Chapter 18, Problem 55QRT

(a)

Interpretation Introduction

Interpretation:

Missing product in the given nuclear fission reaction has to be determined.

    $\text{U}\text{+}\text{n}\stackrel{}{\to }?+\text{S}\text{r}\text{+}3\text{n}$

Concept Introduction:

Nuclear reaction can be written in the form of nuclear equation.  Nuclear equation considers that atomic number and mass number of the elements that is involved in the reaction.  Balanced nuclear equation is the one that has mass number balance and atomic number balance equal on both sides of the equation.  Missing particle can be identified from the mass number balance and atomic number balance in the balanced nuclear equation.

Answer to Problem 55QRT

The missing product is $\text{X}\text{e}$.

Explanation of Solution

Given nuclear equation in the problem statement is,

    $\text{U}\text{+}\text{n}\stackrel{}{\to }?+\text{S}\text{r}\text{+}3\text{n}$

In a balanced nuclear equation, the mass number balance and atomic number balance has to be equal on both sides of the equation.  The unknown element that is formed is represented as $\text{X}$.  Therefore, the equation can be written as,

    $\text{U}\text{+}\text{n}\stackrel{}{\to }\text{X}+\text{S}\text{r}\text{+}3\text{n}$

Sum of mass number in the reactant side has to be equal to the sum of mass number in the product side.

    $\begin{array}{l}235+1=\text{A}\text{+}\text{93}\text{+}\text{3}\\ \text{236}\text{=}\text{A}\text{+}\text{96}\\ \text{A}\text{=}\text{236}-96\\ =140\end{array}$

Therefore, the mass number of the unknown product is found to be 140.

Sum of atomic number in the reactant side has to be equal to the sum of atomic number in the product side.

  $\begin{array}{l}92+0=\text{Z}\text{+}\text{38}\text{+}\text{0}\\ \text{92}\text{=}\text{Z}\text{+}\text{38}\\ \text{Z}\text{=}\text{92}-38\\ =54\end{array}$

Therefore, the atomic number of the unknown product is found to be 54.  The element with atomic number 54 is Xenon.  Therefore, the missing product is found to be $\text{X}\text{e}$.  The complete equation can be given as,

  $\text{U}\text{+}\text{n}\stackrel{}{\to }\text{X}\text{e}+\text{S}\text{r}\text{+}3\text{n}$

(b)

Interpretation Introduction

Interpretation:

Missing product in the given nuclear fission reaction has to be determined.

    $\text{U}\text{+}\text{n}\stackrel{}{\to }?+\text{S}\text{b}\text{+}3\text{n}$

Concept Introduction:

Refer part (a).

Answer to Problem 55QRT

The missing product is $\text{N}\text{b}$.

Explanation of Solution

Given nuclear equation in the problem statement is,

    $\text{U}\text{+}\text{n}\stackrel{}{\to }?+\text{S}\text{b}\text{+}3\text{n}$

In a balanced nuclear equation, the mass number balance and atomic number balance has to be equal on both sides of the equation.  The unknown element that is formed is represented as $\text{X}$.  Therefore, the equation can be written as,

    $\text{U}\text{+}\text{n}\stackrel{}{\to }\text{X}+\text{S}\text{b}\text{+}3\text{n}$

Sum of mass number in the reactant side has to be equal to the sum of mass number in the product side.

    $\begin{array}{l}235+1=\text{A}\text{+}\text{132}\text{+}\text{3}\\ \text{236}\text{=}\text{A}\text{+}\text{135}\\ \text{A}\text{=}\text{236}-135\\ =101\end{array}$

Therefore, the mass number of the unknown product is found to be 101.

Sum of atomic number in the reactant side has to be equal to the sum of atomic number in the product side.

  $\begin{array}{l}92+0=\text{Z}\text{+}\text{51}\text{+}\text{0}\\ \text{92}\text{=}\text{Z}\text{+}\text{51}\\ \text{Z}\text{=}\text{92}-51\\ =41\end{array}$

Therefore, the atomic number of the unknown product is found to be 41.  The element with atomic number 41 is Niobium.  Therefore, the missing product is found to be $\text{N}\text{b}$.  The complete equation can be given as,

  $\text{U}\text{+}\text{n}\stackrel{}{\to }\text{N}\text{b}+\text{S}\text{b}\text{+}3\text{n}$

(c)

Interpretation Introduction

Interpretation:

Missing product in the given nuclear fission reaction has to be determined.

    $\text{U}\text{+}\text{n}\stackrel{}{\to }?+\text{B}\text{a}\text{+}3\text{n}$

Concept Introduction:

Refer part (a).

Answer to Problem 55QRT

The missing product is $\text{K}\text{r}$.

Explanation of Solution

Given nuclear equation in the problem statement is,

    $\text{U}\text{+}\text{n}\stackrel{}{\to }?+\text{B}\text{a}\text{+}3\text{n}$

In a balanced nuclear equation, the mass number balance and atomic number balance has to be equal on both sides of the equation.  The unknown element that is formed is represented as $\text{X}$.  Therefore, the equation can be written as,

    $\text{U}\text{+}\text{n}\stackrel{}{\to }\text{X}+\text{B}\text{a}\text{+}3\text{n}$

Sum of mass number in the reactant side has to be equal to the sum of mass number in the product side.

    $\begin{array}{l}235+1=\text{A}\text{+}\text{141}\text{+}\text{3}\\ \text{236}\text{=}\text{A}\text{+}\text{144}\\ \text{A}\text{=}\text{236}-144\\ =92\end{array}$

Therefore, the mass number of the unknown product is found to be 92.

Sum of atomic number in the reactant side has to be equal to the sum of atomic number in the product side.

  $\begin{array}{l}92+0=\text{Z}\text{+}\text{56}\text{+}\text{0}\\ \text{92}\text{=}\text{Z}\text{+}\text{56}\\ \text{Z}\text{=}\text{92}-56\\ =36\end{array}$

Therefore, the atomic number of the unknown product is found to be 36.  The element with atomic number 36 is Krypton.  Therefore, the missing product is found to be $\text{K}\text{r}$.  The complete equation can be given as,

  $\text{U}\text{+}\text{n}\stackrel{}{\to }\text{K}\text{r}+\text{B}\text{a}\text{+}3\text{n}$