Chapter 10, Problem 10.11E

Interpretation Introduction

(a)

Interpretation:

The balanced equation for the decay reaction of the given isotope is to be stated.

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 10.11E

The balanced equation for the decay reaction of the given isotope is,

${}_{50}^{121}\text{Sn}{\to }_{-1}^0\text{β}+{}_{51}^{121}\text{Sb}$

Explanation of Solution

The given parent nucleus is ${}_{50}^{121}\text{Sn}$. The given decay process is beta emission, that is ${}_{-1}^0\text{β}$ particle is emitted.

The net mass is obtained by subtracting the mass of the emitted particle from the mass of the parent nucleus. Therefore, the net mass is,

$\begin{array}{rll}\text{Net}\text{mass}& =& 121-0\\ & =& 121\end{array}$

The net charge is obtained by subtracting the charge on the emitted particle from the charge on the parent nucleus. Therefore, the net charge is,

$\begin{array}{rll}\text{Net}\text{charge}& =& 50-\left(-1\right)\\ & =& 51\end{array}$

The nucleus that has $+51$ charge and mass equal to $121$ is ${}_{51}^{121}\text{Sb}$. Therefore, the daughter nucleus is ${}_{51}^{121}\text{Sb}$. The balanced equation for the decay reaction of the given isotope is,

${}_{50}^{121}\text{Sn}{\to }_{-1}^0\text{β}+{}_{51}^{121}\text{Sb}$

Conclusion

The balanced equation for the decay reaction of the given isotope is,

${}_{50}^{121}\text{Sn}{\to }_{-1}^0\text{β}+{}_{51}^{121}\text{Sb}$

Interpretation Introduction

(b)

Interpretation:

The balanced equation for the decay reaction of the given isotope is to be stated.

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 10.11E

The balanced equation for the decay reaction of the given isotope is,

${}_{26}^{55}\text{Fe}+{}_{-1}^0\text{e}{\to }_{25}^{55}\text{Mn}$

Explanation of Solution

The given parent nucleus is ${}_{26}^{55}\text{Fe}$. The given decay process is electron capture, that is ${}_{-1}^0\text{e}$ particle is captured.

The net mass is obtained by adding the mass of the captured particle and the mass of the parent nucleus. Therefore, the net mass is,

$\begin{array}{rll}\text{Net}\text{mass}& =& 55+0\\ & =& 55\end{array}$

The net charge is obtained by adding the charge on the emitted particle and the charge on the parent nucleus. Therefore, the net charge is,

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

The nucleus that has $+25$ charge and mass equal to $55$ is ${}_{25}^{55}\text{Mn}$. Therefore, the daughter nucleus is ${}_{25}^{55}\text{Mn}$. The balanced equation for the decay reaction of the given isotope is,

${}_{26}^{55}\text{Fe}+{}_{-1}^0\text{e}{\to }_{25}^{55}\text{Mn}$

Conclusion

The balanced equation for the decay reaction of the given isotope is,

${}_{26}^{55}\text{Fe}+{}_{-1}^0\text{e}{\to }_{25}^{55}\text{Mn}$

Interpretation Introduction

(c)

Interpretation:

The balanced equation for the decay reaction of the given isotope is to be stated.

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 10.11E

The balanced equation for the decay reaction of the given isotope is,

${}_{11}^{22}\text{Na}{\to }_{+1}^0\text{β}+{}_{10}^{22}\text{Ne}$

Explanation of Solution

The given parent nucleus is ${}_{11}^{22}\text{Na}$. The given daughter nucleus is neon-$22$, that is ${}_{10}^{22}\text{Ne}$.

The net mass is obtained by subtracting the mass of the daughter nucleus from the mass of the parent nucleus. Therefore, the net mass is,

$\begin{array}{rll}\text{Net}\text{mass}& =& 22-22\\ & =& 0\end{array}$

The net charge is obtained by subtracting the charge on the daughter nucleus from the charge on the parent nucleus. Therefore, the net charge is,

$\begin{array}{rll}\text{Net}\text{charge}& =& 11-10\\ & =& 1\end{array}$

The nucleus that has $+1$ charge and mass equal to $0$ is a positron, that is, ${}_{+1}^0\text{β}$. Therefore, the emitted particle is ${}_{+1}^0\text{β}$. The balanced equation for the decay reaction of the given isotope is,

${}_{11}^{22}\text{Na}{\to }_{+1}^0\text{β}+{}_{10}^{22}\text{Ne}$

Conclusion

The balanced equation for the decay reaction of the given isotope is,

${}_{11}^{22}\text{Na}{\to }_{+1}^0\text{β}+{}_{10}^{22}\text{Ne}$

Interpretation Introduction

(d)

Interpretation:

The balanced equation for the decay reaction of the given isotope is to be stated.

