Chapter 19, Problem 19.74QP

(a)

Interpretation Introduction

Interpretation:

The equation should be written with proper abbreviation.

Concept Introduction:

• Nuclear reaction can be written in the shorthand notation with the parentheses. Bombarding particle, that is projectile can be represented as first symbol in the parentheses and the emitted particle that is ejectile which can be represented as the second particle in the parentheses.

Parent nucleus and daughter nucleus can be represented in the front part of the parentheses and back part of the parentheses respectively.

$\begin{array}{l}{}_7{N}^{15}\left(p,\alpha \right){}_6{C}^{12}\\ Parentnucleus\left(Projectile,ejectile\right)Daughternucleus\end{array}$

• On accordance with law of conservation of mass, for any chemical reaction, total masses of reactants and products must be equal.
• . Half-life period: The time required to the reduce to half of its original value.
• This is denoted by:${\text{t}}_{\text{1/2}}$
• This term is popularly use in nuclear chemistry, to describe the stability & unstability of atoms or radioactive decay.
• For determination of first order rate constant K in half –life period we use

  $\text{K=}\frac{\text{0}\text{.693}}{{\text{t}}_{\text{1/2}}}$ Equation.

• The half-life in first order does not depends upon the initial concentration.

Formula:

First order rate constant:$\text{K=}\frac{\text{0}\text{.693}}{{\text{t}}_{\text{1/2}}}$

Age calculation:

$\text{ln}\frac{{\text{N}}_{\text{t}}}{{\text{N}}_{\text{0}}}\text{=-kt}$

Answer to Problem 19.74QP

The balance equation is ${}_{\text{19}}^{\text{40}}\text{K}\stackrel{}{\to }{}_{\text{18}}^{\text{40 }}{\text{Ar +}}_{\text{0}}^{\text{-1}}\text{β }$

Explanation of Solution

To find: for the balanced equation write Suitable abbreviation.

Nuclear reaction can be written in the shorthand notation with the parentheses. Bombarding particle, that is projectile can be represented as first symbol in the parentheses and the emitted particle that is ejectile which can be represented as the second particle in the parentheses. Parent nucleus and daughter nucleus can be represented in the front part of the parentheses and back part of the parentheses respectively. So for any nuclear reaction, short hand notation will be in this form that is,

$\begin{array}{l}{}_{\text{19}}^{\text{40}}\text{K}\left({}_{\text{0}}^{\text{-1}}\text{β }\right){}_{\text{18}}^{\text{40 }}\text{Ar}\\ \text{Parent}\text{nucleus}\left(\text{Projectile,}\text{ejectile}\right)\text{Daughter}\text{nucleus}\end{array}$

For the given reaction,

 So the balanced equation can be written as,

${}_{\text{19}}^{\text{40}}\text{K}\stackrel{}{\to }{}_{\text{18}}^{\text{40 }}{\text{Ar +}}_{\text{0}}^{\text{-1}}\text{β }$

Conclusion

The balanced equation is ${}_{\text{19}}^{\text{40}}\text{K}\stackrel{}{\to }{}_{\text{18}}^{\text{40 }}{\text{Ar +}}_{\text{0}}^{\text{-1}}\text{β }$.

(b)

Interpretation Introduction

Interpretation:

The age of the rock should be determine & the equation should be written with proper abbreviation.

Concept Introduction:

• Nuclear reaction can be written in the shorthand notation with the parentheses. Bombarding particle, that is projectile can be represented as first symbol in the parentheses and the emitted particle that is ejectile which can be represented as the second particle in the parentheses.

Parent nucleus and daughter nucleus can be represented in the front part of the parentheses and back part of the parentheses respectively.

$\begin{array}{l}{}_7{N}^{15}\left(p,\alpha \right){}_6{C}^{12}\\ Parentnucleus\left(Projectile,ejectile\right)Daughternucleus\end{array}$

• On accordance with law of conservation of mass, for any chemical reaction, total masses of reactants and products must be equal.
• . Half-life period: The time required to the reduce to half of its original value.
• This is denoted by:${\text{t}}_{\text{1/2}}$
• This term is popularly use in nuclear chemistry, to describe the stability & unstability of atoms or radioactive decay.
• For determination of first order rate constant K in half –life period we use

  $\text{K=}\frac{\text{0}\text{.693}}{{\text{t}}_{\text{1/2}}}$ Equation.

• The half-life in first order does not depends upon the initial concentration.

Formula:

First order rate constant:$\text{K=}\frac{\text{0}\text{.693}}{{\text{t}}_{\text{1/2}}}$

Age calculation:

$\text{ln}\frac{{\text{N}}_{\text{t}}}{{\text{N}}_{\text{0}}}\text{=-kt}$

Answer to Problem 19.74QP

The age of the rock is $\text{t=3}{\text{.0×10}}^{\text{9}}\text{yr}$

Explanation of Solution

To find: The age of the rock determine using the half-life equation.

At first, we have to calculate the rate constant k

$\text{ k=}\frac{\text{0}\text{.693}}{{\text{t}}_{\text{1/2}}}=\frac{\text{0}\text{.693}}{\text{1}\text{.2}\times {\text{10}}^9\text{yr}}\text{ =5}\text{.8 }\times {\text{10}}^{-10}{\text{ yr}}^{-1}$

Then calculate the age of the rock by subtracting K into the following equation (${\text{N}}_{\text{t}}\text{=0}{\text{.18N}}_{\text{0}}$)

$\text{ln}\frac{{\text{N}}_{\text{t}}}{{\text{N}}_{\text{0}}}\text{=-kt}$

Putting the value in the above equation, we can get

$\text{ln }\frac{\text{0}\text{.18}}{\text{1}\text{.00}}\text{= (5}\text{.8 }\times {\text{10}}^{-10}{\text{ yr}}^{-1}\text{)t }$

The age of the rock is found to be

$\text{t=3}{\text{.0×10}}^{\text{9}}\text{yr}$

Conclusion

The age of the rock is determine $\text{t=3}{\text{.0×10}}^{\text{9}}\text{yr}$