Chapter 10, Problem 10.10E

Interpretation Introduction

(a)

Interpretation:

The given equation using appropriate notations and formulas are to be completed.

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 position or an electron is emitted is known as beta decay.

Answer to Problem 10.10E

The given equation using appropriate notations and formulas is,

${}_{82}^{204}\text{Pb}{\to }_{80}^{200}\text{Hg}+{}_2^4\text{α}$

Explanation of Solution

The given equation is,

${}_{82}^{204}\text{Pb}\to ?+{}_2^4\text{α}$

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. Therefore, the net mass is,

$\begin{array}{rll}\text{Net}\text{mass}& =& 204-4\\ & =& 200\end{array}$

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,

$\begin{array}{rll}\text{Net}\text{charge}& =& 82-2\\ & =& 80\end{array}$

The species that has $80$ charge that is atomic number and mass equal to $200$ is mercury, that is, ${}_{80}^{200}\text{Hg}$. Therefore, the given equation using appropriate notations and formulas is,

${}_{82}^{204}\text{Pb}{\to }_{80}^{200}\text{Hg}+{}_2^4\text{α}$

Conclusion

The given equation using appropriate notations and formulas is,

${}_{82}^{204}\text{Pb}{\to }_{80}^{200}\text{Hg}+{}_2^4\text{α}$

Interpretation Introduction

(b)

Interpretation:

The given equation using appropriate notations and formulas are to be completed.

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 10.10E

The given equation using appropriate notations and formulas is,

${}_{35}^{84}\text{Br}{\to }_{36}^{84}\text{Kr}\text{+}{}_{-1}^0\text{β}$

Explanation of Solution

The given equation is,

${}_{35}^{84}\text{Br}\to \text{?}\text{+}{}_{-1}^0\text{β}$

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. Therefore, the net mass is,

$\begin{array}{rll}\text{Net}\text{mass}& =& 84-0\\ & =& 84\end{array}$

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,

$\begin{array}{rll}\text{Net}\text{charge}& =& 35-\left(-1\right)\\ & =& 36\end{array}$

The species that has $36$ charge that is, an atomic number and mass equal to $84$ is krypton, that is, ${}_{36}^{84}\text{Kr}$. Therefore, the given equation using appropriate notations and formulas is,

${}_{35}^{84}\text{Br}{\to }_{36}^{84}\text{Kr}\text{+}{}_{-1}^0\text{β}$

Conclusion

The given equation using appropriate notations and formulas is,

${}_{35}^{84}\text{Br}{\to }_{36}^{84}\text{Kr}\text{+}{}_{-1}^0\text{β}$

Interpretation Introduction

(c)

Interpretation:

The given equation using appropriate notations and formulas are to be completed.

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 10.10E

The given equation using appropriate notations and formulas is,

${}_{20}^{41}\text{Ca}\text{+}{}_{-1}^0\text{e}{\to }_{19}^{41}\text{K}$

Explanation of Solution

The given equation is,

$\text{?}\text{+}{}_{-1}^0\text{e}{\to }_{19}^{41}\text{K}$

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. Therefore, the net mass is,

$\begin{array}{rll}\text{Net}\text{mass}& =& 41-0\\ & =& 41\end{array}$

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,

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

The species that has $20$ charge that is an atomic number and mass equal to $41$ is calcium, that is, ${}_{20}^{41}\text{Ca}$. Therefore, the given equation using appropriate notations and formulas is,

${}_{20}^{41}\text{Ca}\text{+}{}_{-1}^0\text{e}{\to }_{19}^{41}\text{K}$

Conclusion

The given equation using appropriate notations and formulas is,

${}_{20}^{41}\text{Ca}\text{+}{}_{-1}^0\text{e}{\to }_{19}^{41}\text{K}$

Interpretation Introduction

(d)

Interpretation:

The given equation using appropriate notations and formulas are to be completed.

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 10.10E

The given equation using appropriate notations and formulas is,

${}_{62}^{149}\text{Sm}{\to }_{60}^{145}\text{Nd}\text{+}{}_2^4\text{α}$

Explanation of Solution

The given equation is,

${}_{62}^{149}\text{Sm}{\to }_{60}^{145}\text{Nd}\text{+}?$

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. Therefore, the net mass is,

$\begin{array}{rll}\text{Net}\text{mass}& =& 149-145\\ & =& 4\end{array}$

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,

$\begin{array}{rll}\text{Net}\text{charge}& =& 62-60\\ & =& 2\end{array}$

The species that has $+2$ charge that is an 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,

${}_{62}^{149}\text{Sm}{\to }_{60}^{145}\text{Nd}\text{+}{}_2^4\text{α}$

Conclusion

The given equation using appropriate notations and formulas is,

${}_{62}^{149}\text{Sm}{\to }_{60}^{145}\text{Nd}\text{+}{}_2^4\text{α}$

Interpretation Introduction

(e)

Interpretation:

The given equation using appropriate notations and formulas are to be completed.

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 10.10E

The given equation using appropriate notations and formulas is,

${}_{14}^{34}\text{Si}{\to }_{15}^{34}\text{P}\text{+}{}_{-1}^0\text{β}$

Explanation of Solution

The given equation is,

$\text{?}{\to }_{15}^{34}\text{P}\text{+}{}_{-1}^0\text{β}$

The net mass of the species on the left-hand side is obtained by adding the mass number of the species on the right-hand side. Therefore, the net mass is,

$\begin{array}{rll}\text{Net}\text{mass}& =& 34+0\\ & =& 34\end{array}$

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

$\begin{array}{rll}\text{Net}\text{charge}& =& 15+\left(-1\right)\\ & =& 14\end{array}$

The species that has $14$ charge that is an atomic number, and a mass equal to $34$ is silicon, that is, ${}_{14}^{34}\text{Si}$. Therefore, the given equation using appropriate notations and formulas is,

${}_{14}^{34}\text{Si}{\to }_{15}^{34}\text{P}\text{+}{}_{-1}^0\text{β}$

Conclusion

The given equation using appropriate notations and formulas is,

${}_{14}^{34}\text{Si}{\to }_{15}^{34}\text{P}\text{+}{}_{-1}^0\text{β}$

Interpretation Introduction

(f)

Interpretation:

The given equation using appropriate notations and formulas are to be completed.

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 10.10E

The given equation using appropriate notations and formulas is,

${}_8^{15}\text{O}{\to }_{-1}^0\text{β}+{}_7^{15}N$

Explanation of Solution

The given equation is,

${}_8^{15}\text{O}{\to }_{-1}^0\text{β}+?$

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. Therefore, the net mass is,

$\begin{array}{rll}\text{Net}\text{mass}& =& 15-0\\ & =& 15\end{array}$

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,

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

The species that has $+7$ charge that is an atomic number, and a mass equal to $15$ is fluorine, that is, ${}_7^{15}N$. Therefore, the given equation using appropriate notations and formulas is,

${}_8^{15}\text{O}{\to }_{-1}^0\text{β}+{}_7^{15}N$

Conclusion

The given equation using appropriate notations and formulas is,

${}_8^{15}\text{O}{\to }_{-1}^0\text{β}+{}_7^{15}N$