Chapter 21, Problem 21.57QP

(a)

Interpretation Introduction

Interpretation: The amount of Radon $-222$ that must be present in sample of drinking water that was set by US Environmental Protection agency is given. Various questions based on the decay radioactivity of this isotope have to be answered.

Concept introduction: The physical phenomenon that involves the spontaneous emission of radiations is known as radioactivity.

To determine: The number of decay events that would occur per second in a milliliter of water for the given level of Radon radioactivity.

Answer to Problem 21.57QP

Solution

The number of decay events that would occur per second in a milliliter of water for the given level of Radon radioactivity is $\underset{\_}{0.148}$.

Explanation of Solution

Explanation

Given

The level of radioactivity is $4.0\text{pCi}/\text{mL}$.

The conversion of $\text{pCi}$ to $\text{Ci}$ is done as,

$1\text{pCi}={10}^{-12}\text{Ci}$

Hence the conversion of $\text{4}\text{.0}\text{pCi}$ to $\text{Ci}$ is done as,

$4.0\text{pCi}=4.0\times {10}^{-12}\text{Ci}$

The conversion of $\text{Ci}$ to $\text{Bq}$ is done as,

$1\text{Ci}=\text{3}\text{.7}\times {\text{10}}^{\text{10}}\text{Bq}$

Hence the conversion of $4.0\times {10}^{-12}\text{Ci}$ to $\text{Bq}$ is done as,

$\begin{array}{c}4.0\times {10}^{-12}\text{Ci}=4.0\times {10}^{-12}\times \text{3}\text{.7}\times {\text{10}}^{\text{10}}\text{Bq}\\ =0.148\text{Bq}\end{array}$

The relationship between $\text{Bq}$ and $\text{disintegration}/\text{s}$ is given as,

$1\text{Bq}=1\text{disintegration}/\text{s}$

Hence the conversion of $0.148\text{Bq}$ to $\text{disintegration}/\text{s}$ is given as,

$\begin{array}{c}0.148\text{Bq}=0.148\times 1\text{disintegration}/\text{s}\\ =0.148\text{disintegration}/\text{s}\end{array}$

Therefore, the number of decay events that would occur per second in a milliliter of water for the given level of Radon radioactivity is $\underset{\_}{0.148}$.

(b)

Interpretation Introduction

To determine: The number of ${}^{222}\text{Rn}$ atoms that would be present in $1.0\text{mL}$ of water.

Answer to Problem 21.57QP

Solution

The number of ${}^{222}\text{Rn}$ atoms that would be present in $1.0\text{mL}$ of water is $\underset{\_}{7.04\times 10^{4}atoms}$.

Explanation of Solution

Explanation

Given

The half-life is $3.8\text{days}$.

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

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

Hence, the conversion of $3.8\text{days}$ into $\text{h}$ is done as,

$\begin{array}{c}3.8\text{day}=3.8\times 24\text{h}\\ =91.2\text{h}\end{array}$

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

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

Hence, the conversion of $\text{91}\text{.2}\text{h}$ into $\text{s}$ is done as,

$\begin{array}{c}\text{91}\text{.2}\text{h}=91.2\times 3600\text{s}\\ =\text{328,320}\text{s}\end{array}$

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 of the reaction.

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

$\begin{array}{c}{t}_{1/2}=\frac{0.693}{\lambda }\\ \lambda =\frac{0.693}{328,320\text{s}}\\ =2.1\times {10}^{-6}{\text{s}}^{-1}\end{array}$

The equation for radioactive decay is given as,

$-\frac{dN}{dt}=\lambda N$

Where,

• $\lambda$ is the decay constant of the reaction.
• $N$ is the undecayed population.
• $\frac{dN}{dt}$ is the number of decays per unit time.

Substitute the value of $\frac{dN}{dt}$ and $\lambda$ in the above equation as,

$\begin{array}{c}-\frac{dN}{dt}=\lambda N\\ 0.148=2.1\times {10}^{-6}\times N\\ N=\underset{\_}{7.04\times 10^{4}atoms}\end{array}$

The number of ${}^{222}\text{Rn}$ atoms that would be present in $1.0\text{mL}$ of water is $\underset{\_}{7.04\times 10^{4}atoms}$.

Conclusion

1. a. The number of decay events that would occur per second in a milliliter of water for the given level of Radon radioactivity is $\underset{\_}{0.148}$.
2. b. The number of ${}^{222}\text{Rn}$ atoms that would be present in $1.0\text{mL}$ of water is $\underset{\_}{7.04\times 10^{4}atoms}$