Chapter 20, Problem 50E

Interpretation Introduction

(a)

Interpretation:

The nuclear bombardment equation $?+\text{H}\text{e}\to \text{C}+\text{n}$ by supplying the nuclear symbol for the missing species is to be completed.

Concept introduction:

The nuclear reaction involves the spontaneous emission of particles or radiation due to the breaking or combination of an atomic nucleus. An alpha particle, beta particle and gamma rays are emitted during the radioactive decay. An alpha particle is a helium nucleus. In beta particle emission, an atom emits an electron $\left(\text{e}\right)$.

Answer to Problem 50E

The complete equation with the nuclear symbol for the missing species $\text{B}\text{e}$ is shown below.

$\text{B}\text{e}+\text{H}\text{e}\to \text{C}+\text{n}$.

Explanation of Solution

To complete a nuclear bombardment equation, it is required to count the total number of atomic number and atomic mass in both sides of the equation. An incomplete nuclear bombardment equation is shown below.

$?+\text{H}\text{e}\to \text{C}+\text{n}$

In the above equation, the identity of the new nuclei is to be determined.

$\text{Total}\text{atomic}\text{number}\text{of}\text{the}\text{reactant}\text{species}=2$

$\begin{array}{rll}\text{Total}\text{atomic}\text{number}\text{of}\text{the}\text{product}\text{species}& =& 6+0\\ & =& 6\end{array}$

Therefore, the atomic number of missing specie is calculated as shown below.

$\begin{array}{rll}\text{Atomic}\text{number}\text{of}\text{missing}\text{nuclei}& =& 6-2\\ & =& 4\end{array}$

Beryllium $\left(\text{Be}\right)$ has the atomic number of $4$.

$\text{Total}\text{atomic}\text{mass}\text{of}\text{the}\text{reactant}\text{species}=4$

$\begin{array}{rll}\text{Total}\text{atomic}\text{mass}\text{of}\text{the}\text{product}\text{species}& =& 12+1\\ & =& 13\end{array}$

Therefore, the atomic mass of missing specie is calculated as shown below.

$\begin{array}{rll}\text{Atomic}\text{mass}\text{of}\text{missing}\text{nuclei}& =& 13-4\\ & =& 9\end{array}$

Therefore, the missing specie is $\text{B}\text{e}$, and the complete equation is shown below.

$\text{B}\text{e}+\text{H}\text{e}\to \text{C}+\text{n}$

Conclusion

The complete equation with the nuclear symbol for the missing species $\text{B}\text{e}$ is shown below.

$\text{B}\text{e}+\text{H}\text{e}\to \text{C}+\text{n}$.

Interpretation Introduction

(b)

Interpretation:

The nuclear bombardment equation $\text{C}\text{d}+\text{H}\to ?+\text{H}$ by supplying the nuclear symbol for the missing species is to be completed.

Concept introduction:

The nuclear reaction involves the spontaneous emission of particles or radiation due to the breaking or combination of an atomic nucleus. An alpha particle, beta particle and gamma rays are emitted during the radioactive decay. An alpha particle is a helium nucleus. In beta particle emission, an atom emits an electron $\left(\text{e}\right)$.

Answer to Problem 50E

The complete equation with the nuclear symbol for the missing species $\text{C}\text{d}$ is shown below.

$\text{C}\text{d}+\text{H}\to \text{C}\text{d}+\text{H}$

Explanation of Solution

To complete a nuclear bombardment equation, it is required to count the total number of atomic number and atomic mass in both sides of the equation. An incomplete nuclear bombardment equation is shown below.

$\text{C}\text{d}+\text{H}\to ?+\text{H}$

In the above equation, the identity of the new nuclei is to be determined.

$\begin{array}{rll}\text{Total}\text{atomic}\text{number}\text{of}\text{the}\text{reactant}\text{species}& =& 48+1\\ & =& 49\end{array}$

$\text{Total}\text{atomic}\text{number}\text{of}\text{the}\text{product}\text{species}=1$

Therefore, the atomic number of missing specie is calculated as shown below.