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 10.11E

The balanced equation for the decay reaction of the given isotope is,

${}_{78}^{190}\text{Pt}{\to }_2^4\text{α}+{}_{76}^{186}\text{Os}$

Explanation of Solution

The given parent nucleus is ${}_{78}^{190}\text{Pt}$. The given decay process is alpha emission, that is ${}_2^4\text{α}$ particle is emitted.

The net mass is obtained by subtracting the mass of the emitted particle from the mass of the parent nucleus. Therefore, the net mass is,

$\begin{array}{rll}\text{Net}\text{mass}& =& 190-4\\ & =& 186\end{array}$

The net charge is obtained by subtracting the charge on the emitted particle from the charge on the parent nucleus. Therefore, the net charge is,

$\begin{array}{rll}\text{Net}\text{charge}& =& 78-2\\ & =& 76\end{array}$

The nucleus that has $+76$ charge and mass equal to $186$ is ${}_{76}^{186}\text{Os}$. Therefore, the daughter nucleus is ${}_{76}^{186}\text{Os}$. The balanced equation for the decay reaction of the given isotope is,

${}_{78}^{190}\text{Pt}{\to }_2^4\text{α}+{}_{76}^{186}\text{Os}$

Conclusion

The balanced equation for the decay reaction of the given isotope is,

${}_{78}^{190}\text{Pt}{\to }_2^4\text{α}+{}_{76}^{186}\text{Os}$

Interpretation Introduction

(e)

Interpretation:

The balanced equation for the decay reaction of the given isotope is to be stated.

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 10.11E

The balanced equation for the decay reaction of the given isotope is,

${}_{28}^{67}\text{Ni}{\to }_{-1}^0\text{β}+{}_{29}^{67}\text{Cu}$

Explanation of Solution

The given parent nucleus is ${}_{28}^{67}\text{Ni}$. The given decay process is beta emission, that is ${}_{-1}^0\text{β}$ particle is emitted.

The net mass is obtained by subtracting the mass of the emitted particle from the mass of the parent nucleus. Therefore, the net mass is,

$\begin{array}{rll}\text{Net}\text{mass}& =& 67-0\\ & =& 67\end{array}$

The net charge is obtained by subtracting the charge on the emitted particle from the charge on the parent nucleus. Therefore, the net charge is,

$\begin{array}{rll}\text{Net}\text{charge}& =& 28-\left(-1\right)\\ & =& 29\end{array}$

The nucleus that has $+29$ charge and mass equal to $67$ is ${}_{29}^{67}\text{Cu}$. Therefore, the daughter nucleus is ${}_{29}^{67}\text{Cu}$. The balanced equation for the decay reaction of the given isotope is,

${}_{28}^{67}\text{Ni}{\to }_{-1}^0\text{β}+{}_{29}^{67}\text{Cu}$

Conclusion

The balanced equation for the decay reaction of the given isotope is,

${}_{28}^{67}\text{Ni}{\to }_{-1}^0\text{β}+{}_{29}^{67}\text{Cu}$

Interpretation Introduction

(f)

Interpretation:

The balanced equation for the decay reaction of the given isotope is to be stated.

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 10.11E

The balanced equation for the decay reaction of the given isotope is,

${}_{31}^{67}\text{Ga}{\to }_{+1}^0\text{β}+{}_{30}^{67}\text{Zn}$

Explanation of Solution

The given parent nucleus is ${}_{31}^{67}\text{Ga}$. The given daughter nucleus is zinc-$22$, that is ${}_{30}^{67}\text{Zn}$.

The net mass is obtained by subtracting the mass of the daughter nucleus from the mass of the parent nucleus. Therefore, the net mass is,

$\begin{array}{rll}\text{Net}\text{mass}& =& 67-67\\ & =& 0\end{array}$

The net charge is obtained by subtracting the charge on the daughter nucleus from the charge on the parent nucleus. Therefore, the net charge is,

$\begin{array}{rll}\text{Net}\text{charge}& =& 31-30\\ & =& 1\end{array}$

The nucleus that has $+1$ charge and mass equal to $0$ is a positron, that is, ${}_{+1}^0\text{β}$. Therefore, the emitted particle is ${}_{+1}^0\text{β}$. The balanced equation for the decay reaction of the given isotope is,

${}_{31}^{67}\text{Ga}{\to }_{+1}^0\text{β}+{}_{30}^{67}\text{Zn}$

Conclusion

The balanced equation for the decay reaction of the given isotope is,

${}_{31}^{67}\text{Ga}{\to }_{+1}^0\text{β}+{}_{30}^{67}\text{Zn}$