Chapter 20, Problem 20.107QP

(a)

Interpretation Introduction

Interpretation:

Given reaction has to be completed.

Answer to Problem 20.107QP

Completed equation is,

${}_{\text{14}}^{\text{31}}\text{Si}\stackrel{}{\to }{}_{\text{15}}^{\text{31}}{\text{P + }}_{\text{-1}}^{\text{0}}\text{e}$

Explanation of Solution

Given incomplete nuclear reaction is,

${}_{\text{14}}^{\text{31}}\text{Si}\stackrel{}{\to }{}_{\text{15}}^{\text{31}}\text{P + ?}$

This is written as,

${}_{\text{14}}^{\text{31}}\text{Si}\stackrel{}{\to }{}_{\text{15}}^{\text{31}}{\text{P + }}_{\text{Z}}^{\text{A}}\text{X}$

Where, Z is atomic number A is mass number.

$\begin{array}{l}\text{From superscripts:}\\ \text{ 31+A = 31}\\ \text{A = 0}\\ \text{From subscripts:}\\ \text{ 15+Z = 14}\\ \text{Z = -1}\end{array}$

So, the missing particle in the reaction is ${}_{\text{-1}}^{\text{0}}\text{X}$, which is identified as an electron.  The symbol of electron is ${}_{\text{-1}}^{\text{0}}\text{e}$.  Thus the reaction is written as,

${}_{\text{14}}^{\text{31}}\text{Si}\stackrel{}{\to }{}_{\text{15}}^{\text{31}}{\text{P + }}_{\text{-1}}^{\text{0}}\text{e}$

(b)

Interpretation Introduction

Interpretation:

Given reaction has to be completed.

Answer to Problem 20.107QP

Completed reactions are,

${}_{\text{22}}^{\text{44}}{\text{Ti + }}_{\text{-1}}^{\text{0}}\text{e}\stackrel{}{\to }{}_{\text{21}}^{\text{44}}\text{Sc}$

Explanation of Solution

Given incomplete nuclear reaction is,

${}_{\text{22}}^{\text{44}}\text{Ti + ?}\stackrel{}{\to }{}_{\text{21}}^{\text{44}}\text{Sc}$

This is written as,

${}_{\text{22}}^{\text{44}}{\text{Ti + }}_{\text{Z}}^{\text{A}}\text{X}\stackrel{}{\to }{}_{\text{21}}^{\text{44}}\text{Sc}$

Where, Z is atomic number A is mass number.

$\begin{array}{l}\text{From superscripts:}\\ \text{ 44+A = 44}\\ \text{A = 0}\\ \text{From subscripts:}\\ \text{ 22+Z = 21}\\ \text{Z = -1}\end{array}$

So, the missing particle in the reaction is ${}_{\text{-1}}^{\text{0}}\text{X}$, which is identified as an electron.  The symbol of electron is ${}_{\text{-1}}^{\text{0}}\text{e}$.  Thus the reaction is written as,

${}_{\text{22}}^{\text{44}}{\text{Ti + }}_{\text{-1}}^{\text{0}}\text{e}\stackrel{}{\to }{}_{\text{21}}^{\text{44}}\text{Sc}$

(c)

Interpretation Introduction

Interpretation:

Given reaction has to be completed.

Answer to Problem 20.107QP

Completed reactions are,

${}_{\text{98}}^{\text{252}}\text{Cf }\stackrel{}{\to }{}_{\text{56}}^{\text{142}}\text{Ba}{\text{+ }}_{\text{42}}^{\text{106}}{\text{Mo + 4}}_{\text{0}}^{\text{1}}\text{n}$

Explanation of Solution

Given incomplete nuclear reaction is,

${}_{\text{98}}^{\text{252}}\text{Cf }\stackrel{}{\to }{}_{\text{56}}^{\text{142}}\text{Ba}{\text{+ ? + 4}}_{\text{0}}^{\text{1}}\text{n}$

This is written as,

${}_{\text{98}}^{\text{252}}\text{Cf }\stackrel{}{\to }{}_{\text{56}}^{\text{142}}\text{Ba}{\text{+ }}_{\text{Z}}^{\text{A}}{\text{X + 4}}_{\text{0}}^{\text{1}}\text{n}$

Where, Z is atomic number A is mass number.

$\begin{array}{l}\text{From superscripts:}\\ \text{ 142+A +4(1)=252}\\ \text{A = 106}\\ \text{From subscripts:}\\ \text{ 56+Z = 98}\\ \text{Z = 42}\end{array}$

The atomic number is 42 and mass number is 106.  So, the missing element in the reaction is ${}_{42}^{\text{106}}\text{X}$.  The element with atomic number of 42 and mass number of 106 is found to be Molybdenum.  The nuclide symbol of Molybdenum is ${}_{\text{42}}^{\text{106}}\text{Mo}$.  Thus the reaction is written as,

${}_{\text{98}}^{\text{252}}\text{Cf }\stackrel{}{\to }{}_{\text{56}}^{\text{142}}\text{Ba}{\text{+ }}_{\text{42}}^{\text{106}}{\text{Mo + 4}}_{\text{0}}^{\text{1}}\text{n}$