$\begin{array}{rll}\text{Atomic}\text{number}\text{of}\text{missing}\text{nuclei}& =& 49-1\\ & =& 48\end{array}$

Cadmium $\left(\text{Cd}\right)$ has the atomic number of $48$.

$\begin{array}{rll}\text{Total}\text{atomic}\text{mass}\text{of}\text{the}\text{reactant}\text{species}& =& 114+2\\ & =& 116\end{array}$

$\text{Total}\text{atomic}\text{mass}\text{of}\text{the}\text{product}\text{species}=1$

Therefore, the atomic mass of missing specie is calculated as shown below.

$\begin{array}{rll}\text{Atomic}\text{mass}\text{of}\text{missing}\text{nuclei}& =& 116-1\\ & =& 115\end{array}$

Therefore, the missing specie is $\text{C}\text{d}$, and the complete equation is shown below.

$\text{C}\text{d}+\text{H}\to \text{C}\text{d}+\text{H}$

Conclusion

The complete equation with the nuclear symbol for the missing species $\text{C}\text{d}$ is shown below.

$\text{C}\text{d}+\text{H}\to \text{C}\text{d}+\text{H}$.

Interpretation Introduction

(c)

Interpretation:

The nuclear bombardment equation $\text{U}+\text{n}\to ?+\text{S}\text{r}+2\text{n}$ by supplying the nuclear symbol for the missing species is to be completed.

Concept introduction:

The nuclear reaction involves the spontaneous emission of particles or radiation due to the breaking or combination of an atomic nucleus. An alpha particle, beta particle and gamma rays are emitted during the radioactive decay. An alpha particle is a helium nucleus. In beta particle emission, an atom emits an electron $\left(\text{e}\right)$.

Answer to Problem 50E

The complete equation with the nuclear symbol for the missing species $\text{X}\text{e}$ is shown below.

$\text{U}+\text{n}\to \text{X}\text{e}+\text{S}\text{r}+2\text{n}$

Explanation of Solution

To complete a nuclear bombardment equation, it is required to count the total number of atomic number and atomic mass in both sides of the equation. An incomplete nuclear bombardment equation is shown below.

$\text{U}+\text{n}\to ?+\text{S}\text{r}+2\text{n}$

In the above equation, the identity of new nuclei is to be found.

$\begin{array}{rll}\text{Total}\text{atomic}\text{number}\text{of}\text{the}\text{reactant}\text{species}& =& 92+0\\ & =& 92\end{array}$

$\begin{array}{rll}\text{Total}\text{atomic}\text{number}\text{of}\text{the}\text{product}\text{species}& =& 38+0\\ & =& 38\end{array}$

Therefore, the atomic number of missing specie is calculated as shown below.

$\begin{array}{rll}\text{Atomic}\text{number}\text{of}\text{missing}\text{nuclei}& =& 92-38\\ & =& 54\end{array}$

Xenon $\left(\text{Xe}\right)$ has the atomic number of $54$.

$\begin{array}{c}\text{Total}\text{atomic}\text{mass}\text{of}\text{the}\text{reactant}\text{species}=235+1\\ =236\end{array}$

$\begin{array}{rll}\text{Total}\text{atomic}\text{mass}\text{of}\text{the}\text{product}\text{species}& =& 92+2\times 1\\ & =& 94\end{array}$

Therefore, the atomic mass of missing specie is calculated as shown below.

$\begin{array}{rll}\text{Atomic}\text{mass}\text{of}\text{missing}\text{nuclei}& =& 236-94\\ & =& 142\end{array}$

Therefore, the missing specie is $\text{X}\text{e}$, and the complete equation is shown below.

$\text{U}+\text{n}\to \text{X}\text{e}+\text{S}\text{r}+2\text{n}$

Conclusion

The complete equation with the nuclear symbol for the missing species $\text{X}\text{e}$ is shown below.

$\text{U}+\text{n}\to \text{X}\text{e}+\text{S}\text{r}+2\text{n}$