Chapter 10, Problem 10.49E

Interpretation Introduction

(a)

Interpretation:

The complete equation is to be stated.

Concept Introduction:

Radioactive decay is the process that involves the emission of radiation by an unstable atomic nucleus. The atomic nucleus loses it energy. The process is spontaneous. It is also known as nuclear radiation.

Answer to Problem 10.49E

The complete equation is ${}_{92}^{235}\text{U}+{}_0^{1}n\to {}_{62}^{160}\text{Sm}+{}_{30}^{71}\text{Zn}+4_0^{1}n$.

Explanation of Solution

The incomplete reaction is given as shown below.

${}_{92}^{235}\text{U}+{}_0^{1}n\to {}_{62}^{160}\text{Sm}+?+4_0^{1}n$

The element is given as ${}_{92}^{235}\text{U}$. The $n$ alpha particle is given by uranium $\text{-}235$ nucleus. The mass and atomic number of $n$ alpha particle is $1$ and $0$. The mass of missing element is calculated as shown below.

$\begin{array}{rll}235+1& =& 160+\text{Mass}\text{of}\text{missing}\text{element}+4\left(1\right)\\ \text{Mass}\text{of}\text{missing}\text{element}& =& 235+1-\left[160+4\left(1\right)\right]\\ & =& 236-164\\ & =& 71\end{array}$

The atomic number of missing element is calculated as shown below.

$\begin{array}{rll}92+0& =& \left[62+\text{Atomic}\text{number}\text{of}\text{missing}\text{element}+4\left(0\right)\right]\\ \text{Atomic}\text{number}\text{of}\text{missing}\text{element}& =& 92+0-\left[62+4\left(0\right)\right]\\ & =& 92-62\\ & =& 30\end{array}$

The mass and atomic number of missing element is $71$ and $30$ respectively. The atomic number of zinc is $30$. Therefore, the missing element is zinc.

Therefore, the complete equation is ${}_{92}^{235}\text{U}+{}_0^{1}n\to {}_{62}^{160}\text{Sm}+{}_{30}^{71}\text{Zn}+4_0^{1}n$.

Conclusion

The complete equation is ${}_{92}^{235}\text{U}+{}_0^{1}n\to {}_{62}^{160}\text{Sm}+{}_{30}^{71}\text{Zn}+4_0^{1}n$.

Interpretation Introduction

(b)

Interpretation:

The complete equation is to be stated.

Concept Introduction:

Radioactive decay is the process that involves the emission of radiation by an unstable atomic nucleus. The atomic nucleus loses it energy. The process is spontaneous. It is also known as nuclear radiation.

Answer to Problem 10.49E

The complete equation is ${}_{92}^{235}\text{U}+{}_0^{1}n\to {}_{35}^{87}\text{Br}+{}_{57}^{146}\text{La}+3_0^{1}n$.

Explanation of Solution

The incomplete equation is given as shown below.

${}_{92}^{235}\text{U}+{}_0^{1}n\to {}_{35}^{87}\text{Br}+?+3_0^{1}n$

The element is given as ${}_{92}^{235}\text{U}$. The $n$ alpha particle is given by uranium $\text{-}235$ nucleus. The mass and atomic number of $n$ alpha particle is $1$ and $0$. The mass of missing element is calculated as shown below.

$\begin{array}{rll}235+1& =& 87+\text{Mass}\text{of}\text{missing}\text{element}+3\left(1\right)\\ \text{Mass}\text{of}\text{missing}\text{element}& =& 235+1-\left[87+3\left(1\right)\right]\\ & =& 236-90\\ & =& 146\end{array}$

The atomic number of missing element is calculated as shown below.

$\begin{array}{rll}92+0& =& \left[35+\text{Atomic}\text{number}\text{of}\text{missing}\text{element}+3\left(0\right)\right]\\ \text{Atomic}\text{number}\text{of}\text{missing}\text{element}& =& 92+0-\left[35+3\left(0\right)\right]\\ & =& 92-35\\ & =& 57\end{array}$

The mass and atomic number of missing element is $146$ and $57$ respectively. The atomic number of lanthanum is $57$. Therefore, the missing element is lanthanum.

Therefore, the complete equation is ${}_{92}^{235}\text{U}+{}_0^{1}n\to {}_{35}^{87}\text{Br}+{}_{57}^{146}\text{La}+3_0^{1}n$

Conclusion

The complete equation is ${}_{92}^{235}\text{U}+{}_0^{1}n\to {}_{35}^{87}\text{Br}+{}_{57}^{146}\text{La}+3_0^{1}n$.