Chapter 21, Problem 21.56QP

Interpretation Introduction

Interpretation: The information regarding the requirement of certain amount of ${}^{226}\text{Ra}$ for painting the glow-in-the-dark dial of a wristwatch that was made in $1914$ is given. The radioactivity of the watch is to be calculated.

Concept introduction: The average number of the atoms that are undergoing disintegration per second is known as the activity of that substance.

To determine: The radioactivity of the watch.

Answer to Problem 21.56QP

Solution

The radioactivity of Radium is $\underset{\_}{0.004\times 10^{7}Bq}$ and $\underset{\_}{1.08\mu Ci}$.

Explanation of Solution

Explanation

Given

The amount of Radium $-226$ is $1.00\text{μg}$.

The half life of Radium $-226$ is $1.60\times {10}^3\text{years}$.

The conversion of years into days is done as,

$1\text{year}=365\text{days}$

Hence the conversion of $1.60\times {10}^3\text{years}$ into days is done as,

$\begin{array}{c}1.60\times {10}^3\text{years}=1.60\times {10}^3\times 365\text{days}\\ =\text{584}\times {10}^3\text{days}\end{array}$

The conversion of days into $\text{h}$ is done as,

$1\text{day}=24\text{h}$

Hence the conversion of $\text{584}\times {10}^3\text{days}$ into $\text{h}$ is done as,

$\begin{array}{c}\text{584}\times {10}^3\text{days}=\text{584}\times {10}^3\times 24\text{h}\\ =\text{14,016}\times {10}^3\text{h}\end{array}$

The conversion of $\text{h}$ into $\text{s}$ is done as,

$1\text{h}=3600\text{s}$

Hence the conversion of $\text{14,016}\times {10}^3\text{h}$ into $\text{s}$ is done as,

$\begin{array}{c}\text{14,016}\times {10}^3\text{h}=\text{14,016}\times {10}^3\times 3600\text{s}\\ =\text{5}\text{.04}\times {\text{10}}^{\text{10}}\text{s}\end{array}$

The half-life of first order reaction is given as,

$t_{1/2}=\frac{0.693}{\lambda }$

Where,

• $t_{1/2}$ is the half-life period.
• $\lambda$ is the decay constant.

Substitute the value of $t_{1/2}$ in the above equation,

$\begin{array}{c}{t}_{1/2}=\frac{0.693}{\lambda }\\ \text{5}\text{.04}\times {\text{10}}^{\text{10}}\text{s}=\frac{0.693}{\lambda }\\ \lambda =0.138\times {10}^{-10}{\text{s}}^{-1}\end{array}$

The value of $\lambda$ is also written as $0.138\times {10}^{-10}\frac{\text{decay events}}{\text{atom}\text{s}}$

The conversion of $\text{μg}$ to $\text{g}$ is done as,

$1\text{μg}={10}^{-6}\text{g}$

The molar mass of Radon is $226\text{g}/\text{mol}$.

The relation between number of atoms and moles is given as,

$1\text{mol}=6.022\times {10}^{23}\text{atoms}$

The mass of $226\text{g}$ corresponds to $=6.022\times {10}^{23}\text{atoms}$

The mass of $1\text{g}$ corresponds to $=\frac{6.022\times {10}^{23}}{226}\text{atoms}$

The mass of ${10}^{-6}\text{g}$ corresponds to $\begin{array}{c}=\frac{6.022\times {10}^{23}}{226}\times {10}^{-6}\text{atoms}\\ =\text{0}\text{.0266}\times {\text{10}}^{\text{17}}\text{atoms}\end{array}$

The formula for activity is given as,

$A=\lambda \times N$

Where,

• $N$ is the number of parent nuclei.
• $\lambda$ is the decay constant.

Substitute the value of $N$ and $\lambda$ in the above equation as,

$\begin{array}{c}A=\lambda \times N\\ =0.138\times {10}^{-10}\frac{\text{decay events}}{\text{atom}\text{s}}\times \text{0}\text{.0266}\times {\text{10}}^{\text{17}}\text{atoms}\\ =\text{0}\text{.004}\times {\text{10}}^{\text{7}}\text{decay events}/\text{s}\end{array}$

The relation between Becquerel and $\text{decay events/s}$ is given as,

$1\frac{\text{decay event}}{\text{s}}=1\text{Bq}$

Hence, the conversion of $\text{0}\text{.004}\times {\text{10}}^{\text{7}}\text{decay events}/\text{s}$ to $\text{Bq}$ is done as,

$\begin{array}{c}\text{0}\text{.004}\times {\text{10}}^{\text{7}}\frac{\text{decay event}}{\text{s}}=\text{0}\text{.004}\times {\text{10}}^{\text{7}}\times 1\text{Bq}\\ =\underset{\_}{0.004\times 10^{7}Bq}\end{array}$

The relation between Curie $\left(\text{Ci}\right)$ and $\text{decay events/s}$ is given as,

$1\frac{\text{decay event}}{\text{s}}=\frac{1}{3.7\times {10}^{10}}\text{Ci}$

Hence, the conversion of $\text{0}\text{.004}\times {\text{10}}^{\text{7}}\text{decay events}/\text{s}$ to $\text{Ci}$ is done as,

$\begin{array}{c}\text{0}\text{.004}\times {\text{10}}^{\text{7}}\text{decay events}/\text{s}=\text{0}\text{.004}\times {\text{10}}^{\text{7}}\times \frac{1}{3.7\times {10}^{10}}\text{Ci}\\ =\text{1}\text{.08}\times {\text{10}}^{-6}\text{Ci}\end{array}$

The relation between $\text{Ci}$ and $\text{μCi}$ is given as,

$1\text{Ci}={10}^6\text{μCi}$

Hence the conversion of $\text{1}\text{.08}\times {\text{10}}^{-6}\text{Ci}$ to $\text{μCi}$ is done as,

$\begin{array}{c}\text{1}\text{.08}\times {\text{10}}^{-6}\text{Ci}=\text{1}\text{.08}\times {\text{10}}^{-6}\times {10}^6\text{μCi}\\ =\underset{\_}{1.08\mu Ci}\end{array}$

The radioactivity of Radium is $\underset{\_}{0.004\times 10^{7}Bq}$ and $\underset{\_}{1.08\mu Ci}$.

Conclusion

The radioactivity of Radium is $\underset{\_}{0.004\times 10^{7}Bq}$ and $\underset{\_}{1.08\mu Ci}$