Chapter 21, Problem 78AP

Interpretation Introduction

Interpretation:

The number of $\text{R}\text{n}$

isotopes at the beginning and at the end of $31$

days is to be calculated.

Concept Introduction:

Radium is radioactive because although it is a noble gas, its mass is so large that its nucleus is unstable and emits radiation when it decays.

According to the quantum theory, it is impossible to determine when or how an atom will decay as it is a random process of energy emission.

Answer to Problem 78AP

Solution: $\text{1}\text{.8}\text{×}{\text{10}}^{\text{19}}\text{Rn}\text{atoms}$ and $\text{6}{\text{.4×10}}^{\text{16}}\text{ Rn atoms}$.

Explanation of Solution

Given information: The partial pressure of $\text{R}\text{n}$

is $\text{1}{\text{.2×10}}^{\text{-6}}\text{ mmHg}$ and the half-life is $\text{3}\text{.8 days}$.

Dimensions of the basement are $\text{14 m ×10 m }\times \text{3}\text{.0 m}$

First, calculate the number of $\text{R}\text{n}$ atoms.

The volume of basement is calculated as follows:

$\begin{array}{c}V=\text{(14 m ×10 m }\times \text{3}\text{.0 m) }\\ =\text{4}{\text{.2×10}}^{\text{2}}{\text{m}}^{\text{3}}\\ =\text{4}{\text{.2×10}}^{\text{5}}\text{ L}\end{array}$

The number of moles of air is calculated as follows:

$\begin{array}{c}{\text{n}}_{\text{air}}\text{=}\frac{\text{PV}}{\text{RT}}\\ \text{=}\frac{\text{(1}\text{.0}\cancel{\text{atm}}\text{)(4}\text{.2}\text{×}{\text{10}}^{\text{5}}\cancel{\text{L}}\text{)}}{\text{(0}\text{.0821}\cancel{\text{L}}\text{×}\cancel{\text{atm}}\text{/mol×}\cancel{\text{K}}\text{)(273}\cancel{\text{K}}\text{)}}\\ \text{=}\text{1}\text{.9}\text{×}{\text{10}}^{\text{4}}\text{mol}\text{air}\end{array}$

The number of moles of radon is calculated as follows:

$\begin{array}{c}{\text{n}}_{\text{Rn}}\text{=}\frac{{\text{P}}_{\text{Rn}}}{{\text{P}}_{\text{air}}}\text{×}\text{(1}\text{.9}\text{×}{\text{10}}^{\text{4}}\text{)}\\ \text{=}\frac{\text{1}\text{.2}\text{×}{\text{10}}^{\text{-6}}\cancel{\text{mmHg}}}{\text{760}\cancel{\text{mmHg}}}\text{×}\text{(1}\text{.9}\text{×}{\text{10}}^{\text{4}}\text{mol)}\\ \text{=}\text{3}\text{.0}\text{×}{\text{10}}^{\text{-5}}\text{mol}\text{Rn}\end{array}$

The number of $\text{R}\text{n}$

atoms at the beginning is calculated as follows:

$\begin{array}{c}\text{n}=\text{(3}\text{.0}\text{×}{\text{10}}^{\text{-5}}\cancel{\text{mol}\text{Rn}}\text{)}\text{×}\left(\frac{\text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}\text{Rn}\text{atoms}}{\text{1}\cancel{\text{mol}\text{Rn}}}\right)\\ =\text{1}\text{.8}\text{×}{\text{10}}^{\text{19}}\text{Rn}\text{atoms}\end{array}$

The rate constant is calculated as follows:

$k\text{=}\frac{\text{0}\text{.693}}{\text{3}\text{.8}\text{d}}\text{=}\text{0}\text{.182}{\text{d}}^{\text{-1}}$

Consider the number of radon atoms after time $t$ to be $x$.

From the equation of first order kinetics, the value of $x$ is calculated as follows:

$\begin{array}{c}\text{ln}\frac{{\text{[A]}}_t}{{\text{[A]}}_0}=-kt\\ \text{ln}\frac{x}{\text{1}\text{.8}\text{×}{\text{10}}^{\text{19}}}=-\text{(0}\text{.182}{\text{day}}^{-\text{1}}\text{)(31}\text{day)}\\ x=\text{ 6}{\text{.4×10}}^{\text{16}}\text{ Rn atoms}\end{array}$

Conclusion

The numbers of $\text{R}\text{n}$

isotopes at the beginning and at the end of $31$

days are $\text{1}\text{.8}\text{×}{\text{10}}^{\text{19}}\text{Rn}\text{atoms}$

and $\text{6}{\text{.4×10}}^{\text{16}}\text{ Rn atoms}$.