Chapter 21, Problem 21.28QE

(a)

Interpretation Introduction

Interpretation:

Given equation has to be completed and balanced.

    $\text{P}\text{u}\to \alpha +\text{\_\_\_\_\_}$

Explanation of Solution

Given equation is written as shown below.

    $\text{P}\text{u}\to \alpha +\text{\_\_\_\_\_}$

Sum of the atomic numbers on the reactant side has to be equal to the sum of atomic numbers on the product side.  Sum of mass number on the reactant side has to be equal to the sum of mass number on the product side.

Sum of atomic number on the reactant side is 94.  Therefore, the missing element on the product side has an atomic number of 92.  The element that has atomic number as 92 is uranium.

Sum of mass number on the reactant side is 242.  Mass number of the missing element is found to be 238, by finding the difference between the mass number on reactant side and product side.  Therefore, the element uranium has a mass number of 238.  Complete equation can be given as shown below.

    $\text{P}\text{u}\to \alpha +\text{U}$

(b)

Interpretation Introduction

Interpretation:

Given equation has to be completed and balanced.

    $\_\_\_\_\_\to \text{S}\text{ + }\beta$

Explanation of Solution

Given equation is written as shown below.

    $\_\_\_\_\_\to \text{S}\text{ + }\beta$

Sum of the atomic numbers on the reactant side has to be equal to the sum of atomic numbers on the product side.  Sum of mass number on the reactant side has to be equal to the sum of mass number on the product side.

Sum of atomic number on the product side is 15.  Atomic number of the missing element is found to be 15, by finding the difference between the atomic number on reactant side and product side.  The element with atomic number 15 is phosphorus.

Sum of mass number on the product side is 32.  Mass number of the missing element is found to be 32, by finding the difference between the mass number on reactant side and product side.  Therefore, the element phosphorus has a mass number of 32.  Complete equation can be given as shown below.

    $\text{P}\to \text{S}\text{ + }\beta$

(c)

Interpretation Introduction

Interpretation:

Given equation has to be completed and balanced.

    $\text{C}\text{f}\text{+ }\_\_\_\_\_\to 3\text{n}\text{ + }\text{L}\text{r}$

Explanation of Solution

Given equation is written as shown below.

    $\text{C}\text{f}\text{+ }\_\_\_\_\_\to 3\text{n}\text{ + }\text{L}\text{r}$

Sum of the atomic numbers on the reactant side has to be equal to the sum of atomic numbers on the product side.  Sum of mass number on the reactant side has to be equal to the sum of mass number on the product side.

Sum of atomic number on the product side is 103.  Atomic number of the missing element is found to be 5, by finding the difference between the atomic number on reactant side and product side.  The element with atomic number 5 is boron.

Sum of mass number on the product side is 262.  Mass number of the missing element is found to be 10, by finding the difference between the mass number on reactant side and product side.  Therefore, the element boron has a mass number of 10.  Complete equation can be given as shown below.

    $\text{C}\text{f}\text{+ }\text{B}\to 3\text{n}\text{ + }\text{L}\text{r}$

(d)

Interpretation Introduction

Interpretation:

Given equation has to be completed and balanced.

    $\text{F}\text{e}\text{+ }\_\_\_\_\_\to \text{M}\text{n}$

Explanation of Solution

Given equation is written as shown below.

    $\text{F}\text{e}\text{+ }\_\_\_\_\_\to \text{M}\text{n}$

Sum of the atomic numbers on the reactant side has to be equal to the sum of atomic numbers on the product side.  Sum of mass number on the reactant side has to be equal to the sum of mass number on the product side.

Sum of atomic number on the product side is 25.  Atomic number of the missing element is found to be $-1$, by finding the difference between the atomic number on reactant side and product side.

Sum of mass number on the product side is 55.  Mass number of the missing element is found to be 0, by finding the difference between the mass number on reactant side and product side.  Therefore, the particle that has atomic number of $-1$ and mass number 0 is beta particle.  Complete equation can be given as shown below.

    $\text{F}\text{e}\text{+ }\beta \to \text{M}\text{n}$

(e)

Interpretation Introduction

Interpretation:

Given equation has to be completed and balanced.

    $\text{O}\to \text{ \_\_\_\_\_+}\beta$

Explanation of Solution

Given equation is written as shown below.

    $\text{O}\to \text{ \_\_\_\_\_+}\beta$

Sum of the atomic numbers on the reactant side has to be equal to the sum of atomic numbers on the product side.  Sum of mass number on the reactant side has to be equal to the sum of mass number on the product side.

Sum of atomic number on the reactant side is 8.  Therefore, the missing element on the product side has an atomic number of 7.  The element that has atomic number as 7 is nitrogen.

Sum of mass number on the reactant side is 15.  Mass number of the missing element is found to be 15, by finding the difference between the mass number on reactant side and product side.  Therefore, the element nitrogen has a mass number of 15.  Complete equation can be given as shown below.

    $\text{O}\to \text{N}\text{ +}\beta$