Chapter 10, Problem 10.39E

Interpretation Introduction

(a)

Interpretation:

The given reactions are to be completed by identifying the missing nuclear symbols.

Concept introduction:

The type of radioactive decay in which an alpha particle is emitted by the nucleus of an atom such that an atom of another element is produced after decay is known as alpha decay. An alpha particle is a helium nucleus. The radioactive decay in which a positron or an electron is emitted is known as beta decay.

Answer to Problem 10.39E

The complete reaction is,

${}_4^9\text{Be}{+}_2^4\text{α}{\to }_6^{12}\text{C}+{}_0^{1}n$

Explanation of Solution

The given reaction is,

${}_4^9\text{Be}+\text{?}{\to }_6^{12}\text{C}+{}_0^{1}n$

The net mass is obtained by subtracting the mass number of the species on the left-hand side from the mass number of the species on the right-hand side. The right-hand side consists of carbon atom and a neutron and the left hand side consists of beryllium atom. Therefore, the net mass is,

$\begin{array}{rll}\text{Net}\text{mass}& =& \left(12+1\right)-9\\ & =& 4\end{array}$

The net charge is obtained by subtracting the charge of the species on the left-hand side from the charge of the species on the right-hand side. Therefore, the net charge is,

$\begin{array}{rll}\text{Net}\text{charge}& =& \left(6+0\right)-4\\ & =& 2\end{array}$

The species that has $+2$ charge, that is atomic number, and a mass equal to $4$ is an alpha particle, that is, ${}_2^4\text{α}$. Therefore, the given equation using appropriate notations and formulas is,

${}_4^9\text{Be}{+}_2^4\text{α}{\to }_6^{12}\text{C}+{}_0^{1}n$

Conclusion

The complete reaction is,

${}_4^9\text{Be}{+}_2^4\text{α}{\to }_6^{12}\text{C}+{}_0^{1}n$

Interpretation Introduction

(b)

Interpretation:

The given reactions are to be completed by identifying the missing nuclear symbols.

Concept introduction:

The type of radioactive decay in which an alpha particle is emitted by the nucleus of an atom such that an atom of another element is produced after decay is known as alpha decay. An alpha particle is a helium nucleus. The radioactive decay in which a positron or an electron is emitted is known as beta decay.

Answer to Problem 10.39E

The complete reaction is,

${}_{13}^{27}\text{Al}{+}_2^4\text{α}{\to }_{15}^{30}\text{P}+{}_0^{1}n$

Explanation of Solution

The given reaction is,

${}_{13}^{27}\text{Al}+\text{?}{\to }_{15}^{30}\text{P}+{}_0^{1}n$

The net mass is obtained by subtracting the mass number of the species on the left-hand side from the mass number of the species on the right-hand side. The right-hand side consists of phosphorus atom and a neutron and the left hand side consists of aluminum atom. Therefore, the net mass is,

$\begin{array}{rll}\text{Net}\text{mass}& =& \left(30+1\right)-27\\ & =& 4\end{array}$

The net charge is obtained by subtracting the charge of the species on the left-hand side from the charge of the species on the right-hand side. Therefore, the net charge is,

$\begin{array}{rll}\text{Net}\text{charge}& =& \left(15+0\right)-13\\ & =& 2\end{array}$

The species that has $+2$ charge, that is atomic number, and a mass equal to $4$ is an alpha particle, that is, ${}_2^4\text{α}$. Therefore, the given equation using appropriate notations and formulas is,

${}_{13}^{27}\text{Al}{+}_2^4\text{α}{\to }_{15}^{30}\text{P}+{}_0^{1}n$

Conclusion

The complete reaction is,

${}_{13}^{27}\text{Al}{+}_2^4\text{α}{\to }_{15}^{30}\text{P}+{}_0^{1}n$

Interpretation Introduction

(c)

Interpretation:

The given reactions are to be completed by identifying the missing nuclear symbols.

Concept introduction:

The type of radioactive decay in which an alpha particle is emitted by the nucleus of an atom such that an atom of another element is produced after decay is known as alpha decay. An alpha particle is a helium nucleus. The radioactive decay in which a positron or an electron is emitted is known as beta decay.

Answer to Problem 10.39E

The complete reaction is,

${}_{13}^{27}\text{Al}{+}_1^2\text{H}{\to }_{12}^{25}\text{Mg}+{}_2^4\text{α}$

Explanation of Solution

The given reaction is,

${}_{13}^{27}\text{Al}{+}_1^2\text{H}\to ?+{}_2^4\text{α}$

The net mass is obtained by subtracting the mass number of the species on the right-hand side from the mass number of the species on the left-hand side. The left-hand side consists of aluminum atom and a hydrogen atom and the right-hand side consists of an alpha particle. Therefore, the net mass is,

$\begin{array}{rll}\text{Net}\text{mass}& =& \left(27+2\right)-4\\ & =& 25\end{array}$

The net charge is obtained by subtracting the charge of the species on the right-hand side from the charge of the species on the left-hand side. Therefore, the net charge is,

$\begin{array}{rll}\text{Net}\text{charge}& =& \left(13+1\right)-2\\ & =& 12\end{array}$

The species that has $+12$ charge, that is atomic number, and a mass equal to $25$ is an isotope of magnesium, that is, ${}_{12}^{25}\text{Mg}$. Therefore, the given equation using appropriate notations and formulas is,

${}_{13}^{27}\text{Al}{+}_1^2\text{H}{\to }_{12}^{25}\text{Mg}+{}_2^4\text{α}$

Conclusion

The complete reaction is,

${}_{13}^{27}\text{Al}{+}_1^2\text{H}{\to }_{12}^{25}\text{Mg}+{}_2^4\text{α}$

Interpretation Introduction

(d)

Interpretation:

The given reactions are to be completed by identifying the missing nuclear symbols.

Concept introduction:

The type of radioactive decay in which an alpha particle is emitted by the nucleus of an atom such that an atom of another element is produced after decay is known as alpha decay. An alpha particle is a helium nucleus. The radioactive decay in which a positron or an electron is emitted is known as beta decay.

Answer to Problem 10.39E

The complete reaction is,

${}_7^{14}\text{N}{+}_2^4\text{α}{\to }_8^{17}\text{O}+{}_1^1\text{H}$

Explanation of Solution

The given reaction is,

${}_7^{14}\text{N}{+}_2^4\text{α}\to ?+{}_1^1\text{H}$

The net mass is obtained by subtracting the mass number of the species on the right-hand side from the mass number of the species on the left-hand side. The left-hand side consists of nitrogen atom and an alpha particle and the right-hand side consists of an atom of hydrogen. Therefore, the net mass is,

$\begin{array}{rll}\text{Net}\text{mass}& =& \left(14+4\right)-1\\ & =& 17\end{array}$

The net charge is obtained by subtracting the charge of the species on the right-hand side from the charge of the species on the left-hand side. Therefore, the net charge is,

$\begin{array}{rll}\text{Net}\text{charge}& =& \left(7+2\right)-1\\ & =& 8\end{array}$

The species that has $+8$ charge, that is atomic number, and a mass equal to $17$ is an isotope of oxygen, that is, ${}_8^{17}\text{O}$. Therefore, the given equation using appropriate notations and formulas is,

${}_7^{14}\text{N}{+}_2^4\text{α}{\to }_8^{17}\text{O}+{}_1^1\text{H}$

Conclusion

The complete reaction is,

${}_7^{14}\text{N}{+}_2^4\text{α}{\to }_8^{17}\text{O}+{}_1^1\text{H}$