\right)\\ & =& 20-1\\ & =& 19\end{array}$

The element having atomic number $19$ is potassium $\left(\text{K}\right)$. Therefore, the nuclide $\text{X}$ is represented as $\text{K}$.

The resulting nuclear equation is shown below.

$\text{K}{\to }_{20}^{43}\text{Ca}+e$

Conclusion

The nuclide $\text{X}$ which on beta decay gives $\text{C}\text{a}$ is $\text{K}$.

Interpretation Introduction

(c)

Interpretation:

The nuclide $\text{X}$ that decays by positron emission to give $\text{B}\text{r}$ 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 11E

The nuclide $\text{X}$ that decays by positron emission and gives $\text{B}\text{r}$ is $\text{K}\text{r}$.

Explanation of Solution

It is given that $\text{X}$ on positron emission gives $\text{B}\text{r}$. 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

$\text{X}{\to }_{35}^{73}\text{Br}+e$

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

$\begin{array}{rll}\text{m}& =& 73+0\\ & =& 73\end{array}$

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

$\begin{array}{rll}\text{x}& =& 35+1\\ & =& 36\end{array}$

The element having atomic number $36$ is krypton $\left(\text{Kr}\right)$. Therefore, the nuclide $\text{X}$ is represented as $\text{K}\text{r}$.

The resulting nuclear equation is shown below.

$\text{K}\text{r}{\to }_{35}^{73}\text{Br}+e$

Conclusion

The nuclide $\text{X}$ on positron decay gives $\text{B}\text{r}$ is $\text{K}\text{r}$.

Interpretation Introduction

(d)

Interpretation:

The nuclide $\text{X}$ that decays by electron capture to give $\text{C}\text{s}$ 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 11E

The nuclide $\text{X}$ that decays by electron capture and gives $\text{C}\text{s}$ is $\text{B}\text{a}$.

Explanation of Solution

It is given that $\text{X}$ on electron capture gives $\text{C}\text{s}$. 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

$\text{X}+e{\to }_{55}^{133}\text{Cs}$

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

$\begin{array}{rll}\text{m}+\text{0}& =& 133\\ \text{m}& =& 133\end{array}$

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

$\begin{array}{rll}\text{x+}\left(-\text{1}\right)& =& 55\\ \text{x}-1& =& 55\end{array}$

Rearrange the above equation as shown below.

$\begin{array}{rll}\text{x}& =& 55+1\\ & =& 56\end{array}$

The element having atomic number $56$ is barium $\left(\text{Ba}\right)$. Therefore, the nuclide $\text{X}$ is represented as $\text{B}\text{a}$.

The resulting nuclear equation is shown below.

$\text{B}\text{a}+e{\to }_{55}^{133}\text{Cs}$

Conclusion

The nuclide $\text{X}$ which on electron capture gives $\text{C}\text{s}$ is $\text{B}\text{a}$.